Kapoor (mk9499) – oldmidterm 01 – Turner – (60230)
1
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at (85 cm, 0.0). The Coulomb constant is
9 × 109 N m2 /C2 .
Calculate the magnitude of the force on the
charge at the origin.
Correct answer: 0.13928 N.
001 10.0 points
Two spheres, fastened to “pucks”, are riding on a frictionless airtrack. Sphere “1” is
charged with 1 nC, and sphere “2” is charged
with 3 nC. Both objects have the same mass.
1 nC is equal to 1 × 10−9 C.
As they repel,
Explanation:
1. sphere “2” accelerates 3 times as fast as
sphere “1”.
2. sphere “1” accelerates 9 times as fast as
sphere “2”.
3. sphere “2” accelerates 9 times as fast as
sphere “1”.
4. they have the same magnitude of acceleration. correct
5. they do not accelerate at all, but rather
separate at constant velocity.
6. sphere “1” accelerates 3 times as fast as
sphere “2”.
Explanation:
The force of repulsion exerted on each mass
is determined by
F =
1 Q1 Q2
= ma
4 π ǫ0 r 2
where r is the distance between the centers of
the two spheres.
~ 12 k = kF
~ 21 k
kF
Since both spheres have the same mass and
are subject to the same force, they have the
same acceleration.
002 10.0 points
Three equal charges of 3 µC are in the x-y
plane. One is placed at the origin, another is
placed at (0.0, 99 cm), and the last is placed
Let : q = 3 µC = 3 × 10−6 C ,
(x1 , y1 ) = (0, 99 cm) = (0, 0.99 m) ,
(x2 , y2 ) = (85 cm, 0) = (0.85 m, 0) and
(x0 , y0 ) = (0, 0) .
The electric fields are
ke q
+ y12
9 × 109 N m2 /C2 3 × 10−6 C
=
0 + (0.99 m)2
= 27548.2 N/C , and
ke q
Ex = 2
x2 + y22
9 × 109 N m2 /C2 3 × 10−6 C
=
(0.85 m)2 + 0
= 37370.2 N/C ,
Thus q
Ey =
E=
x21
Ex2 + Ey2
and
q
Ex2 + Ey2
= 3 × 10−6 C
q
· (27548.2 N/C)2 + (37370.2 N/C)2
F = qE = q
= 0.13928 N .
003 10.0 points
Two identical small charged spheres hang in
equilibrium with equal masses as shown in
the figure. The length of the strings are equal
and the angle (shown in the figure) with the
vertical is identical.
The acceleration of gravity is 9.8 m/s2
and the value of Coulomb’s constant is
8.98755 × 109 N m2 /C2 .
Kapoor (mk9499) – oldmidterm 01 – Turner – (60230)
vertical directions must separately add up to
zero:
X
Fx = T sin θ − Fe = 0
X
Fy = T cos θ − m g = 0 .
0. 1
5m
5◦
0.03 kg
0.03 kg
Find the magnitude of the charge on each
sphere.
Correct answer: 4.4233 × 10−8 C.
Explanation:
Let : L = 0.15 m ,
m = 0.03 kg ,
θ = 5◦ .
and
L
q
m
θ
a
m
q
From the right triangle in the figure above,
we see that
a
sin θ = .
L
Therefore
a = L sin θ
= (0.15 m) sin(5◦ )
= 0.0130734 m .
The separation of the spheres is r = 2 a =
0.0261467 m . The forces acting on one of the
spheres are shown in the figure below.
Fe = m g tan θ
= (0.03 kg) 9.8 m/s2 tan(5◦ )
= 0.0257217 N ,
for the electric force.
From Coulomb’s law, the electric force between the charges has magnitude
|q|2
|Fe | = ke 2 ,
r
where |q| is the magnitude of the charge on
each sphere.
Note: The term |q|2 arises here because the
charge is the same on both spheres.
This equation can be solved for |q| to give
s
|Fe | r 2
|q| =
ke
s
(0.0257217 N) (0.0261467 m)2
=
(8.98755 × 109 N m2 /C2 )
= 4.4233 × 10−8 C .
004 10.0 points
A circular arc has a uniform linear charge
density of 9 nC/m.
The value of the Coulomb constant is
8.98755 × 109 N · m2 /C2 .
y
θ
◦
Fe
From the second equation in the system
mg
above, we see that T =
, so T can be
cos θ
eliminated from the first equation if we make
this substitution. This gives a value
74
T
T cos θ
2
2. 5
θ
T sin θ
mg
Because the sphere is in equilibrium, the
resultant of the forces in the horizontal and
m
x
Kapoor (mk9499) – oldmidterm 01 – Turner – (60230)
What is the magnitude of the electric field
at the center of the circle along which the arc
lies?
Correct answer: 38.9437 N/C.
Explanation:
Let : λ = 9 nC/m = 9 × 10−9 C/m ,
∆θ = 74◦ , and
r = 2.5 m .
θ is defined as the angle in the counterclockwise direction from the positive x axis
as shown in the figure below.
◦
r
37
consider the right-half of the arc (beginning
on the positive y axis and extending towards
the positive x axis) and multiply the answer
by 2.
Note: The upper angular limit θ = 90◦ .
The lower angular limit θ = 90◦ − 37◦ = 53◦ ,
is the angle from the positive x axis to the
right-hand end of the arc.
!
Z ◦
λ 90
E = −2 ke
sin θ dθ ̂
r 53◦
= −2 ke
θ
~
E
First, position the arc symmetrically
around the y axis, centered at the origin. By
symmetry (in this rotated configuration) the
field in the x direction cancels due to charge
from opposites sides of the y-axis, so
Ex = 0 .
For a continuous linear charge distribution,
Z
~ = ke dq r̂
E
r2
In polar coordinates
dq = λ (r dθ) ,
where λ is the linear charge density. The
positive y axis is θ = 90◦ , so the y component
of the electric field is given by
dEy = dE sin θ .
Note: By symmetry, each half of the arc
about the y axis contributes equally to the
electric field at the origin. Hence, we may just
λ
[cos (53◦ ) − cos (90◦ )] ̂ .
r
Since
ke
37 ◦
3
λ
= (8.98755 × 109 N · m2 /C2 )
r
(9 × 10−9 C/m)
×
(2.5 m)
= 32.3552 N/C ,
E = −2 (32.3552 N/C)
× [(0.601815) − (0)] ̂
= −38.9437 N/C ̂
~ = 38.9437 N/C .
kEk
~ in
Alternate Solution: Just solve for kEk
a straight forward manner, positioning the
beginning of the arc on the positive x axis (as
shown in the original figure in the question).
θ is still defined as the angle in the counterclockwise direction from the positive x axis.
!
Z ◦
ke λ 74
Ex = −
cos θ dθ ı̂
r 0◦
ke λ
[sin (74◦ ) − sin (0◦ )] ı̂
r
= −(32.3552 N/C)
× [(0.961262) − 0.0] ı̂
= [−31.1018 N/C] ı̂ ,
!
Z ◦
ke λ 74
Ey = −
sin θ dθ ̂
r 0◦
=−
ke λ
[cos (0◦ ) − cos (74◦ )] ̂
r
= −(32.3552 N/C)
× [1.0 − (0.275637)] ̂
= [−23.4369 N/C] ̂ ,
=−
Kapoor (mk9499) – oldmidterm 01 – Turner – (60230)
q
~ = E2 + E2
kEk
y
x
h
= (−31.1018 N/C)2
+ (−23.4369 N/C)2
= 38.9437 N/C .
i1/2
005 10.0 points
From the electric field vector at a point, one
can determine which of the following?
I) the direction of the electrostatic force on
a test charge of known sign at that point;
II) the magnitude of the electrostatic force
exerted per unit charge on a test charge
at that point;
III) the electrostatic charge at that point.
1. I and II only correct
2. I only
3. I, II and III
4. III only
Let : m = 9.1 × 10−5 g = 9.1 × 10−8 kg ,
Ex = 7.2 N/C ,
Ey = Ez = 0 ,
vy = 4 × 105 m/s ,
vx = vz = 0 , and
t = 0.7 s .
According to Newton’s second law and the
definition of an electric field,
~ = m~a = q E
~.
F
Since the electric field has only an x component, the particle accelerates only in the x
direction
q Ex
ax =
.
m
To determine the x component of the final
velocity, vxf , use the kinematic relation
vxf = vxi + a (tf − ti ) = a tf .
Since ti = 0 and vxi = 0,
006 10.0 points
A particle of mass 9.1 × 10−5 g and charge
18 mC moves in a region of space where the
electric field is uniform and is 7.2 N/C in the
x direction and zero in the y and z direction.
If the initial velocity of the particle is given
by vy = 4 × 105 m/s, vx = vz = 0, what is the
speed of the particle at 0.7 s?
Correct answer: 1.07418 × 106 m/s.
Explanation:
q E x tf
m
(0.018 C) (7.2 N/C)(0.7 s)
=
(9.1 × 10−8 kg)
= 9.96923 × 105 m/s .
vxf =
5. II and III only
Explanation:
The definition of the electrostatic force is
~
~ = F . In another way, F
~ = q E.
~ This means
E
q
~ is in the same direction of, or opposite diF
~ depending on the sign of the
rection to E,
charge. And if we only consider the magnitude F = qE for a unit charge, the force on it
F
is just
= E. This is statement II.
q
4
vxf
No external force acts on the particle in the y
direction so vyi = vyf = 4 × 105 m/s. Hence
the final speed is given by
q
2 + v2
vf = vyf
xf
2
= 4 × 105 m/s
2
+ 9.96923 × 10 m/s
5
= 1.07418 × 106 m/s .
1/2
Note: This is analogous to a particle in a
gravitational field with the coordinates roπ
tated clockwise by (90◦ ).
2
007 (part 1 of 2) 10.0 points
Kapoor (mk9499) – oldmidterm 01 – Turner – (60230)
A dipole (electrically neutral) is placed in an
external field.
(a)
−
+
−
+
(b)
+
−
−
(c)
+
5
Gauss’ law states
I
~ · dA
~ = Q.
ΦS = E
ǫ0
Solutions: The electric dipole consists of
two equal and opposite charges separated by
a distance. In either situation (c) or (d), the
electric field is uniform everywhere between
the parallel infinite plates. Thus, the electric
force on one charge is equal but opposite to
that on another so that the net force on the
whole dipole is zero. By contrast, electric
fields are nonuniform for situations both (a)
and (b).
(d)
For which situation(s) shown above is the
net force on the dipole equal to zero?
1. (b) and (d)
008 (part 2 of 2) 10.0 points
For which situation(s) shown above is the net
torque on the dipole equal to zero?
1. (a) and (c) correct
2. None of these
3. (c) only
2. None of these
3. (a) only
4. (c) and (d) correct
5. (a) and (c)
4. (b) and (d)
5. (c) only
6. Another combination
7. (a) only
Explanation:
Basic Concepts: Field patterns of point
charge and parallel plates of infinite extent.
The force on a charge in the electric field is
given by
~ = qE
~
F
and the torque is defined as
~ = ~r × F
~
T
~ = k ∆q r̂
∆E
r2
X
~ =
~i.
E
∆E
Symmetry of the configuration will cause
some component of the electric field to be
zero.
6. (c) and (d)
7. (a) and (b)
8. other combination
Explanation:
A electric dipole can be regarded as a pair
of charges of opposite sign. Only in figures
(a) and (c), the electric fields are along the
direction of ~r, where ~r is the vector between
~ is
the pair of charges. Therefore the force F
also along ~r . This will lead to zero torque,
since
~ = ~r × F
~ ∝ ~r × ~r = 0 .
T
For figures (b) and (d), the torque on both
charges are nonzero and the resultant torques
are also nonzero.
009
10.0 points
Kapoor (mk9499) – oldmidterm 01 – Turner – (60230)
Five charges are placed in a closed box. Each
charge (except the first) has a magnitude
which is twice that of the previous one placed
in the box. All charges have the same sign
and (after all the charges have been placed in
the box) the net electric flux through the box
is 9.5 × 107 N · m2 /C.
What is the magnitude of the smallest
charge in the box?
Correct answer: 27.1338 µC.
6
r4
r3
r2
r1
q3
q2
q1
R1
R2 R3 R4 R5
Explanation:
Let : Φ = 9.5 × 107 N · m2 /C
Let the first charge in the box be q. Then
the other four charges have magnitude 2 q, 4 q,
8 q, and 16 q. We can figure out a value for q
by considering Gauss’ law,
Φ=
qtotal
.
ǫ0
Our closed box is a Gaussian surface, and
we know the total flux through this closed
surface. Hence,
Φ=
(1 + 2 + 4 + 8 + 16) q
31 q
=
ǫ0
ǫ0
and solving for q,
Hint: Under static conditions, the charge
on a conductor resides on the surface of the
conductor.
What is the charge Qr3 on the inner surface
of the larger spherical conducting shell?
1. Qr3 = +q1 + q2 + q3
2. Qr3 = +q1
3. Qr3 = +q1 + q2
4. Qr3 = −q1 + q2
5. Qr3 = +q1 − q2
6. Qr3 = 0
q=
1
(9.5 × 107 N · m2 /C)
31
× (8.85419 × 10−12 C2 /N · m2 )
1 × 106 µC
×
1C
= 27.1338 µC .
010 10.0 points
A point charge q1 is concentric with two spherical conducting thick shells, as shown in the
figure below. The smaller spherical conducting shell has a net charge of q2 and the larger
spherical conducting shell has a net charge of
q3 .
7. Qr3 = −q1
8. Qr3 = −q1 − q2 correct
9. Qr3 = −q1 − q2 − q3
10. Qr3 = −q1 − q2 + q3
Explanation:
The net charge inside a Gaussian surface
located at r = R4 must be zero, since the field
in the conductor must be zero. Therefore,
the charge on the inner surface of the large
spherical conducting shell must be −(q1 + q2 ).
011
10.0 points
Kapoor (mk9499) – oldmidterm 01 – Turner – (60230)
1
.
4 π ǫ0
A uniformly charged sphere (an insulating
sphere with radius R) is shown in the figure
below.
Given: k =
Q is the total
charge inside
the sphere.
R
p
R
2
If Q = 8.2 × 10−6 C, the magnitude of the
~ electric field at r = R is given by
kEk
2
~ = kQ .
1. kEk
8 R2
~ = k Q . correct
2. kEk
2 R2
~ = kQR.
3. kEk
~ = kQ .
4. kEk
32 R2
~ = 4kQR.
5. kEk
~ = 8kQR.
6. kEk
~ = kQ.
7. kEk
R2
~ = kQ .
8. kEk
4 R2
~ = kQ .
9. kEk
16 R2
~ = 2kQR.
10. kEk
Explanation:
Pick aspherical Gaussian
surface with ra
R
dius r, where r =
concentric with the
2
sphere of charge.
From Gauss’ law, we find
Φ = E · 4 π r2 .
ρ=
4
3
Q
, so the net charge inside the Gausπ R3
7
R
sian surface of radius
is
2
4
3
πr
qen = ρ
3
Q
4
3
=
πr
4
3
π R3
3
r3
=Q 3.
R
Therefore the electric field is
Φc
E=
4 π r2
Q r3 1
=
ǫ0 R3 4 π r 2
Qr
=
,
4 π ǫ0 R3
at r. At r =
R
we have
2
Q
1
E=
4 π ǫ0 2 R2
kQ
.
=
2 R2
012 (part 1 of 2) 10.0 points
Consider the figure
+Q
A −Q
+
− y
+
−
+
−
+
−
+
−
x
+
−
+ C D −
+
−
+
−
+
−
+
−
#1
#2
B
Of the following elements, identify all that
correspond to an equipotential line or surface.
1. line AB only correct
2. line CD only
3. both AB and CD
4. neither AB nor CD
Explanation:
Kapoor (mk9499) – oldmidterm 01 – Turner – (60230)
Consider the electric field
+Q
A −Q
+
− y
+
−
+
−
+
−
+
−
x
+ C D −
+
−
+
−
+
−
+
−
+
−
#1
#2
B
An equipotential line or surface (AB) is
normal to the electric field lines.
013 (part 2 of 2) 10.0 points
Consider the figure
C
B
+
+q
A
−
−q
D
Of the following elements, identify all that
correspond to an equipotential line or surface.
1. line CD only correct
8
Through what potential difference would an
electron need to be accelerated for it to
achieve a speed of 4.5 % of the speed of light
(2.99792 × 108 m/s), starting from rest?
Correct answer: 517.387 V.
Explanation:
Let : s = 4.5 % = 0.045 ,
c = 2.99792 × 108 m/s ,
me = 9.10939 × 10−31 kg ,
qe = 1.60218 × 10−19 C .
and
The speed of the electron is
v = 0.045 c
= 0.045 2.99792 × 108 m/s
= 1.34907 × 107 m/s ,
By conservation of energy
1
me v 2 = −(−qe ) ∆V
2
v2
∆V = me
2 qe
= 9.10939 × 10−31 kg
2
1.34907 × 107 m/s
×
2 (1.60218 × 10−19 C)
2. both AB and CD
= 517.387 V .
3. neither AB nor CD
4. line AB only
Explanation:
Consider the electric field:
C
A
B
−
+
015 (part 1 of 3) 10.0 points
Consider a solid conducting sphere with a
radius a and charge Q1 on it. There is a
conducting spherical shell concentric to the
sphere. The shell has an inner radius b (with
b > a) and outer radius c and a net charge
Q2 on the shell. Denote the charge on the
inner surface of the shell by Q′2 and that on
the outer surface of the shell by Q′′2 .
b , Q′2
Q1 , a
D
P
Q2
An equipotential line or surface (CD) is
normal to the electric field lines.
014
10.0 points
Q′′2 , c
Q1
Kapoor (mk9499) – oldmidterm 01 – Turner – (60230)
Find the charge Q′′2 .
1. Q′′2 =
Q1 − Q2
2
2. Q′′2 = Q1 − Q2
3. Q′′2 = 2 (Q1 + Q2 )
4. Q′′2 =
Q2 − Q1
2
5. Q′′2 = Q1 + Q2 correct
6. Q′′2 = Q2 − Q1
8.
Q1 + Q2
=
2
2 ke Q 2
(a + b)2
4 ke (Q1 + Q2 )
7. EP =
(a + b)2
2 ke (Q1 − Q2 )
8. EP =
(a + b)2
2 ke (Q1 + Q2 )
9. EP =
(a + b)2
6. EP =
9. Q′′2 = 2 (Q2 − Q1 )
Explanation:
Basic Concepts: Gauss’ Law
Sketch a concentric Gaussian surface S
(dashed line) within the shell.
r
Since the electrostatic field in a conducting
medium is zero, according to Gauss’s Law,
ΦS =
Find the magnitude of
the electric field at
~ P k ≡ EP , where the distance
point P kE
a+b
from P to the center is r =
.
2
4 ke Q 1
1. EP =
correct
(a + b)2
4 ke Q 2
2. EP =
(a + b)2
2 ke Q 1
3. EP =
(a + b)2
4 ke (Q1 − Q2 )
4. EP =
(a + b)2
5. EP = 0
7. Q′′2 = 2 (Q1 − Q2 )
Q′′2
9
Q1 + Q′2
=0
ǫ0
Q′2 = −Q1
But the net charge on the shell is
Q2 = Q′2 + Q′′2 ,
so the charge on the outer surface of the shell
is
Q′′2 = Q2 − Q′2
= Q2 + Q1 .
016 (part 2 of 3) 10.0 points
Explanation:
Choose as your Gaussian surface concentric
with the spherical surface S, which passes
through P . Here,
Z
~ ·A
~ = 4 π r 2 EP
E
Q1
ǫ0
Q1
EP =
4 π ǫ0 r 2
ke Q 1
=
r2
4 ke Q 1
=
.
(a + b)2
=
017 (part 3 of 3) 10.0 points
Assume: The potential at r = ∞ is zero.
Find the potential VP at point P .
2 ke Q1 ke Q1 ke (Q1 + Q2 )
−
+
cor1. VP =
a+b
b
c
rect
Kapoor (mk9499) – oldmidterm 01 – Turner – (60230)
Q
+
2 ke (Q1 − Q2 )
2. VP =
a+b
d
3. VP = 0
4. VP =
5. VP =
6. VP =
7. VP =
8. VP =
9. VP =
10
2 ke Q 1
a+b
2 ke Q1 ke Q1 ke (Q1 − Q2 )
+
−
a+b
b
c
ke Q 1 ke Q 2
−
a+b
b
2 ke Q 1 ke Q 2
−
a+b
c
2 ke Q 1 2 ke Q 2
−
a+b
b
2 ke Q 1 ke Q 2
+
a+b
c
Explanation:
Using the superposition principle, adding
the 3 concentric charge distributions; i.e., Q1
at a, −Q1 at b and Q1 + Q2 at c, gives
+
Q
P
What is the direction of the net electric field
at point P ?
1.
2.
correct
3.
4.
5.
VP = −
=−
=−
Z
Z
Z
~ r · d~r ,
E
by symmetry,
Er dr
c
Z∞c
Er dr −
Z
c
b
0 dr −
Z
(a+b)/2
Er dr
b
k (Q1 + Q2 )
dr
r2
∞
Z b
Z (a+b)/2
k (Q1 + Q2 )
dr
−
0 dr −
r2
c
b
c
1
= −k (Q1 + Q2 )
r ∞
(a+b)/2
1
−0 − k (Q1 + Q2 )
r b
=−
=
2 ke Q1 ke Q1 ke (Q1 + Q2 )
.
−
+
a+b
b
c
018 (part 1 of 2) 10.0 points
The figure below shows two particles, each
with a charge of +Q, that are located at the
opposite corners of a square of side d.
Explanation:
At point P , the electric fields due to the two
positive charged particles have the same magnitude. One points downward, and the other
horizontally to the left. Thus the direction of
the net electric field points left and downward,
forming a 45◦ angle with the horizontal.
019 (part 2 of 2) 10.0 points
What is the potential energy of a particle of
charge q that is held at point P ?
√
2 qQ
1. U =
2 π ǫ0 d
2. U = 0
1 qQ
correct
2 π ǫ0 d
1 qQ
4. U =
4 π ǫ0 d
3. U =
Kapoor (mk9499) – oldmidterm 01 – Turner – (60230)
√
2 qQ
4 π ǫ0 d
Explanation:
The potential energy of the particle charged
+q is the sum of the potential energies due to
the two +Q charged particles.
5. U =
U = U1 + U2
1 qQ
1 qQ
=
+
4 π ǫ0 d
4 π ǫ0 d
1 qQ
.
=
2 π ǫ0 d
020 10.0 points
The two charges Q are fixed at the vertices of
an equilateral triangle with sides of length a
as shown.
q
a
W = U2 − U1 = 2 k
V = 2 x − y 2 − cos(z) ,
what is the y-component of the electric field
at the point P = (x′ , y ′ , z ′ )?
1. Ey = cos(z ′ )
2. Ey = − sin(z ′ )
3. Ey = 2
4. Ey = 2 x′
5. Ey = sin(z ′ )
a
a
Q
The work required to move a charge q from
the other vertex to the center of the line joining the fixed charges is
1. W =
2. W =
3. W =
4. W =
5. W =
4kQq
a
kQq
a
2kQq
correct
a
6kQq
a
√
2kQq
a
6. W = 0
Explanation:
Qq
Qq
Qq
+k
= +2 k
a
a
a
Qq
Qq
Qq
U2 = +k a + k a = +4 k
a
2
2
U1 = +k
Qq
.
a
021 10.0 points
If the potential in a region is given by the
function
6. Ey =
Q
11
y ′3
3
7. Ey = 2 y ′ correct
8. Ey =
x′2
4
9. Ey = y ′2
10. Ey = −2 y ′
Explanation:
The electric field is the gradient of the potential, so the y-component of the E-field evaluated at P is
∂V
∂y
∂ =−
2 x − y 2 − cos(x)
∂y
= − [−2 y]
= 2y.
Ey = −
022 10.0 points
A wire that has a uniform linear charge density of 2.1 µC/m is bent into the shape as
shown below, with radius 4.8 m.
Kapoor (mk9499) – oldmidterm 01 – Turner – (60230)
Hence the electric potential at p is
4.8 m
p
9.6 m
9.6 m
The Coulomb constant is 8.98755 ×
109 N · m2 /C2 .
Find the electrical potential at point p.
Correct answer: 1.00764 × 105 V.
Explanation:
Let :
λ = 2.1 µC/m = 2.1 × 10−6 C/m ,
R = 4.8 m ,
2 R = 9.6 m , and
ke = 8.98755 × 109 N · m2 /C2 .
R
p
2R
2R
Let p be the origin. Consider the potential
due to the line of charge to the right of p.
Vright =
Z
= ke
= ke
dV
Z
Z
dq
r
3R
R
λdx
x
3R
= ke λ ln x
= ke λ ln 3 .
R
By symmetry, the contribution from the line
of charge to the left of p is the same. The
contribution from the semicircle is
Z π
λRdθ
Vsemi = ke
R
0
Z π
= ke λ
dθ
0
π
= ke λ θ
0
= ke λ π .
Vp = Vright + Vlef t + Vsemi
= 2 ke λ ln 3 + ke λ π
= ke λ (2 ln 3 + π)
= (8.98755 × 109 N · m2 /C2 )
× (2.1 × 10−6 C/m)
× (2 ln 3 + π)
= 1.00764 × 105 V .
12
Version 047/AACDD – midterm 01 – Turner – (60230)
This print-out should have 18 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
1
FT,x = FT sin θ
FT,y = FT cos θ
Each sphere is in equilibrium horizontally
001 10.0 points
Three identical point charges hang from three
strings, as shown.
The Coulomb constant is 8.98755 ×
109 N · m2 /C2 , and the acceleration of gravity
is 9.81 m/s2 .
Felectric − FT,x = 0
Felectric − FT sin θ = 0
and vertically
FT,y − Fg = 0
FT cos θ − Fg = 0
FT =
45◦ 45◦
20.0 cm
20.0 cm
+q
+q
Fg
+q
+
+
+
0.10 kg
0.10 kg
0.10 kg
What is the value of q?
1. 2.44481e-06
2. 1.98228e-06
3. 1.58582e-06
4. 1.9162e-06
5. 2.31266e-06
6. 1.78405e-06
7. 8.58988e-07
8. 1.85013e-06
9. 1.6519e-06
10. 1.32152e-06
Correct answer: 1.32152 × 10−6 C.
Explanation:
Let :
m = 0.10 kg ,
L = 20.0 cm ,
θ = 45◦ , and
ke = 8.98755 × 109 N · m2 /C2 .
r = 2 L sin θ
√
√
2
=L 2
= 2 L sin 45 = 2 L
2
◦
Fg
.
cos θ
From the horizontal equilibrium,
Fg
sin θ
Felectric =
cos θ
Felectric = Fg tan θ = Fg (tan 45◦ ) = Fg .
For either of the end charges,
Felectric = ke
q2
q2
+
k
e
r 2
r2
2
q2
q2
= ke 2 + 4 ke 2
r
r
2
q
= 5 ke 2
r
5 ke
q2
= mg.
r2
Thus
s
r2 m g
5 ke
s √
(L 2)2 m g
=
5 ke
r
2mg
= L·
5 ke
1m
= (20 cm)
100 cm
s
2(0.1 kg)(9.81 m/s2 )
×
5(8.98755 × 109 N · m2 /C2 )
|q| =
= 1.32152 × 10−6 C .
Version 047/AACDD – midterm 01 – Turner – (60230)
002 10.0 points
An insulating sphere of radius 15 cm has a
uniform charge density throughout its volume.
15 cm
p
29.0129 cm
6.7517 cm
If the magnitude of the electric field at
a distance of 6.7517 cm from the center is
22769.2 N/C, what is the magnitude of the
electric field at 29.0129 cm from the center?
1. 2835.34
2. 17185.7
3. 12383.8
4. 13521.5
5. 131170.0
6. 27046.9
7. 42930.6
8. 60133.9
9. 82917.4
10. 62637.4
Correct answer: 13521.5 N/C.
Explanation:
Let : R = 15 cm ,
E1 = 22769.2 N/C ,
r1 = 6.7517 cm ,
r2 = 29.0129 cm ,
4
V = π R3 , and
3
Q
ρ= .
V
Method 1: We know the magnitude of
the electric field at a radius r1 = 6.7517 cm
(corresponding to a smaller sphere with surface area A1 and volume V1 ): the magnitude
is E1 = 22769.2 N/C.
We want to find the magnitude E2 at a
radius r2 = 29.0129 cm, corresponding to a
sphere with surface area A2 and volume V2
that is larger than the insulating sphere. From
Gauss’s Law, we know that since the flux is
constant over the sphere,
Q1
E 1 A 1 = Φ1 =
ǫ0
relating the flux through the Gaussian sphere
of radius r1 to the charge enclosed, Q1 . We
also know Q1 = ρ V1 . For the outer sphere
(radius r2 = 29.0129 cm),
Q
E 2 A 2 = Φ2 =
ǫ0
with Q = ρ V (Not ρ V2 , as the Gaussian surface is larger than the actual physical sphere,
and no charge is outside of the sphere.), so
ρV
V
E 2 A2
ǫ
= ◦ =
.
ρ
V
E 1 A1
V1
1
ǫ◦
We know the surface area of a sphere is proportional to the radius squared, and the volume is proportional to the cube of the radius
4
(in particular: A = 4 π R2 and V = π R3 ),
3
so
E 2 A2
E2 r22
=
E 1 A1
E1 r12
and
V
R3
= 3 .
V1
r1
Combining, we get
E2 r22
R3
=
E1 r12
r13
R
p
r2
2
r1
R3
· E1
E2 =
r1 r22
(15 cm)3
(22769.2 N/C)
=
(6.7517 cm) (29.0129 cm)2
= 13521.5 N/C .
Version 047/AACDD – midterm 01 – Turner – (60230)
Note that in this solution, we did not actually
need to remember the specific formulae for
the surface area and volume of a sphere as the
constants cancelled.
Method 2:
First, calculate the charge density ρ inside the
insulating sphere. Select a spherical Gaussian
surface with r = 6.7517 cm, concentric with
the charge distribution. To apply Gauss’ law
in this situation, we must know the charge qin
within the Gaussian surface of volume V ′ . To
calculate qin , we use the fact that qin = ρ V ′ ,
where ρ is the charge per unit volume and
V ′ is the volume enclosed by the Gaussian
4
surface, given by V ′ = π r 3 for a sphere.
3
Therefore,
4
qin = ρ V ′ = ρ
π r3 .
3
Gauss’ law in the region r < R
E (4 π r 2 ) =
4 π r3
qin
=
ρ.
ǫ0
3 ǫ0
Therefore,
3 ǫ0 E
ρ=
r
For r = r1 = 6.7517 cm, R = 15 cm, we know
E = 22769.2 N/C. So
ρ=
3 ǫ0 E
.
r1
Now we can calculate the magnitude of the
electric field at 29.0129 cm from the center.
In this situation, r > R. Therefore, we choose
a spherical Gaussian surface of radius r and
concentric with the insulating sphere. The
net charge inside the Gaussian surface is the
total charge of the insulating sphere. That is
4
Qin = ρ V = ρ π R3
3
3 ǫ0 E 4
3
=
πR
r1
3
4 π ǫ0 E R3
.
=
r1
The electric field on the Gaussian surface is
uniform and normal to the surface. From
3
Gauss’ law, the electric field Er at radius r
can be found through
I
~~
~ = Er · 4 π r 2 = Qin
E
· dA
ǫ0
4 π ǫ0 E R3
Qin
r1
=
E2 =
2
4 π ǫ0 r2
4 π ǫ0 r22
R3
E
=
r1 r 2
(15 cm)3
=
(22769.2 N/C)
(6.7517 cm) (29.0129 cm)2
= 13521.5 N/C .
003 10.0 points
A dipole (electrically neutral) is placed in an
external field.
(a)
−
+
−
+
(c)
(b)
+
−
−
+
(d)
For which situation(s) shown above is the
net force on the dipole equal to zero?
1. (c) only
2. (a) and (c)
3. (c) and (d) correct
4. None of these
5. (b), (c), and (d)
6. (a) and (d)
Version 047/AACDD – midterm 01 – Turner – (60230)
4
y (m)
7. (a), (b), and (c)
8. Another combination
9. (a) only
10. (b) and (d)
Explanation:
Basic Concepts: Field patterns of point
charge and parallel plates of infinite extent.
The force on a charge in the electric field is
given by
~ = qE
~
F
and the torque is defined as
~ = ~r × F
~
T
~ =
∆E
~ =
E
k ∆q
r̂
r2
X
~i.
∆E
Symmetry of the configuration will cause
some component of the electric field to be
zero.
Gauss’ law states
ΦS =
I
~ · dA
~ = Q.
E
ǫ0
Solutions: The electric dipole consists of
two equal and opposite charges separated by
a distance. In either situation (c) or (d), the
electric field is uniform and parallel everywhere. Thus, the electric force on one charge
is equal but opposite to that on another so
that the net force on the whole dipole is zero.
By contrast, electric fields are nonuniform for
situations both (a) and (b).
004 10.0 points
Three charges are arranged in the (x, y)
plane (as shown in the figure below, where
the scale is in meters).
10
9
8
7
6
5
4
3
2
1
0
6 nC
1 nC
8 nC
0 1 2 3 4 5 6 7 8 9 10
x
(m)
Select the figure showing the direction of
the resultant force on the 1 nC charge at the
origin.
1. None of these figures is correct.
y (m)
θ ≈ 334◦
10
8
6
4
2
θ
x
2.
0
(m)
−2
F
−4
−6
−8
−10
−10 −6 −2 0 2 4 6 8 10
y (m)
θ ≈ 206◦
10
8
θ
6
4
2
x cor3.
0
(m)
−2
F
−4
−6
−8
−10
−10 −6 −2 0 2 4 6 8 10
rect
Version 047/AACDD – midterm 01 – Turner – (60230)
y (m)
θ ≈ 154◦
10
8
θ
6
4
F
2
x
4.
0
(m)
−2
−4
−6
−8
−10
−10 −6 −2 0 2 4 6 8 10
y (m)
θ ≈ 26◦
10
8
6
4
F θ
2
x
5.
0
(m)
−2
−4
−6
−8
−10
−10 −6 −2 0 2 4 6 8 10
Explanation:
Let : qo = 1 × 10−9 C ,
(xo , yo ) = (0 m, 0 m) ,
qa = 8 × 10−9 C ,
(xa , ya ) = (4 m, 0 m) ,
qb = 6 × 10−9 C , and
(xb , yb ) = (0 m, 5 m) .
Applying Coulomb’s Law to qo and qa ,
qo qa
Foax = −ke 2 cos θoa
roa
qo qa xoa
= −ke 2
roa roa
= −(8.98755 × 109 N C2 /m2 )
(1 × 10−9 C) (8 × 10−9 C)
×
(4 m)2
(4 m)
×
(4 m)
= −4.49378 × 10−9 N .
Applying Coulomb’s Law to qo and qb ,
qo qa
Foby = −ke 2 sin θob
roa
= −ke
5
qo qb yob
2
rob
rob
= −(8.98755 × 109 N C2 /m2 )
(1 × 10−9 C) (6 × 10−9 C)
×
(5 m)2
(5 m)
×
(5 m)
= −2.15701 × 10−9 N .
y (m)
10
8
6
4
2
x
0
(m)
−2
−4
−6
−8
−10
−10 −6 −2 0 2 4 6 8 10
The direction θ as measured in a counterclockwise direction from the positive x axis
is
Fy
θ = arctan
Fx
−2.15701 × 10−9 N
= arctan
−4.49378 × 10−9 N
= 205.641◦ .
005 10.0 points
A point charge q1 is concentric with two spherical conducting thick shells, as shown in the
figure below. The smaller spherical conducting shell has a net charge of q2 and the larger
spherical conducting shell has a net charge of
q3 .
Version 047/AACDD – midterm 01 – Turner – (60230)
r4
r3
r2
r1
q3
q2
q1
R1
R2 R3 R4 R5
Hint: Under static conditions, the charge
on a conductor resides on the surface of the
conductor.
What is the charge Qr3 on the inner surface
of the larger spherical conducting shell?
1. Qr3 = +q1 + q2 + q3
2. Qr3 = −q1
3. Qr3 = −q1 − q2 − q3
4. Qr3 = −q1 + q2
5. Qr3 = −q1 − q2 + q3
A charge of 6 µC is on the y axis at y = 1 cm,
and a second charge of −6 µC is on the y axis
at y = −1 cm.
y
5
4
3
2
1 6 µC 5 µC
x
0
1 2µC3 4 5 6 7 8 9
−1 −6
−2
−3
−4
−5
Find the force on a charge of 5 µC on the x
axis at x = 4 cm. The value of the Coulomb
constant is 8.98755 × 109 N · m2 /C2 .
1. -143.61
2. -8.52634
3. -7.05933
4. -95.2012
5. -89.7565
6. -138.057
7. -76.9342
8. -5.68423
9. -53.3833
10. -7.14552
Correct answer: −76.9342 N.
6. Qr3 = +q1 − q2
Explanation:
7. Qr3 = +q1 + q2
8. Qr3 = 0
9. Qr3 = +q1
10. Qr3 = −q1 − q2 correct
Explanation:
The net charge inside a Gaussian surface
located at r = R4 must be zero, since the field
in the conductor must be zero. Therefore,
the charge on the inner surface of the large
spherical conducting shell must be −(q1 + q2 ).
006
10.0 points
6
Let : q1 = 6 µC = 6 × 10−6 C ,
q2 = −6 µC = −6 × 10−6 C ,
q3 = 5 µC = 5 × 10−6 C ,
(x1 , y1 ) = (0 cm, 1 cm) = (0 m, 0.01 m) ,
(x2 , y2 ) = (0 cm, −1 cm)
= (0 m, −0.01 m) , and
(x3 , y3 ) = (4 cm, 0 cm) = (0.04 m, 0 m) .
Version 047/AACDD – midterm 01 – Turner – (60230)
y
5
4
3
2
1
0
−1
−2
−3
−4
−5
q1
q3
x
7
007 10.0 points
A cubic box of side a, oriented as shown, contains an unknown charge. The vertically directed electric field has a uniform magnitude
E at the top surface and 2 E at the bottom
surface.
F1
3
q12 2F 2 3 F43 5 36 7 8 9
E
a
The force that q1 exerts on q3 is
~ 1,3 = F cos θ î − F sin θ ĵ
F
2E
and the force that q2 exerts on q3 is
~ 2,3 = −F cos θ î − F sin θ ĵ .
F
Since
F = ke
q3 q1
r2
How much charge Q is inside the box?
1. insufficient information
2. Qencl = 6 ǫ0 E a2
q3 q1
(x1 − x3 )2 + (y1 − y3 )2
= (8.98755 × 109 N · m2 /C2 )
(5 × 10−6 C) (6 × 10−6 C)
×
(0.01 m)2 + (0.04 m)2
= 158.604 N
= ke
3. Qencl = 2 ǫ0 E a2
4. Qencl = ǫ0 E a2 correct
5. Qencl = 3 ǫ0 E a2
1
ǫ0 E a2
2
E
=
ǫ0 a2
E
=3
ǫ0 a2
6. Qencl =
and
y1
θ = tan
x
3
−1 0.01 m
= tan
0.04 m
◦
= 14.0362 ,
−1
7. Qencl
8. Qencl
9. Qencl = 0
10. Qencl = 2
~3 = F
~ 1,3 + F
~ 2,3
F
E
ǫ0 a2
= (F cos θ) î − (F sin θ) ĵ − (F cos θ) î
− (F sin θ) ĵ
= −2 F sin θ ĵ
= −2 (158.604 N) (sin 14.0362◦ ) ĵ
= (−76.9342 N) ĵ .
Explanation:
Electric flux through a surface S is, by convention, positive for electric field lines going
out of the surface S and negative for lines
going in.
Version 047/AACDD – midterm 01 – Turner – (60230)
Here the surface is a cube and no flux goes
through the vertical sides. The top receives
Φtop = −E a2
(inward is negative) and the bottom
2
Φbottom = 2 E a .
The total electric flux is
ΦE = −E a2 + 2 E a2 = E a2 .
Using Gauss’s Law, the charge inside the box
is
Qencl = ǫ0 ΦE = ǫ0 E a2 .
008 10.0 points
A charge of 0.4794 nC is placed at the center
of a cube that measures 3.384 m along each
edge.
What is the electric flux through one face
of the cube?
1. 35.4576
2. 20.2712
3. 9.02825
4. 34.258
5. 27.3258
6. 28.9266
7. 24.1431
8. 38.4087
9. 13.5669
10. 30.5461
keywords:
009 10.0 points
A uniform line charge of density λ lies on the
x axis between x = 0 and x = L. Its total
charge is 10 nC. The electric field at x = 2 L
is (440 N/C)ı̂.
Find the electric field at x = 3 L.
1. 150.0
2. 176.667
3. 246.667
4. 223.333
5. 256.667
6. 216.667
7. 183.333
8. 146.667
9. 200.0
10. 143.333
Correct answer: 146.667 N/C.
Explanation:
Let : Q = 10 nC ,
E(2 L) = (440 N/C)ı̂ , and
k = 8.99 × 109 N · m2 /C2 .
Correct answer: 9.02825 N · m2 /C.
Explanation:
Let : q = 0.4794 nC
The electric field at x > L produced by this
line charge is
= 4.794 × 10−10 C and
a = 3.384 m .
~ · dA
~ = qen .
E
ǫ0
By symmetry, the electric flux through each
face of the cube is equal, so
q
Φ =
6 ǫ0
4.794 × 10−10 C
=
6 (8.85 × 10−12 C2 /N · m2 )
I
= 9.02825 N · m2 /C .
L
λ dx′
′ 2
0 (x − x )
1
1
= kλ
−
x−L x
kQ
kλL
=
=
x(x − L)
x(x − L)
E(x) = k
By Gauss’ law,
Φc =
8
Z
Thus
E(2 L) =
kQ
= 440 N/C ,
2 L2
and
Version 047/AACDD – midterm 01 – Turner – (60230)
kQ
1
= E(2 L)
2
6L
3
1
= (440 N/C) = 146.667 N/C .
3
E(3 L) =
010 10.0 points
An electron moves at 1.6 × 106 m/s into a
uniform electric field of magnitude 1227 N/C.
The
charge
on
an
−19
electron is 1.60218 × 10
C and the mass
of an electron is 9.10939 × 10−31 kg .
The field is parallel to the electron’s velocity
and acts to decelerate the electron.
How far does the electron travel before it is
brought to rest?
1. 6.60702
2. 0.305269
3. 2.11268
4. 3.26564
5. 4.4659
6. 1.48683
7. 3.39441
8. 0.593122
9. 3.66202
10. 1.62309
Correct answer: 0.593122 cm.
Explanation:
Let : v = 1.6 × 106 m/s ,
qe = 1.60218 × 10−19 C ,
m = 9.10939 × 10
E = 1227 N/C .
−31
kg ,
1
m v2
x= 2
qe E
1
9.10939 × 10−31 kg
= 2
1.60218 × 10−19 C
2
1.6 × 106 m/s
×
1227 N/C
= 0.593122 cm .
011 10.0 points
Three equal charges 7.1 µC are located in
the xy-plane, one at (0 m, 59 m), another at
(67 m, 0 m), and the third at (26 m, −58 m).
The value of Coulomb’s constant is
8.98755 × 109 N · m2 /C2 .
Find the magnitude of the electric field at
the origin due to these three charges.
1. 21.955
2. 40.8809
3. 13.335
4. 44.4801
5. 40.1681
6. 61.8366
7. 11.7528
8. 10.7683
9. 21.0441
10. 50.9775
Correct answer: 21.0441 N/C.
and
Explanation:
The kinetic energy
1
m v2
2
is depleted by the amount of work done by
the electric force F = qe E on the particle:
Z
W = F dx = F x = qe E x
K=
since the force is constant. When the electron
comes to rest, all its kinetic energy has been
converted, so
1
m v 2 = qe E x .
2
9
Let : q = 7.1 µC ,
(x1 , y1 ) = (0 m, 59 m) ,
(x2 , y2 ) = (67 m, 0 m) , and
(x3 , y3 ) = (26 m, −58 m) .
~ ik = k Q
E i ≡ kE
ri2
Version 047/AACDD – midterm 01 – Turner – (60230)
y3
gives the magnitude of the electric field due
~
Ey,3 = kE3 k
th
to the i charge. Thus
r3
~ 1 k = ke Q = ke Q
kE
r12
y12
= 8.98755 × 109 N · m2 /C2
7.1 × 10−6 C
(59 m)2
= 18.3314 N/C ,
×
7.1 × 10−6 C
×
(67 m)2
= 14.2151 N/C ,
−58 m
= 15.795 N/C
63.561 m
= −14.413 N/C .
~ are
The x and y components of E
Ex =
3
X
Ex,i
i=1
~ 2 k = ke Q = ke Q
kE
r22
x22
= 8.98755 × 109 N · m2 /C2
10
= Ex,1 + Ex,2 + Ex,3
= 0 N/C + 14.2151 N/C
+ 6.46102 N/C
= 20.6761 N/C and
3
X
Ey =
Ey,i
i=1
and
~ 3 k = ke
kE
Q
Q
= ke 2
2
r3
x3 + y32
= 8.98755 × 109 N · m2 /C2
7.1 × 10−6 C
(26 m)2 + (−58 m)2
= 15.795 N/C .
×
The contributions add like vectors, so we have
to take components
~ i k xi
Ex,i = kE
r
i
~ i k yi .
Ey,i = kE
ri
The x and y components of E1 , E2 , and E3
are
Ex,1
Ey,1
Ex,2
Ey,2
Ex,3
= 0 N/C
= 18.3314 N/C ,
= 14.2151 N/C
= 0 N/C , and
~ 3 k x3
= kE
r3
26 m
= 15.795 N/C
63.561 m
= 6.46102 N/C ,
= Ey,1 + Ey,2 + Ey,3
= 18.3314 N/C + 0 N/C
−14.413 N/C
= 3.91836 N/C .
The net magnitude of the electric field is
q
~ = E2 + E2
kEk
x
y
q
= (20.6761 N/C)2 + (3.91836 N/C)2
= 21.0441 N/C .
012 10.0 points
Two electrostatic point charges of +52.0 µC
and +59.0 µC exert a repulsive force on each
other of 184 N.
The Coulomb constant is 8.98755 ×
109 N · m2 /C2 .
What is the distance between the two
charges?
1. 0.413235
2. 0.419764
3. 0.397161
4. 0.472022
5. 0.447326
6. 0.436577
7. 0.363934
8. 0.372499
Version 047/AACDD – midterm 01 – Turner – (60230)
11
y
++
++
++
Explanation:
++
M.
++
Correct answer: 0.387114 m.
++
9. 0.387114
10. 0.408644
x
−−−−−−
y
−
−
q1 q2
Felectric = kC 2
s r
kC q1 q2
r=
Felectric
q
= 8.98755 × 109 N · m2 /C2
r
(5.2 × 10−5 C) (5.9 × 10−5 C)
×
184 N
= 0.387114 m .
S.
−
q1 = 52.0 µC ,
q2 = 59.0 µC ,
Felectric = 184 N , and
kC = 8.98755 × 109 N · m2 /C2 .
−
Let :
+
+
+
+
x
y
+++++
L.
+
+
+
+
+
+++++
−
−
−
− x
−
For which configuration(s) does the total
electric field vector at the origin have nonzero components in the x direction as well as
the y direction (i.e., both x and y components
are non-zero)?
1. Configurations G, L and M only
013
2. Configurations S, P and M only
10.0 points
y
++++
3. Configuration L only
4. Configurations S and G only
G.
x
6. Configurations G and M only
++++
7. Configuration G only
y
++++
P.
8. Configurations S and M only
x
−−−−
5. Configuration S only
9. Configurations S, G and M only
10. Configuration P only correct
Explanation:
Version 047/AACDD – midterm 01 – Turner – (60230)
y
+++++
L
+
+
+
+
+
Configuration M: It is symmetric about
the y-axis, so the x component of the total
field must vanish.
y
−
−
−
−
++++
x
−−−−
Configuration L: It is symmetric about the
x-axis, so the y component of the total field
must vanish.
++
++
−−−−−−
014 10.0 points
A circular arc has a uniform linear charge
density of 8 nC/m.
The value of the Coulomb constant is
8.98755 × 109 N · m2 /C2 .
y
◦
P
x
64
Configuration P: It is anti-symmetric by
rotation of a 180◦ , so the total field has nonzero components in both x and y directions,
just like the field generated by just one piece
of charge.
y
++++
++
x
++
G
M
++
x
Configuration G: It is symmetric by a rotation of 180◦ , so the electric fields generated
by these two pieces have opposite directions;
therefore the total field is zero.
y
++++
−
−
−
− x
−
+++++
++
k∆q
Basic Concepts: ∆E = 2 r̂ and E =
r
X
∆E . Symmetry of the configuration will
cause some component of the electric field to
be zero.
Solution: Configuration S: It is antisymmetric about the y-axis (opposite sign
of charges), so the electric field has no ycomponent.
y
+
+
+
S
+
12
2. 3
m
x
What is the magnitude of the electric field
at the center of the circle along which the arc
lies?
1. 21.4131
2. 12.8692
3. 40.6483
4. 46.9173
5. 30.6393
6. 33.1316
7. 29.0072
8. 25.5445
9. 16.644
Version 047/AACDD – midterm 01 – Turner – (60230)
10. 41.6814
Correct answer: 33.1316 N/C.
Explanation:
Let : λ = 8 nC/m = 8 × 10−9 C/m ,
∆θ = 64◦ , and
r = 2.3 m .
θ is defined as the angle in the counterclockwise direction from the positive x axis
as shown in the figure below.
◦
on the positive y axis and extending towards
the positive x axis) and multiply the answer
by 2.
Note: The upper angular limit θ = 90◦ .
The lower angular limit θ = 90◦ − 32◦ = 58◦ ,
is the angle from the positive x axis to the
right-hand end of the arc.
!
Z ◦
λ 90
E = −2 ke
sin θ dθ ̂
r 58◦
= −2 ke
θ
~
E
First, position the arc symmetrically
around the y axis, centered at the origin. By
symmetry (in this rotated configuration) the
field in the x direction cancels due to charge
from opposites sides of the y-axis, so
Ex = 0 .
For a continuous linear charge distribution,
Z
dq
~
E = ke
r̂
r2
In polar coordinates
dq = λ (r dθ) ,
where λ is the linear charge density. The
positive y axis is θ = 90◦ , so the y component
of the electric field is given by
dEy = dE sin θ .
Note: By symmetry, each half of the arc
about the y axis contributes equally to the
electric field at the origin. Hence, we may just
consider the right-half of the arc (beginning
λ
[cos (58◦ ) − cos (90◦ )] ̂ .
r
Since
ke
32 ◦
r
32
13
λ
= (8.98755 × 109 N · m2 /C2 )
r
(8 × 10−9 C/m)
×
(2.3 m)
= 31.261 N/C ,
E = −2 (31.261 N/C)
× [(0.529919) − (0)] ̂
= −33.1316 N/C ̂
~ = 33.1316 N/C .
kEk
~ in
Alternate Solution: Just solve for kEk
a straight forward manner, positioning the
beginning of the arc on the positive x axis (as
shown in the original figure in the question).
θ is still defined as the angle in the counterclockwise direction from the positive x axis.
!
Z ◦
ke λ 64
cos θ dθ ı̂
Ex = −
r 0◦
ke λ
[sin (64◦ ) − sin (0◦ )] ı̂
r
= −(31.261 N/C)
× [(0.898794) − 0.0] ı̂
= [−28.0972 N/C] ı̂ ,
!
Z ◦
ke λ 64
Ey = −
sin θ dθ ̂
r 0◦
=−
ke λ
[cos (0◦ ) − cos (64◦ )] ̂
r
= −(31.261 N/C)
× [1.0 − (0.438371)] ̂
= [−17.5571 N/C] ̂ ,
=−
Version 047/AACDD – midterm 01 – Turner – (60230)
q
~ = E2 + E2
kEk
y
x
h
= (−28.0972 N/C)2
+ (−17.5571 N/C)2
= 33.1317 N/C .
equilibrium, so the electric force must cancel
the force of gravity Fe = Fg , we have
Correct answer: −2.53371 µC/m2 .
Explanation:
m = 7.3 g ,
q = −0.5 µC ,
2
g = 9.8 m/s ,
Fg = m g
Fe = q E
σ
mg = q
,
2 ǫ0
i1/2
015 10.0 points
A 7.3 g piece of Styrofoam carries a net charge
of −0.5 µC and floats above the center of a
very large horizontal sheet of plastic that has
a uniform charge density on its surface.
The acceleration of gravity is 9.8 m/s2
and the permittivity of free space is
8.85419 × 10−12 C2 /N/m2 .
What is the charge per unit area on the
plastic sheet?
1. -5.24097
2. -5.13685
3. -1.45775
4. -2.84609
5. -4.72034
6. -3.78322
7. -4.65093
8. -3.2973
9. -4.89389
10. -2.53371
Let :
14
so
mg
q
= 2 (8.85419 × 10−12 C2 /N/m2 )
(0.0073 kg) (9.8 m/s2 ) 106 µC
×
·
−5 × 10−7 C
1C
σ = 2 ǫ0
= −2.53371 µC/m2 .
016 10.0 points
A charge of 9 pC is uniformly distributed
throughout the volume between concentric
spherical surfaces having radii of 1.1 cm and
3.1 cm.
Let: ke = 8.98755 × 109 N · m2 /C2 .
What is the magnitude of the electric field
1.5 cm from the center of the surfaces?
1. 25.8195
2. 6.49101
3. 21.0549
4. 19.477
5. 19.789
6. 16.8386
7. 18.861
8. 24.2635
9. 12.8113
10. 14.0578
Correct answer: 25.8195 N/C.
and
ǫ0 = 8.85419 × 10−12 C2 /N/m2 .
Explanation:
Let :
qtot = 9 pC
The field due to a nonconducting infinite
sheet of charge is the same as that very close
to any plane uniform charge distribution. The
field is
σ
E=
,
2 ǫ0
where σ is the surface charge density (charge
per unit area) of the plastic sheet. Call the
charge on the styrofoam q, and its mass m.
Since the styrofoam is floating, it must be in
= 9 × 10−12 C ,
r1 = 1.1 cm ,
r2 = 3.1 cm , and
r = 1.5 cm
= 0.015 m .
By Gauss’ law,
Φc =
I
~ · dA
~ = qin
E
ǫ0
Version 047/AACDD – midterm 01 – Turner – (60230)
The tricky part of this question is to determine the charge enclosed by our Gaussian
surface, which by symmetry considerations is
chosen to be a concentric sphere with radius
r. Since the charge q is distributed uniformly
within the solid, we have the relation
8. 506672.0
9. 523974.0
10. 337735.0
Correct answer: 1.0001 × 106 N · m2 /C.
Explanation:
Vin
qin
=
qtot
Vtot
where qin and Vin are the charge and volume
enclosed by the Gaussian surface. Therefore
Vin
qin = qtot
Vtot
3
r − r13
= qtot 3
r2 − r13
(1.5 cm)3 − (1.1 cm)3
= (9 pC) ×
(3.1 cm)3 − (1.1 cm)3
= 0.646381 pC
= 6.46381 × 10−13 C .
And by Gauss’s Law,
qin
r2
= 8.98755 × 109 N · m2 /C2
6.46381 × 10−13 C
×
(0.015 m)2
15
Let : E = 31000 N/C ,
ℓ = 7.7 m ,
w = 4.3 m , and
θ = 13◦ .
By Gauss’ law,
~ ·A
~
Φ=E
The flux through the bottom of the car is
Φ = E A cos θ
= E ℓ w cos θ
= (31000 N/C) (7.7 m)
× (4.3 m) cos(13◦ )
= 1.0001 × 106 N · m2 /C .
E = ke
018 10.0 points
1) Two uncharged metal balls, X and Y,
stand on glass rods and are touching.
Y
= 25.8195 N/C .
017 10.0 points
An electric field of magnitude 31000 N/C and
directed upward perpendicular to the Earth’s
surface exists on a day when a thunderstorm
is brewing. A truck that can be approximated
as a rectangle 7.7 m by 4.3 m is traveling
along a road that is inclined 13 ◦ relative to
the ground.
Determine the electric flux through the bottom of the truck.
1. 446640.0
2. 452033.0
3. 75450.2
4. 1000100.0
5. 161696.0
6. 397754.0
7. 495662.0
X
2) A third ball, carrying a positive charge, is
brought near the first two.
+
Y
X
3) Then the first two balls are separated from
each other,
+
Y
X
Version 047/AACDD – midterm 01 – Turner – (60230)
4) and the third ball is finally removed.
Y
X
When this is all four steps are done, it is
found out that
1. Balls X and Y are both positive.
2. Ball X is positive and ball Y is negative.
correct
3. Balls X and Y are both negative.
4. Ball X is negative and ball Y is positive.
5. Balls X and Y are still uncharged.
Explanation:
Basic Concept: Electric induction caused
by nearby charges.
Solution: When a positive ball is moved
near a metallic object (X and Y), the positive
charge will attract negative charges, causing
X to have excess positive charge and Y to
have excess negative charge (X and Y are in
contact, so the total net charge on X and Y
should be zero).
+
−
+
Later, X and Y are separated, retaining
their charges, so when the third ball is finally
removed, X will have net positive charge and
Y will have net negative charge.
16
Kapoor (mk9499) – oldmidterm 02 – Turner – (60230)
This print-out should have 21 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
A capacitor network is shown in the following
figure.
4.44 µF
5.5 µF
a
b
13.6 µF
A coaxial cable with length ℓ has an inner
conductor that has a radius a and carries a
charge of Q. The surrounding conductor has
an inner radius b and a charge of −Q.
Assume the region between the conductors
Q
is air. The linear charge density λ ≡ .
ℓ
Explanation:
radius = b
−Q
E=
2.
E=
3.
E=
4.
E=
Since C1 and C2 are in series they carry the
same charge
5.
E=
C1 V1 = C2 V2 ,
6.
E=
and their voltages add up to V , voltage of the
battery
7.
E=
8.
E=
9.
E=
10.
E=
C1
C2
C3
V
= 4.44 µF,
= 5.5 µF,
= 13.6 µF,
= 18.2 V.
and
V1 + V2 = V
C2 V2
+ V2 = V
C1
C2 V2 + C1 V2 = V C1
V C1
V2 =
C1 + C2
(18.2 V)(4.44 µF)
=
4.44 µF + 5.5 µF
= 8.12958 V .
002 (part 1 of 2) 10.0 points
b
What is the electric field halfway between
the conductors?
1.
Let :
ℓ
radius = a
+Q
18.2 V
What is the voltage across the 5.5 µF upper
right-hand capacitor?
Correct answer: 8.12958 V.
1
λ
π ǫ0 r 2
λ
2 π ǫ0 r 2
Q
π ǫ0 r
Q
4 π ǫ0 r
Q
π ǫ0 r 2
λ
π ǫ0 r
λ
4 π ǫ0 r
Q
2 π ǫ0 r 2
Q
2 π ǫ0 r
λ
correct
2 π ǫ0 r
Explanation:
The electric field of a cylindrical capacitor
is given by
λ
.
E=
2 π r ǫ0
Kapoor (mk9499) – oldmidterm 02 – Turner – (60230)
C=
2.
C=
3.
C=
4.
C=
5.
C=
6.
C=
7.
C=
8.
C=
9.
10.
C=
C=
c
µF
a
55.3 µF
1.
2ℓ
a
ke ln
b
ℓ
correct
b
2 ke ln
a
ℓ
b
ke ln
a
2ℓ
b
ke ln
a
ke ℓ
b
ln
a
ℓ
a
ke ln
b
ℓ
ke
ke ℓ
b
2 ln
a
ℓ
2 ke
ℓa
2 ke b
5
17.
33.1 µF
003 (part 2 of 2) 10.0 points
What is the capacitance C of this coaxial
cable?
2
93.4 V
b
1
73.
d
µF
Find the capacitance between points a and
b of the entire capacitor network.
Correct answer: 111.306 µF.
Explanation:
Let :
C1
C2
C3
C4
E
C1
= 17.5 µF ,
= 33.1 µF ,
= 55.3 µF ,
= 73.1 µF ,
= 93.4 V .
c
and
C2
a
b
C3
E
C4
d
A good rule of thumb is to eliminate junctions connected by zero capacitance.
Explanation:
C1
a
C2
C=
C4
b
C3
ℓ
.
b
2 ke ln
a
004 (part 1 of 4) 10.0 points
Four capacitors are connected as shown in the
figure.
Q
The definition of capacitance is C ≡ .
V
The series connection of C2 and C3 gives
the equivalent capacitance
1
C23 =
1
1
+
C2 C3
C2 C3
=
C2 + C3
Kapoor (mk9499) – oldmidterm 02 – Turner – (60230)
(33.1 µF) (55.3 µF)
33.1 µF + 55.3 µF
= 20.7062 µF .
=
The total capacitance Cab between a and b can
be obtained by calculating the capacitance in
the parallel combination of the capacitors C1 ,
C4 , and C23 ; i.e.,
Cab = C1 + C4 + C23
= 17.5 µF + 73.1 µF + 20.7062 µF
= 111.306 µF .
005 (part 2 of 4) 10.0 points
What is the charge on the 33.1 µF centeredupper capacitor?
Correct answer: 1933.96 µC.
Explanation:
The voltages across C2 and C3 , respectively,
(the voltage between a and b) are Vab = V23 =
93.4 V , and we have
Q23 = Q3 = Q2
= Vab C23
= (93.4 V)(20.7062 µF)
= 1933.96 µC .
006 (part 3 of 4) 10.0 points
A dielectric with dielectric constant 3.1 is
inserted into the 55.3 µF capacitor (lowercentered capacitor) while the battery is connected.
What is the charge on the 73.1 µF lowerright capacitor?
Correct answer: 6827.54 µC.
Explanation:
Since the battery is still connected, the voltage will remain the same. Thus the charge is
simply
Q4 = Vab C4
= (93.4 V) (73.1 µF)
= 6827.54 µC .
007 (part 4 of 4) 10.0 points
3
If the battery is removed before the dielectric
in the above question is inserted, what will be
the charge on the 73.1 µF lower-right capacitor?
Correct answer: 6421.55 µC.
Explanation:
After the battery is removed, as the dielectric is inserted into C3 , there will be a
redistribution of charge.
Note: The total charge is unchanged, so
Qab = Q′ab .
The primed quantities correspond to those
after the insertion of the dielectric. Before the
battery was disconnected,
Qab = Vab Cab
C2 C3
= Vab C1 + C4 +
.
C2 + C3
from Part 1. After the battery was disconnected and the dielectric was inserted,
′
Vab
=
Qab
′ .
Cab
The equivalent capacitance of the two capacitors C2 and C3 is now
′
C23
=
C2 κ C3
,
C2 + κ C3
′
so the new total equivalent capacitance Cab
is
C2 κ C3
C2 + κ C3
= 17.5 µF + 73.1 µF
(33.1 µF) (3.1)(55.3 µF)
+
33.1 µF + (3.1)(55.3 µF)
= 118.343 µF .
′
Cab
= C1 + C4 +
′
The new voltage Vab
between a and b is therefore
Qab
′
Cab
C
= Vab ab
′ ,
Cab
′
Vab
=
Kapoor (mk9499) – oldmidterm 02 – Turner – (60230)
since Q′ab = Qab = Vab Cab . The new charge
Q′4 on C4 is now
′
Q′4 = Vab
C4
Cab
= Vab ′ C4
Cab
(111.306 µF)
(73.1 µF)
= (93.4 V)
(118.343 µF)
= 6421.55 µC .
008 10.0 points
A parallel-plate capacitor is charged by connecting it to a battery.
If the battery is disconnected and the separation between the plates is increased, what
will happen to the charge on the capacitor
and the electric potential across it?
1. The charge remains fixed and the electric
potential increases. correct
2. The charge increases and the electric potential decreases.
3. The charge and the electric potential decrease.
4. The charge remains fixed and the electric
potential decreases.
5. The charge decreases and the electric potential increases.
6. The charge decreases and the electric potential remains fixed.
7. The charge and the electric potential remain fixed.
8. The charge increases and the electric potential remains fixed.
9. The charge and the electric potential increase.
Explanation:
Charge is conserved, so it must remain constant since it is stuck on the plates. With the
4
battery disconnected, Q is fixed.
C = ǫ0
A
d
A larger d makes the fraction smaller, so C is
Q
smaller. Thus the new potential V ′ = ′ is
C
larger.
009 10.0 points
A resistor is made from a hollow cylinder of
length l, inner radius a, and outer radius b.
The region a < r < b is filled with material of
resistivity ρ. Current runs along the axis of
the cylinder.
The resistance R of this component is
ρ ℓ b2
1. R =
π a4
ρ ℓ a2
2. R =
π b4
2ρ ℓ
3. R =
π b2
ρℓ
4. R =
correct
2
π (b − a2 )
ρa
5. R =
πℓ b2
2πρ ℓ
6. R = 2
(b − a2 )
ρℓ
7. R =
π b2
π b2 − a2 ρ
8. R =
ℓ
2
πb ρ
9. R =
ℓ
ρℓ
10. R =
π a2
Explanation:
By definition
R=ρ
ℓ
A
ℓ
− π a2
ρℓ
.
=
2
π (b − a2 )
=ρ
010
π b2
10.0 points
Kapoor (mk9499) – oldmidterm 02 – Turner – (60230)
The emf of a battery is E = 15 V . When the
battery delivers a current of 0.2 A to a load,
the potential difference between the terminals
of the battery is 13 V volts.
Find the internal resistance of the battery.
Correct answer: 10 Ω.
Explanation:
Given : E = 15 V ,
Vload = 13 V , and
I = 0.2 A .
The potential difference across the internal
resistance is E − Vload , so the internal resistance is given by
E − Vload
I
15 V − 13 V
=
0.2 A
= 10 Ω .
5
27 Ω
1 + [0.007 (◦ C)−1 ] (25◦ C)
= 22.9787 Ω .
=
Using Eq. (1), with resistance R2 ,
R2
−1
R0
∆T2 =
α
160 Ω
−1
22.9787
Ω
=
0.007 (◦ C)−1
= 851.852◦ C .
(3)
Or directly substituting Eq. (1) into (2), we
have
r=
∆T2 =
=
011 10.0 points
An incandescent bulb has a resistance of 27 Ω
when it is at room temperature (25 ◦ C) and
160 Ω when it is hot and delivering light to
the room. The temperature coefficient of resistivity of the filament is 0.007 (◦ C)−1 , where
the base resistance R0 is determined at 0◦ C.
What is the temperature of the bulb when
in use?
Correct answer: 851.852◦C.
=
=
=
=
R2
−1
R0
α
R2
[1 + α ∆T1 ] − 1
R1
α
R2 + α R2 ∆T1 − R1
α
R2
R2 − R1
+
∆T1
(4)
α R1
R1
160 Ω − 27 Ω
[0.007 (◦ C)−1 ] (27 Ω)
160 Ω
(25◦ C)
+
27 Ω
◦
851.852 C .
Explanation:
Let : R1 = 27 Ω ,
R2 = 160 Ω , and
α = 0.007 (◦ C)−1 .
Temperature dependence of resistance is
RT = R0 [1 + α (T − 0◦ C)]
= R0 [1 + α ∆T ] .
96 Ω
R1
[1 + α ∆T1 ]
18 Ω
(1)
Using Eq. (1), at room temperature; e.g.,
R1 = 27 Ω and ∆T1 = 25◦ C
R0 =
012 10.0 points
The equivalent resistance of the circuit in
the figure is Req = 80.0 Ω .
18 Ω
96 Ω
(2)
R
Find the value of R.
Correct answer: 23 Ω.
E
S
Kapoor (mk9499) – oldmidterm 02 – Turner – (60230)
6
Explanation:
R4
R1
R2
E
R
Let :
R1
R2
R3
R4
Req
= 96.0 Ω ,
= 18.0 Ω ,
= 18.0 Ω ,
= 96.0 Ω ,
= 80.0 Ω .
34 Ω
30 Ω
2Ω
S
25 µF
R3
62 Ω
32 V
S
What is the magnitude of the electric potential across the capacitor?
Correct answer: 14 V.
and
Explanation:
t
R2
R1
1
Req,p
=
a
It
R3
1
1
+
Ra Rb
E
For resistors in series,
b
It
C
Basic Concepts: For resistors in parallel,
R4
Ib
Ib
S1
b
Req,s = Ra + Rb
Let :
R12 = R1 + R2
= 96 Ω + 18 Ω
= 114 Ω , and
R34 = R3 + R4
= 18 Ω + 96 Ω
= 114 Ω , and
−1
1
1
+
R1234 =
R12 R34
−1
1
1
=
+
114 Ω 114 Ω
= 57 Ω , and
R = Req − R1234
= 80 Ω − 57 Ω
R1
R2
R3
R4
C
= 30 Ω ,
= 34 Ω ,
= 2 Ω,
= 62 Ω , and
= 25 µF .
After a “long time” implies that the capacitor C is fully charged and therefore the
capacitor acts as an open circuit with no current flowing to it. The equivalent circuit is
It
It
a
R1
R3
Ib
R2
b
R4
Ib
= 23 Ω .
Rt = R1 + R2 = 30 Ω + 34 Ω = 64 Ω
013 (part 1 of 2) 10.0 points
The circuit has been connected as shown in
the figure for a “long” time.
Rb = R3 + R4 = 2 Ω + 62 Ω = 64 Ω
It =
32 V
E
=
= 0.5 A
Rt
64 Ω
Kapoor (mk9499) – oldmidterm 02 – Turner – (60230)
Ib =
E
32 V
= 0.5 A
=
Rb
64 Ω
7
and
Across R1
Req =
E 1 = It R 1
= (0.5 A) (30 Ω)
= 15 V .
1
1
+
Rℓ Rr
−1
1
1
+
=
32 Ω 96 Ω
= 24 Ω .
Across R3
−1
Therefore the time constant τ is
E 3 = Ib R 3
= (0.5 A) (2 Ω)
= 1 V.
τ ≡ Req C = (24 Ω) (25 µF) = 600 µs .
The equation for discharge of the capacitor is
Qt
= e−t/τ , or
Q0
Et
1
= e−t/τ = .
E0
e
Since E1 and E3 are “measured” from the same
point “a”, the potential across C must be
EC = E3 − E1
= 1 V − 15 V
= −14 V
|EC | = 14 V .
014 (part 2 of 2) 10.0 points
If the battery is disconnected, how long does it
1
Et
=
take for the capacitor to discharge to
E0
e
of its initial voltage?
Correct answer: 600 µs.
Explanation:
With the battery removed, the circuit is
Ir
Iℓ
R2
= 600 µs .
015
10.0 points
A network below consists of with three batteries, each having an internal resistance, and
five resistors.
18 V
7Ω
5Ω
1Ω
R4
Iℓ
Ir
Req
C
r
Ieq
R1
R3
C
ℓ
Taking the logarithm of both sides, we have
1
t
− = ln
τ
e
t = −τ (− ln e)
= −(600 µs) (−1)
where
Rℓ = R1 + R3 = 30 Ω + 2 Ω = 32 Ω,
Rr = R2 + R4 = 34 Ω + 62 Ω = 96 Ω
28 V
a
b
4Ω
35 V
1Ω
6Ω
1Ω
7Ω
Find the magnitude of the potential difference between points a and b.
Correct answer: 1.16 V.
Explanation:
Kapoor (mk9499) – oldmidterm 02 – Turner – (60230)
E1
R1
r1
R3
i
We can then find the magnitude of the potential difference between a and b, |Vab |, by using
the top loop; i.e.,
i
a
b
R2
E2
E3
r2
r3
R5
R4
i
i
Let : E1
E2
E3
R1
R2
R3
R4
R5
r1
r2
r3
= 18 V ,
= 28 V ,
= 35 V ,
= 7 Ω,
= 4 Ω,
= 5 Ω,
= 7 Ω,
= 6 Ω,
= 1 Ω,
= 1 Ω , and
= 1 Ω.
Basic Concepts: Kirchhoff’s Laws:
X
V = 0 around a closed loop.
X
I = 0 at a circuit junction.
Solution: There is no current flow between
a and b. Therefore, applying the loop rule to
the outside path in the figure above, we have
E1 − E3 − i (r1 + R1 + R2 + r3 + R4 + R3 ) = 0 ,
Since
R1234 = 1 Ω + 7 Ω + 4 Ω
+1 Ω+7Ω+5Ω
= 25 Ω ,
We have
E1 − E3
r1 + R 1 + R 2 + r3 + R 4 + R 3
E1 − E3
=
R1234
18 V − 35 V
=
25 Ω
= −0.68 A .
i=
8
Vab = E1 − E2 − i (R1 + r1 + R3 )
= (18 V) − (28 V)
− (−0.68 A) (7 Ω + 1 Ω + 5 Ω)
= −1.16 V
|Vab | = 1.16 V ,
where the current through r2 and R5 is zero.
016 10.0 points
In the circuit shown, the capacitor is initially
uncharged. At t1 = 0, the switch S is moved
to position “a”.
R2
C
R1
S b
V0
a
Find VR1 , the voltage drop across R1 , as a
function of time t1 .
1. VR1 = V0 et1 /[(R1 +R2 )C]
o
n
2. VR1 = V0 1 − et1 /[(R1 +R2 ) C]
n
o
−t1 /(R2 C)
3. VR1 = V0 1 − e
4. VR1 = V0 e−[t1 (R1 +R2 )]/R1 R2 C
o
n
−t1 /(R1 C)
5. VR1 = V0 1 − e
6. VR1 = V0 e−t1 /(R2 C)
7. VR1 = V0 e−t1 /(R1 C) correct
n
o
[−t1 (R1 +R2 )]/R1 R2 C
8. VR1 = V0 1 − e
Kapoor (mk9499) – oldmidterm 02 – Turner – (60230)
Right-hand rule for cross-products.
~
b ≡ F ; i.e., a unit vector in the F direcF
~k
kF
tion.
~ = q ~v × B.
~
Solution: The force is F
I = I0 e−t1 /(RC)
V0 −t1 /(R1 C)
=
e
.
R1
Since I R1 = VR1 ,
~
B
~v
q
~
F
VR1 = V0 e−t1 /(R1 C) .
10.0 points
A positively charged particle moving parallel to the y-axis enters a magnetic field (pointing toward the left-hand side of the page), as
shown in the figure below.
z
~
B
x
+q
y
v
~
B
Figure: ı̂ is in the x-direction, ̂ is
in the y-direction, and k̂ is in the
z-direction.
= B (−̂) ,
= v (+̂) , and
> 0 , therefore,
~
= +|q| ~v × B
= +|q| v B [(+̂) × (−̂)]
= +|q| v B (0)
b = 0 no deflection .
F
This is the second of eight versions of the
problem.
018 10.0 points
A wire carrying a current 20 A has a length
0.1 m between the pole faces of a magnet at
an angle 60 ◦ (see the figure). The magnetic
field is approximately uniform at 0.5 T. We
ignore the field beyond the pole pieces.
I
B
θ
ℓ
Explanation:
For an “RC” circuit,
017
What is the initial direction of deflection?
b = −k̂
1. F
b = +̂
2. F
b = +ı̂
3. F
b = −ı̂
4. F
b = −̂
5. F
~ = 0 ; no deflection correct
6. F
b = +k̂
7. F
9
Explanation:
Basic Concepts: Magnetic Force on a
Charged Particle:
~ = q ~v × B
~
F
What is the force on the wire?
Correct answer: 0.866025 N.
Explanation:
Let : I = 20 A ,
ℓ = 0.1 m ,
θ = 60 ◦ , and
B = 0.5 T .
we use F = I ℓ B sin θ, so
F = I ℓ B sin θ
= (20 A) (0.1 m) (0.5 T) sin 60 ◦
= 0.866025 N .
Kapoor (mk9499) – oldmidterm 02 – Turner – (60230)
019 10.0 points
A square loop of wire carries a current and is
located in a uniform magnetic field.
The left side of the loop is aligned and
attached to a fixed axis (dashed line in figure).
y
x
0.44 m
→ 5.4 A →
← axis of rotation
~ = 0.054 T
B
10
= d×I ℓB
= (0.44 m) (5.4 A) (0.44 m) (0.054 T)
= 0.0564538 N m .
020 10.0 points
What is the radius of the smallest possible circular orbit that a proton with energy
0.47 MeV can have in a 2 T magnetic field?
The mass of a proton is 1.67 × 10−27 kg and
its charge is 1.609 × 10−19 C.
Correct answer: 0.0493872 m.
Explanation:
0.44 m
~ = 0.054 T
B
When the plane of the loop is parallel to the
magnetic field in the position shown, what is
the magnitude of the torque exerted on the
loop about the axis of rotation, which is along
the left side of the square as indicated by the
dashed line in the figure?
Correct answer: 0.0564538 N m.
Let : m = 1.67 × 10−27 kg ,
Q = 1.609 × 10−19 C ,
B = 2 T , and
E = 0.47 MeV .
The energy of a proton is E =
Explanation:
← axis of rotation
~
B
v=
y
x
d
ℓ
→ I →
=
r
s
1
m v 2 , so
2
2E
m
2 (0.47 MeV) (1.609 × 10−19 J/eV)
(1 × 10−6 MeV/eV) (1.67 × 10−27 kg)
= 9.51664 × 106 m/s .
If the proton is shot into the magnetic field
with a velocity at right angles to the direction
of the field, we will get the smallest radius
~
B
mv
Bq
(1.67 × 10−27 kg) (9.51664 × 106 m/s)
=
(2 T) (1.609 × 10−19 C)
r=
Let : d = 0.44 m ,
ℓ = 0.44 m ,
B = 0.054 T ,
I = 5.4 A .
and
Only the right side of the loop contributes to
the torque. By definition, the torque is
~k
τ = k~d × F
= 0.0493872 m .
021 10.0 points
Consider the setup of a velocity selector for
~ is pointing
the case where the electric field E
downward along y axis.
Kapoor (mk9499) – oldmidterm 02 – Turner – (60230)
y
−
v
E
O
x
z
Choose the direction of the magnetic field
~
B such that a negatively charged particle,
moving at an appropriate speed in the positive
~ and B
~
x-direction (ı̂), passes through the E
region undeflected.
b = −ı̂
1. B
b = −k̂ correct
2. B
b = √1 (−̂ + k̂)
3. B
2
b = √1 (−̂ − k̂)
4. B
2
1
b = √ (̂ − k̂)
5. B
2
b = +̂
6. B
b = −̂
7. B
b = √1 (̂ + k̂)
8. B
2
b = +k̂
9. B
b = +ı̂
10. B
Explanation:
The negatively charged particle passes
through the selector undeflected when the
electric force is equal and opposite to the magnetic force. Since the electric field is pointing
in the negative y direction, and the particle’s
charge is negative, the particle feels an electric
force in the positive y direction.
~ E = |q| E ̂
F
where E is the magnitude of the electric field.
We must choose the magnetic field such that
the magnetic force is in the negative y direction. The general equation for the magnetic
force is given by
~ B = q (~v × B)
~
F
11
~ B to point opposite to F
~ E , i.e., in
We want F
the (−̂) direction to have a chance of cancelling the net force. Considering only the
unit vectors along which the above vectors
point, we must satisfy
−̂ = −(ı̂ × r̂)
(1)
where r̂ denotes the unknown direction in
~ must point. Since
which B
ı̂ × (−k̂) = ̂
we see that equation (1) for r̂ is satisfied by
r̂ = −k̂
so the magnetic field must point into the page.
Version 147/ACBAD – midterm 02 – Turner – (60230)
This print-out should have 18 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 (part 1 of 2) 10.0 points
The circuit has been connected as shown in
the figure for a “long” time.
11 µF
2Ω
C = 11 µF .
After a “long time” implies that the capacitor C is fully charged and therefore the
capacitor acts as an open circuit with no current flowing to it. The equivalent circuit is
It
It
a
R1
R3
18 Ω
14 Ω
1
R2
b
R4
Ib
Ib
30 Ω
Rt = R1 + R2 = 14 Ω + 18 Ω = 32 Ω
32 V
S
What is the magnitude of the electric potential across the capacitor?
1. 20.0
2. 19.0
3. 3.0
4. 12.0
5. 16.0
6. 40.0
7. 1.0
8. 2.0
9. 14.0
10. 6.0
Correct answer: 12 V.
Explanation:
t
It
R3
E
R4
Ib
b
Let :
b
It
C
a
R1
R2
R3
R4
= 14 Ω ,
= 18 Ω ,
= 2 Ω,
= 30 Ω ,
Ib
S1
and
E 1 = It R 1
= (1 A) (14 Ω)
= 14 V .
Across R3
E 3 = Ib R 3
= (1 A) (2 Ω)
= 2 V.
Since E1 and E3 are “measured” from the same
point “a”, the potential across C must be
R2
R1
Rb = R3 + R4 = 2 Ω + 30 Ω = 32 Ω
E
32 V
It =
=
=1A
Rt
32 Ω
E
32 V
Ib =
=1A
=
Rb
32 Ω
Across R1
EC = E3 − E1
= 2 V − 14 V
= −12 V
|EC | = 12 V .
002 (part 2 of 2) 10.0 points
If the battery is disconnected, how long does it
1
Et
=
take for the capacitor to discharge to
E0
e
of its initial voltage?
1. 91.0
Version 147/ACBAD – midterm 02 – Turner – (60230)
2
Taking the logarithm of both sides, we have
t
1
− = ln
τ
e
t = −τ (− ln e)
= −(132 µs) (−1)
2. 132.0
3. 147.0
4. 504.0
5. 1020.0
6. 456.0
7. 510.0
8. 462.0
9. 196.0
10. 374.0
= 132 µs .
003
10.0 points
Correct answer: 132 µs.
Explanation:
With the battery removed, the circuit is
Ir
Iℓ
R1
R3
R2
8V
1Ω
r
C
ℓ
In the figure below the battery has an emf of
8 V and an internal resistance of 1 Ω . Assume
there is a steady current flowing in the circuit.
R4
Iℓ
11 Ω
4Ω
Ieq
C
Req
Ir
where
Rℓ = R1 + R3 = 14 Ω + 2 Ω = 16 Ω,
Rr = R2 + R4 = 18 Ω + 30 Ω = 48 Ω
and
Req =
1
1
+
Rℓ Rr
−1
1
1
+
=
16 Ω 48 Ω
= 12 Ω .
−1
Therefore the time constant τ is
τ ≡ Req C = (12 Ω) (11 µF) = 132 µs .
The equation for discharge of the capacitor is
Qt
= e−t/τ , or
Q0
1
Et
= e−t/τ = .
E0
e
2 µF
Find the charge on the 2 µF capacitor.
1. 15.5556
2. 9.88235
3. 4.0
4. 32.4
5. 11.375
6. 7.22222
7. 27.6923
8. 40.5
9. 33.3333
10. 7.35294
Correct answer: 4 µC.
Explanation:
Let :
R1
R2
rin
V
C
= 11 Ω ,
= 4 Ω,
= 1 Ω,
= 8 V , and
= 2 µF .
The equivalent resistance of the three resistors
in series is
Req = R1 + R2 + rin
Version 147/ACBAD – midterm 02 – Turner – (60230)
= (11 Ω) + (4 Ω) + (1 Ω)
= 16 Ω ,
V
so the current in the circuit is I =
, and
Req
the voltage across R2 is
V2 = I R2
R2
V
=
Req
(4 Ω)
=
(8 V)
(16 Ω)
= 2 V.
3
∂V
∂y
2
= − α x (2 y) + δ (3 y 2) z
h
= − (3 V/m4 ) (2.8 m)2 [2 (−1 m)]
i
4
2
+ (6.1 V/m ) 3 (−1 m) (2.6 m)
Ey = −
= −0.54 V/m .
Since R2 and C are parallel, the potential
difference across each is the same. Hence the
charge on the capacitor is
Q = C V2
= (2 µF) (2 V)
005 10.0 points
A wire that has a uniform linear charge density of 2.9 µC/m is bent into the shape as
shown below, with radius 4.8 m.
= 4 µC .
004
4.8 m
10.0 points
Given:
9.6 m
2 2
2
2
p
9.6 m
3
V = α x y + β z (x − γ) + δ y z ,
where α = 3 V/m4 , β = 9.5 V/m4 , γ =
4.7 m2 , and δ = 6.1 V/m4 .
What is the y component of the electric
field Ey at (2.8 m, −1 m, 2.6 m)?
1. 29.92
2. -1520.55
3. -684.8
4. -19.71
5. -0.54
6. -656.32
7. 60.39
8. -10.5
9. -306.52
10. 79.88
Correct answer: −0.54 V/m.
Explanation:
α = 3 V/m4 ,
β = 9.5 V/m4 ,
γ = 4.7 m2 ,
δ = 6.1 V/m4 , and
(x, y, z) = (2.8 m, −1 m, 2.6 m) .
The Coulomb constant is 8.98755 ×
109 N · m2 /C2 .
Find the electrical potential at point p.
1. 143949.0
2. 57579.5
3. 47982.9
4. 52781.2
5. 62377.8
6. 124756.0
7. 71974.3
8. 139150.0
9. 115159.0
10. 129554.0
Correct answer: 1.3915 × 105 V.
Explanation:
Let :
Let : λ = 2.9 µC/m = 2.9 × 10−6 C/m ,
R = 4.8 m ,
2 R = 9.6 m , and
ke = 8.98755 × 109 N · m2 /C2 .
Version 147/ACBAD – midterm 02 – Turner – (60230)
R
p
2R
2R
Let p be the origin. Consider the potential
due to the line of charge to the right of p.
Z
Vright = dV
Z
dq
= ke
r
Z 3R
λdx
= ke
x
R
3R
= ke λ ln x
= ke λ ln 3 .
R
By symmetry, the contribution from the line
of charge to the left of p is the same. The
contribution from the semicircle is
Z π
λRdθ
Vsemi = ke
R
0
Z π
= ke λ
dθ
0
2. 52373600000000.0
3. 2660850000000.0
4. 79269500000000.0
5. 36783600000000.0
6. 46767200000000.0
7. 28332800000000.0
8. 87190800000000.0
9. 13072800000000.0
10. 21286800000000.0
Correct answer: 2.66085 × 1012 m3 .
Explanation:
Let : C = 1 µF = 1 × 10−6 F and
ǫ0 = 8.85419 × 10−12 C2 /N · m2 .
Let the radii be a and 2 a. Let us put a
charge Q in the inner conductor and −Q in
the outer conductor. The potential of the
ke Q
while that
inner conductor is Vinner =
a
ke Q
of the outer is Vouter =
. The potential
2a
difference between the conductors is
∆V = Vinner − Vouter =
π
= ke λ θ
0
= ke λ π .
= 1.3915 × 105 V .
006 10.0 points
A 1 µF spherical capacitor is composed of two
metal spheres, one having a radius twice as
large as the other.
If the region between the spheres is a vacuum, determine the volume of this region.
1. 24642200000000.0
ke Q
2a
from which we obtain
C=
Hence the electric potential at p is
Vp = Vright + Vlef t + Vsemi
= 2 ke λ ln 3 + ke λ π
= ke λ (2 ln 3 + π)
= (8.98755 × 109 N · m2 /C2 )
× (2.9 × 10−6 C/m)
× (2 ln 3 + π)
4
Q
= 8 π ǫ0 a .
∆V
Thus,
a=
C
8 π ǫ0
1 × 10−6 F
8 π (8.85419 × 10−12 C2 /N · m2 )
= 4493.78 m .
=
The intervening volume is
4
4
π (2 a)3 − π (a)3
3
3
4
= π 7a3
3
4 = π 7 (4493.78 m)3
3
∆Vol =
= 2.66085 × 1012 m3 .
Version 147/ACBAD – midterm 02 – Turner – (60230)
5
Since
007 10.0 points
At 30 ◦ C, the resistance of a segment of gold
wire is 77 Ω. When the wire is placed in a
liquid bath, the resistance increases to 190 Ω.
The temperature coefficient is 0.0034 (◦ C)−1
at 20◦ C.
What is the temperature of the bath?
1. 504.546
2. 462.917
3. 746.849
4. 551.716
5. 720.588
6. 476.303
7. 682.801
8. 317.07
9. 247.676
10. 795.517
Correct answer: 476.303◦C.
Explanation:
Let : R1 = 77 Ω ,
R2 = 190 Ω ,
T0 = 20◦ C ,
T1 = 30◦ C , and
α = 0.0034(◦ C)−1 .
Neglecting change in the shape of the wire,
we have
R1 = R0 [1 + α(T1 − T0 )]
R1
R0 =
1 + α (T1 − T0 )
and
R2 = R0 [1 + α (T2 − T0 )] ,
where T0 = 20◦ C .
Thus
R2 =
R1 [1 + α (T2 − T0 )]
1 + α (T1 − T0 )
R2 + α R2 (T1 − T0 ) = R1 + α R1 (T2 − T0 )
α R1 T2 = R2 − R1 + α R2 (T1 − T0 ) + α R1 T0
so that
T2 =
R2 − R1 + α R2 (T1 − T0 ) + α R1 T0
.
α R1
R2 − R1 + α R2 (T1 − T0 ) + R1 α T0
= 190 Ω − 77 Ω
+ [0.0034 (◦ C)−1 ]
× (190 Ω) (30◦ C − 20◦ C)
(77 Ω) [0.0034(◦ C)−1 ] (20◦ C)
= 124.696 Ω ,
then
T2 =
124.696 Ω
(77 Ω) (0.0034(◦ C)−1 )
= 476.303◦C .
008 10.0 points
A 62 m length of coaxial cable has a solid
cylindrical wire inner conductor with a diameter of 3.921 mm and carries a charge of
7.76 µC. The surrounding conductor is a
cylindrical shell and has an inner diameter of
10.761 mm and a charge of −7.76 µC.
Assume the region between the conductors
is air. The Coulomb constant is 8.98755 ×
109 N · m2 /C2 .
What is the capacitance of this cable?
1. 9.10287
2. 2.62668
3. 0.811357
4. 3.24664
5. 1.87607
6. 1.73136
7. 0.917751
8. 2.26507
9. 2.83943
10. 3.41648
Correct answer: 3.41648 nF.
Explanation:
Let :
ke = 8.98755 × 109 N · m2 /C2 ,
Q = 7.76 µC ,
ℓ = 62 m ,
a = 3.921 mm , and
b = 10.761 mm .
Version 147/ACBAD – midterm 02 – Turner – (60230)
The charge per unit length is λ ≡
Z
Q
.
ℓ
b
~ · d~s
E
Z b
dr
= −2 ke λ
a r Q
b
= −2 ke ln
.
ℓ
a
V =−
2.
a
3.
4.
The capacitance of a cylindrical capacitor
is given by
Q
C≡
V
ℓ
=
2 ke
5.
6.
7.
1
b
ln
a
62 m
=
2 (8.98755 × 109 N · m2 /C2 )
1
1 × 109 nF
·
× 10.761 mm
1F
ln
3.921 mm
= 3.41648 nF
009 10.0 points
Consider two cylindrical conductors made of
the same ohmic material. Conductor 1 has a
radius r1 and length ℓ1 while conductor 2 has
a radius r2 and length ℓ2 .
Denote: The currents of the two conductors
as I1 and I2 , the potential differences between
the two ends of the conductors as V1 and V2 ,
and the electric fields within the conductors
as E1 and E2 .
~E 1
1.
~E 2
I1
V1
I2
V2
b
b
ℓ1
r1
ℓ2
8.
9.
10.
R2
R1
R2
R1
R2
R1
R2
R1
R2
R1
R2
R1
R2
R1
R2
R1
R2
R1
R2
R1
=
6
2
3
=4
=
1
3
=3
=
=
=
=
=
1
4
1
2
3
2
4
3
3
correct
4
=2
Explanation:
The relation between resistance and resistivity is given by
R=
ρℓ
ρℓ
=
.
A
π r2
Thus the ratio of the resistances is
R2
ρ ℓ2 π r12
=
R1
π r22 ρ ℓ1
ℓ2 r12
=
ℓ1 r22
3 ℓ1 r12
=
ℓ1 (2 r1 )2
=
3
.
4
r2
If ρ2 = ρ1 , r2 = 2 r1 , ℓ2 = 3 ℓ1 and V2 =
R2
of the resistances.
V1 , find the ratio
R1
010 10.0 points
A capacitor network with air-filled capacitors
as shown below.
Version 147/ACBAD – midterm 02 – Turner – (60230)
85.4 µF
85.4 µF
c
a
85.4 µF
40 V
85.4 µF
b
d
When the top right-hand capacitor is filled
with a material of dielectric constant κ, the
charge on this capacitor is increases by a factor of 1.47.
Find the dielectric constant κ of the material inserted into the top right-hand capacitor.
1. 2.7037
2. 2.92157
3. 1.5974
4. 2.77358
5. 2.33333
6. 2.07692
7. 3.0
8. 2.50877
9. 1.8169
10. 2.0303
do with this problems. In addition, the capacitances are all equal and their specific values
are immaterial. Furthermore, the electric potential of the battery is not required.
C1 = C2 = C3 = C4 , where Q and Q′
are the initial and final charges on C2 and
Q′
≡ α=ratio of final to initial charge on C2 .
Q
We know the charges on C1 and C2 are the
same. Initially,
Vab = V1 + V2
Q
Q
=
+
C1 C2
Q Q
= +
C
C
Q
=2 .
C
1
Vab C .
2
After the dielectric material is inserted in C2 ,
the capacitance becomes C2′ = κ C. Therefore,
Q=
Vab = V1′ + V2′
Q′
Q′
=
+ ′
C1 C2
Q′
Q′
+
=
C
κC
κ + 1 Q′
=
,
κ C
Explanation:
= C = 85.4 µF ,
= C = 85.4 µF ,
= C = 85.4 µF ,
= C = 85.4 µF ,
= 40 V , and
= 1.5 Q .
C1
and using Eq. (1) and solving for Q′ , we have
κ + 1 Q′
2Q
=
C
κ C
κ
′
Q =
Vab C
κ+1
κ
=
2Q
κ+1
Q′
2κ
≡α=
= 1.47 .
Q
κ+1
C2
c
a
EB
C3
C4
(1)
Therefore
Correct answer: 2.77358.
Let : C1
C2
C3
C4
EB
Q′
7
b
d
The capacitors C3 and C4 have nothing to
Solving for κ, we have
κ=
1.47
α
=
= 2.77358 .
2−α
2 − 1.47
011
10.0 points
Version 147/ACBAD – midterm 02 – Turner – (60230)
Four charges are fixed at the corners of a
square centered at the origin as follows: q at
(−a, +a); 2 q at (+a, +a); −3 q at (+a, −a);
and 6 q at (−a, −a). A fifth charge +q with
mass m is placed at the origin and released
from rest.
Find the speed when it is a great distance
from the origin, where the potential energy of
the fifth charge due to the four point charges
is negligible.
s
√
6 2k
1. k~vk = q
correct
ma
s
√
3 5k
2. k~vk = q
ma
s
√
3 3k
3. k~vk = q
ma
s
√
2 2k
4. k~vk = q
ma
s
√
6 5k
5. k~vk = q
ma
s
√
6 3k
6. k~vk = q
ma
s
√
3 6k
7. k~vk = q
ma
s
√
3 2k
8. k~vk = q
ma
s
√
6 6k
9. k~vk = q
ma
s
√
2 5k
10. k~vk = q
ma
The initial energy of the charge is
Ei = Ki + Ui
= Ui
2kq
(−3 q) k
6kq
kq
+√
+ √
+√
=q √
2a
2a
2a
2a
2
6kq
.
= √
2a
The final energy is
Ef =
a
Ei = Ef
6 k q2
1
√
= m v2
2
2a
s √
6 2k
.
v=q
ma
012 (part 1 of 2) 10.0 points
Consider a cubic resistor network shown,
where all the resistors are the same. Each
resistor has a resistance r. A current I comes
into the network at A. The same current
leaves the network at B.
r
√
2a
I
r
C
+6 q
−3 q
r
r
r
r
r
r
r
x
a
D
r
A
+q, m
B
r
I
+2 q
1
m v2 .
2
From energy conversation, we have
r
Explanation:
+q
8
Determine ICD ; i.e., the current between
points C and D.
1. ICD =
I
5
Version 147/ACBAD – midterm 02 – Turner – (60230)
2. ICD
3. ICD
4. ICD
I
=
9
I
=
4
I
= correct
6
5. ICD = I
I
3
I
7. ICD =
2
Explanation:
Ther are 3 paths leaving A. Since all resistors are equal, the paths are symmetric, and
the current entering at A splits into thirds:
I
IAC = .
3
Now we are at C. The current coming in from
A has two possible choices to exit, and there
is complete symmetry between the two paths.
Thus the current going through C splits in
half:
I
IAC
=
ICD =
.
2
6
6. ICD =
013 (part 2 of 2) 10.0 points
Determine RAB ; i.e., the effective resistance
between points A and B.
1. RAB =
2. RAB =
3. RAB =
4. RAB =
5. RAB =
6. RAB =
7. RAB =
8. RAB =
3r
2
r
2
2r
3
r
3
5r
correct
6
7r
6
4r
3
r
6
9. RAB = r
9
Explanation:
Finally, we calculate the equivalent resistance.
We can step through the path ACDB to find
the potential VAB between A and B. Kirchoff’s Laws tell us that each time we cross a
resistor (moving with the current) we drop a
potential V = R I. Therefore
VA − VB = r IAC + r ICD + r IDB
Now note that IAC = IDB by symmetry, so
I
5
I I
VA − VB = |VAB | = r
= rI
+ +
3 6 3
6
Now |VAB | = RAB I, so
RAB =
5
r .
6
014 10.0 points
Two conducting spheres with diameters of
0.46 m and 1.2 m are separated by a distance
that is large compared to the diameters. The
spheres are connected by a thin wire and are
charged to 4.5 µC.
The permittivity of a vacuum is 8.85419 ×
10−12 C2 /N · m2 .
What is the potential of the system of
spheres when the reference potential is taken
to be 0 at ∞?
1. 131.714
2. 48.7277
3. 127.516
4. 150.067
5. 129.421
6. 96.484
7. 42.7979
8. 65.5446
9. 43.6895
10. 78.6411
Correct answer: 48.7277 kV.
Explanation:
Let : ǫ0 = 8.85419 × 10−12 C2 /N · m2 ,
Q = 4.5 µC ,
d1 = 0.46 m , and
d2 = 1.2 m .
Version 147/ACBAD – midterm 02 – Turner – (60230)
Since the spheres are connected by a wire,
their potentials will be equal. Since they are
separated by a distance much larger than their
diameters,
Q2
Q1
= V2 =
.
C1
C2
V1 =
which there is a vertical uniform electric field
(in the −̂ direction, due to the potential difference 1000 V across the distance 1.9 cm)
and a 0.3 T uniform magnetic field (aligned
with the ±k̂-direction) as shown in the figure
by the shaded area. (There is magnetic field
in both Region I and Region II).
Because C = 4 π ǫ0 R = 2 π ǫ0 d, we obtain
Q1
Q2
=
,
d1
d2
Q2 = Q − Q1
= 4.5 µC − 1.24699 µC
= 3.25301 µC .
The potential of the system is
Q1
2 π ǫ0 d1
1.24699 × 10−6 C 1 kV
·
=
2 π ǫ0 (0.46 m)
1000 V
V1 = V2 =
= 48.7277 kV .
015
10.0 points
A device (“source”) emits a bunch of
charged ions (particles) with a range of velocities (see figure). Some of these ions pass
through the left slit and enter “Region I” in
Region of
Magnetic
Field
0.3 T
q
m
1.9 cm
y
cm
Q − Q1
Q1
=
d1
d2
d2
= Q − Q1
Q1
d1
Q
Q1 =
d2
1+
d1
4.5 µC
=
1.2 m
1+
0.46 m
= 1.24699 µC .
+1000 V
43
where d1 and d2 are the diameters of the
spheres. Since the total charge is Q = Q1 +
Q2 ,
10
x
z
Region I
Region II
Figure: ı̂ is in the direction +x
(to the right), ̂ is in the direction
+y (up the page), and k̂ is in the
direction +z (out of the page).
The ions that make it into “Region II” are
observed to be deflected downward and then
follow a circular path with a radius of r =
0.43 m.
The charge on each ion is 4.4 × 10−18 C.
What is the mass of the ions?
1. 3.6047e-25
2. 1.33754e-25
3. 3.41775e-25
4. 1.30693e-25
5. 3.23532e-24
6. 5.04e-26
7. 6.27455e-25
8. 7.52578e-25
9. 4.94795e-25
10. 1.728e-25
Correct answer: 3.23532 × 10−24 kg.
Explanation:
To obtain a straight orbit, the upward and
downward forces need to cancel. The force on
a charged particle is
~ =F
~E + F
~ B = q (E
~ + ~v × B)
~ .
F
Version 147/ACBAD – midterm 02 – Turner – (60230)
For the force to be zero, we need
~E + F
~B = 0 ,
F
Therefore, the forces are equal and opposite
and the magnitude of forces are equal; i.e.,
~ E k = kF
~Bk .
kF
The force due to the magnetic field provides
the centripetal force that causes the negative
ions to move in the semicircle.
As the positively charged ion exits the re~ B = q ~v × B,
~ so by
gion of the electric field, F
the right-hand rule the
magnetic field must
point out of the page or in the +z-direction
~ is in the direction up
+k̂ , since the force F
the page; i.e., “+̂ ”
qE = qvB
E
52631.6 N/C
v=
=
= 1.75439 × 105 m/s .
B
0.3 T
Let :
r = 0.43 m = 0.43 m
q = 4.4 × 10−18 C .
and
The radius of a circular path taken by a
charged particle in a magnetic field is given
by
mv
.
qB
Br
m=q
v
= (4.4 × 10−18 C)
(0.3 T)(0.43 m)
×
1.75439 × 105 m/s
r=
= 3.23532 × 10−24 kg .
and
the vector product
ı̂ × k̂ = −̂ , and since
~ = q ~v × B
~ = kF
~ k (−̂)
F
#
"
~
~
F
q
~v
B
=
×
~k
~
|q| k~vk kBk
kF
h
i
= + (+ı̂) × +k̂
= −̂ ,
~
B
= +k̂
~
kBk
Since the electric and magnetic forces on the
ion are equal,
or
~ E = −F
~B .
F
q
=+
|q|
~v
= +ı̂
k~v k
~
B
=?
~
kBk
~B
F
= −̂ ,
~B k
kF
11
consequently
is correct.
Let : B = 0.3 T , and
(1000 V)
V
=
E≡
d
(1.9 cm)
= 52631.6 N/C .
016 10.0 points
An 12.23 m long copper wire with a crosssectional area of 6.31 × 10−5 m2 , in the shape
of a square loop, is connected to a 0.1517 V
battery. The resistivity of copper is 2.8 ×
10−8 Ω m.
If the loop is placed in a uniform magnetic
field of magnitude 0.219 T, what is the maximum torque that can act on it?
1. 8.08982
2. 71.1179
3. 191.702
4. 113.233
5. 38.2864
6. 26.4157
7. 57.2279
8. 43.1942
9. 61.1678
10. 176.326
Correct answer: 57.2279 N m.
Explanation:
Version 147/ACBAD – midterm 02 – Turner – (60230)
12
Ut
of energy stored in
Ub
the top portion to the bottom portion.
Determine the ratio
Let :
L = 12.23 m ,
A = 6.31 × 10−5 m2 ,
V = 0.1517 V ,
B = 0.219 T , and
ρ = 2.8 × 10−8 Ω m .
1.
2.
First find the current that flows in the wire:
ρL
R=
A
(2.8 × 10−8 Ω m)(12.23 m)
=
6.31 × 10−5 m2
= 0.00542694 Ω ,
4.
so
6.
0.1517 V
V
=
R
0.00542694 Ω
= 27.9531 A.
I=
Each side of the square loop formed by the
wire is 12.23 m m long so the enclosed area is
2
12.23 m
2
= 9.34831 m2 .
S=L =
4
Then the maximum torque on the loop is
τmax = I S B
= (27.9531 A) (9.34831 m2 ) (0.219 T)
3.
5.
7.
8.
9.
10.
Ut
Ub
Ut
Ub
Ut
Ub
Ut
Ub
Ut
Ub
Ut
Ub
Ut
Ub
Ut
Ub
Ut
Ub
Ut
Ub
1
κ
1
= 2
κ
=
=κ+1
=
1
2κ
= κ correct
=
1
2κ + 1
= κ2
= 2κ + 1
= 2κ
=
1
κ+1
Explanation:
For a capacitor with dielectric κ,
= 57.2279 N m .
017 10.0 points
A capacitor is constructed from two metal
plates. The bottom portion is filled with air
and the top portion is filled with material of
dielectric constant κ. The plate area in the
top region is the same as that in the bottom
region. Neglect edge effects.
κ
C=
κǫ0 A
d
Energy stored in a charged capacitor:
U=
1
1 Q2
1
QV = V2C =
.
2
2
2 C
Ct
Cb
E
The capacitance of a capacitor is determined solely by the dielectric constant of the
Version 147/ACBAD – midterm 02 – Turner – (60230)
material and geometric configuration of the
capacitor.
The bottom portion and the top portion
may be treated as two capacitors connected
in parallel. Here Vb = Vt = V . Therefore,
R6
i
i
R1
R2
1 2
V Ct
Ut
2
=
1 2
Ub
V Cb
2
Ct
=
Cb
= κ.
Let : R1
R2
R3
R4
R5
R6
R7
10.0 points
R7
i
= 3 Ω,
= 1.9 Ω ,
= 4.1 Ω ,
= 9.3 Ω ,
= 7.9 Ω ,
= 2 Ω , and
= 4.7 Ω .
4.1 Ω
• Ohm’s Law.
2 Ω
b
What is the resistance between point a and
point b?
1. 6.94554
2. 8.03982
3. 7.32645
4. 7.08374
5. 8.40939
6. 7.16595
7. 7.68918
8. 6.81295
9. 7.52074
10. 7.8355
Explanation:
Let’s redraw the figure
R5
• Equivalent resistance.
There are two rules for adding up resistances. If the resistances are in series, then
Rseries = R1 + R2 + R3 + · · · + Rn .
4.7 Ω
Correct answer: 6.81295 Ω.
i
b
Basic Concepts:
7.9 Ω
1.9 Ω
9.3 Ω
3 Ω
i
R4
Consider the combination of resistors
shown in the figure.
a
i
R3
i
a
018
13
If the resistances are parallel, then
1
1
1
1
1
+
+
+···+
.
=
Rparallel
R1 R2 R3
Rn
Solution: The key to a complex arrangements of resistors like this is to split the problem up into smaller parts where either all the
resistors are in series, or all of them are in
parallel. It is easier to visualize the problem
if you redraw the circuit each time you add
them.
R367
a
i
i
R1
R2
i
i
i
R4
R5
b
Version 147/ACBAD – midterm 02 – Turner – (60230)
Step 1: The three resistors on the right are
all in series, so
R367 = R3 + R6 + R7
= (4.1 Ω) + (2 Ω) + (4.7 Ω)
= 10.8 Ω .
Step 5: Finally, R1 and R236754 are in series, so the equivalent resistance of the circuit
is
Req = R1 + R236754
= 3 Ω + 3.81295 Ω
= 6.81295 Ω .
a
i
i
i
R1
R2
i
b
R3675
R4
Step 2: R5 and R367 are connected parallel, so
R3675
a
−1
1
1
=
+
R5 R367
R5 R367
=
R5 + R367
(7.9 Ω) (10.8 Ω)
=
18.7 Ω
= 4.56257 Ω .
i
i
R1
R36752
i
b
R4
Step 3: R2 and R3675 are in series, so
R23675 = R2 + R3675
= (1.9 Ω) + (4.56257 Ω)
= 6.46257 Ω .
Step 4: R23675 and R4 are parallel, so
R236754
a
−1
1
1
=
+
R4 R23675
R4 R23675
=
R4 + R23675
(9.3 Ω) (6.46257 Ω)
=
15.7626 Ω
= 3.81295 Ω .
i
i
R1
R367524
b
14
Kapoor (mk9499) – oldmidterm 03 – Turner – (60230)
This print-out should have 16 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
A copper strip (8.47 × 1022 electrons per cubic centimeter) 7.4 cm wide and 0.02 cm thick
is used to measure the magnitudes of unknown magnetic fields that are perpendicular
to the strip.
The charge on the electron is 1.6 × 10−19 C.
Find the magnitude of B when the current
is 24 A and the Hall voltage is 1.8 µV.
Correct answer: 0.20328 T.
Explanation:
Let : n = 8.47 × 1022 cm−3 ,
= 8.47 × 1028 m−3 ,
q = 1.6 × 10−19 C ,
t = 0.02 cm = 0.0002 m ,
w = 7.4 cm = 0.074 m ,
I = 24 A , and
VH = 1.8 µV = 1.8 × 10−6 V .
current of 1.2 A. When the plane of the coil
makes an angle of 36◦ with a uniform magnetic
field, the torque on the coil is 0.05 N m.
What is the magnitude of the magnetic
field?
Correct answer: 0.256361 T.
Explanation:
Let : N = 49 turns ,
I = 1.2 A ,
θ = 36◦ ,
A = 41 cm2 = 0.0041 m2 ,
τ = 0.05 N m .
The Hall voltage is
VH = vd w B
VH
B=
vd w
n q t VH
B=
I
(8.47 × 1028 m−3 ) (1.6 × 10−19 C)
=
24 A
× (0.0002 m) (1.8 × 10−6 V)
= 0.20328 T .
002 10.0 points
A small rectangular coil composed of 49 turns
of wire has an area of 41 cm2 and carries a
and
The magnetic force on the current is
~ = I ~ℓ × B
~
F
and the torque is
~,
~τ = ~r × F
so the torque on the loop due to the magnetic
field is
τ = 2 F r cos θ
= (N I ℓ B) w cos θ
= N I B (ℓ w) cos θ
= N I B A cos θ ,
The current in the metal strip is
I = n q vd A = n q vd (w t)
I
vd w =
nqt
1
where A is the area of the loop and θ is the
angle between the plane of the loop and the
magnetic field.
The magnetic field from above is
B=
=
τ
N I A cos θ
0.05 N m
(49 turns) (1.2 A) (0.0041 m2 ) cos(36◦ )
= 0.256361 T .
003 10.0 points
Given: Assume the bar and rails have negligible resistance and friction.
In the arrangement shown in the figure,
the resistor is 6 Ω and a 7 T magnetic field
is directed into the paper. The separation
Kapoor (mk9499) – oldmidterm 03 – Turner – (60230)
B 2 ℓ2 v 2
R
(7 T)2 (5 m)2 (5 m/s)2
=
(6 Ω)
= 5104.17 W .
between the rails is 5 m . Neglect the mass of
the bar.
An applied force moves the bar to the left
at a constant speed of 5 m/s .
7T
2
=
5 m/s
m≪1 g
6Ω
5m
I
Note: First of four versions.
7T
At what rate is energy dissipated in the
resistor?
Correct answer: 5104.17 W.
Explanation:
Basic Concept: Motional E
E = Bℓv.
004 10.0 points
A coil is wrapped with 582 turns of wire on
the perimeter of a circular frame (of radius
37 cm). Each turn has the same area, equal
to that of the frame. A uniform magnetic
field is directed perpendicular to the plane of
the coil. This field changes at a constant rate
from 22 mT to 64 mT in 54 ms.
What is the magnitude of the induced E in
the coil at the instant the magnetic field has
a magnitude of 49 mT?
Correct answer: 194.685 V.
Explanation:
Basic Concepts:
Ohm’s Law
E = −N
V
I= .
R
ΦB ≡
Solution: The motional E induced in the
circuit is
Solution:
E = Bℓv
= (7 T) (5 m) (5 m/s)
= 175 V .
From Ohm’s law, the current flowing through
the resistor is
E
R
Bℓv
=
R
(7 T) (5 m) (5 m/s)
=
R
= 29.1667 A .
I=
The power dissipated in the resistor is
P = I2 R
B 2 ℓ2 v 2
R
=
R2
Z
d ΦB
dt
~ · dA
~ =B·A
B
d ΦB
dt
∆B
= −N A
∆t
(B2 − B1 )
= −N π r 2
∆t
= −(582) π (37 cm)2
(64 mT) − (22 mT)
×
54 ms
= −194.685 V
|E| = 194.685 V .
E = −N
005 (part 1 of 2) 10.0 points
A horizontal circular wire loop of radius 0.3 m
lies in a plane perpendicular to a uniform
magnetic field pointing from above into the
plane of the loop, has a magnitude of 0.51 T.
Kapoor (mk9499) – oldmidterm 03 – Turner – (60230)
If in 0.14 s the wire is reshaped from a circle
into a square, but remains in the same plane,
what is the magnitude of the average induced
emf in the wire during this time?
Correct answer: 0.221038 V.
Explanation:
3. does not arise.
Explanation:
The deformation causes the flux through
the loop to decrease since the area of the loop
is reduced. By Lenz’s law, the induced emf
will cause the current to flow in the loop so
as to induce a magnetic field that attempts
to resist the change of magnetic flux through
the loop. A clockwise flow of current, when
viewed from above tends to increase the existing downward magnetic field through the
loop, thereby resisting the decrease of magnetic flux through the loop.
and
The average induced emf E is given by
∆Φ
∆Φ
=
hEi = N
∆t
∆t
since N = 1, and
∆Φ = B (Acircle − Asquare )
= B (π r 2 − Asquare ) .
Also, the circumference of the circle is 2 π r,
so each side of the square has a length
2πr
πr
=
,
4
2
so
Asquare = L2 =
Thus
2. flows in a direction that cannot be determined from given information.
4. flows clockwise when viewed from above.
correct
Let : r = 0.3 m ,
b = 0.51 T ,
∆t = 0.14 s .
L=
3
π r 2
2
π r 2 2
∆Φ = B π r −
2
"
= 0.51 T π (0.3 m)2 −
007 10.0 points
A magnetic dipole is falling in a conducting
metallic tube. Consider the induced current
in an imaginary current loop when the magnet
is moving away from the upper loop and when
the magnet is moving toward the lower loop.
.
?
π (0.3 m) 2
2
!#
and the average induced emf is
−0.0309454 T · m2
hEi = −
= 0.221038 V .
0.14 s
006 (part 2 of 2) 10.0 points
The current in the loop during the deformation
1. flows counter-clockwise when viewed from
above.
Npole
y
= −0.0309454 T · m2 .
Iabove
x
v
N
dipole
magnet
S
Spole
?
Ibelow
z
Determine the directions of the induced
currents Iabove and Ibelow in an imaginary loop
Kapoor (mk9499) – oldmidterm 03 – Turner – (60230)
s
2. 2
v/
re
3 × 10−5 T
9
1. no current flow
magnetic field is into the plane of the paper.
1.
shown in the figure, as viewed from above,
when the loop is above the falling magnet and
when the loop is below the falling magnet.
4
m
2. Iabove = counter-clockwise and
Ibelow = counter-clockwise
3. Iabove = clockwise and
Ibelow = counter-clockwise
3 × 10−5 T
4. Iabove = clockwise and
Ibelow = clockwise
What is the magnitude of the emf E induced
between the blade tip and the central hub?
Correct answer: 0.000866703 V.
5. Iabove = counter-clockwise and
Ibelow = clockwise correct
Explanation:
Iabove
Npole
y
x
v
N
dipole
magnet
S
Spole
z
Ibelow
008 (part 1 of 2) 10.0 points
The figure below shows one of the blades of
a helicopter which rotates around a central
hub. The vertical component of the Earth’s
B
f
Explanation:
When the falling magnet is below the up→
per loop, −
µ ind must be up to attract the
falling magnet and slow it down; i.e., counterclockwise as viewed from above.
→
Before reaching the lower loop, −
µ ind must
be down to oppose the falling magnet; i.e.,
clockwise as viewed from above.
ℓ
B
Let : ℓ = 2.2 m ,
f = 1.9 rev/s , and
B = 3 × 10−5 T .
For a point on the blade, the velocity with
which the point moves changes linearly with
the distance from the point to the center of
the hub. Then the effective velocity for the
whole blade is the mean velocity,
ω·ℓ
2
2πf · ℓ
=
2
2π (1.9 rev/s) (2.2 m)
=
2
= 13.1319 m/s ,
veff =
and the induced emf in the blade is
E = B ℓ veff
Kapoor (mk9499) – oldmidterm 03 – Turner – (60230)
1
B ω ℓ2
2
1
= (3 × 10−5 T)
2
× (11.9381 rad/s) (2.2 m)2
= 0.000866703 V .
=
Alternative Solution: An alternative
dA
d ΦB
=B
, where
method is to calculate
dt
dt
A is the area enclosed in the diagram below
and ΦB the enclosed flux in the figure.
ω
ℓ
θ
The area inside the triangle is A =
ℓ2 d θ
ω ℓ2
dA
=
=
.
and
dt
2 dt
2
Therefore,
ℓ2
2
θ,
d ΦB
dt
dA
= −B
dt
ℓ2
= −B ω
2
1
|E| = B ω ℓ2 ,
2
5
The magnetic force on a charge carrier of
~ where ~v is the local veloccharge q is q ~v × B
ity of the blade. If q is positive, a force (use
the cross product rule) is induced in the direction from the hub to the tip of the blade.
Hence, positive charges move to the tip (leaving negative charges at the hub).
Eventually an electric field induced by this
charge separation points from the positively
charged tip to the hub and the total force
balance between magnetic and electric forces
is established on the charge carriers in the
blade.
The net result is the tip is charged positively. If the charge q is negative, the magnetic force is towards the hub; negative charge
accumulates there, leaving positive charge at
the tip, with the same conclusion that the tip
is positively charged.
010 10.0 points
A square piece of copper is pulled through
a magnetic field B (into the page ⊗, out of
the page ⊙). Shown below are different charge
configurations associated with this procedure.
Select the figure with an acceptable charge
distribution.
E =−
B
B
−−−−−
1.
v
+++++
B
B
B
B
the same result as obtained above.
009 (part 2 of 2) 10.0 points
The tip of the blade is
1. charged negative.
2.
−
−
−
−
−
v
2. charged, but sign cannot be determined.
3. uncharged.
4. charged positive. correct
Explanation:
B
+
+
+
+
+
B
Kapoor (mk9499) – oldmidterm 03 – Turner – (60230)
B
B
+
+
+
+
+
3.
v
v
+++++
B
B
B
−−−−−
4.
B
011 10.0 points
A circular coil enclosing an area of 95.1 cm2
is made of 192 turns of copper wire as shown
schematically in the figure. Initially, a 0.868 T
uniform magnetic field points perpendicularly
left-to-right through the plane of the coil. The
direction of the field then reverses to right-toleft. The field reversal takes 1.0 ms.
B
v
+++++
B
B
−−−−−
−
−
−
−
−
B
B
6
B
correct
B
B
Magnetic
Field B(t)
+++++
5.
R
v
−−−−−
B
B
B
Explanation:
From Faraday’s Law for Solenoids
B
+++++
6.
During the time the field is changing its
direction, how much charge flows through the
coil if the resistance is 4.31 Ω?
Correct answer: 0.735452 C.
E = −N
v
and Ohm’s Law
−−−−−
B
dΦB
dt
B
V
,
R
the current through R is
I=
V
R
E
=
R
N d ΦB
=
R dt
dBA 1
=N
dt R
dB A
.
=N
dt R
I=
Explanation:
Using the right-hand-rule, the only acceptable charge distribution is
Kapoor (mk9499) – oldmidterm 03 – Turner – (60230)
Integrating both sides of the equation above
yields
Z t
Z t
A dB
N
I dt =
dt
R dt
t0
t0
Z B
A
=
N dB
R
−B
A
= N ∆B
R
A
= N 2B.
R
The left hand side of the above equation is
just the charge flowing through the R during
this period of time! So,
Z t
I dt
Q=
Faraday’s Law for solenoid
dΦ
dt
d (BA)
.
= −N
dt
Magnetic field induced by solenoid
E = −N
B = µ0 n I .
Faraday’s Law for solenoid
dΦ
dt
d (BA)
.
= −N
dt
Magnetic field induced by inner solenoid
E = −N
B = µ0 n I .
t0
A
2B
R
= 0.735452 C .
=N
012 10.0 points
A coil with N1 = 12.8 turns and radius r1 =
6.45 cm surrounds a long solenoid of radius
N2
= n2 = 1300 m−1 (see
r2 = 1.21 cm and
ℓ2
the figure below). The current in the solenoid
changes as I = I0 sin(ω t), where I0 = 5 A
and ω = 120 rad/s.
ℓ2
ℓ1
A1
a
R
b
A2
E
Inside solenoid has N2 turns
Outside solenoid has N1 turns
What is the magnitude of the induced emf,
EAB , across the 12.8 turn coil at t = 1000 s?
Correct answer: 0.00480987 V.
Explanation:
7
So the induced emf is
dΦ
E = −N1
dt
d (B A2 )
= −N1
dt
dB
= −N1 A2
dt
d (µ0 n2 I)
= −N1 A2
dt
dI
= −µ0 N1 n2 A2
dt
= 0.00480987 V .
013 10.0 points
In a series RLC AC circuit, the resistance is
18 Ω, the inductance is 10 mH, and the capacitance is 24 µF. The maximum potential is
149 V, and the angular frequency is 100 rad/s.
Calculate the maximum current in the circuit.
Correct answer: 0.358125 A.
Explanation:
Let : R = 18 Ω ,
L = 10 mH = 0.01 H ,
C = 24 µF = 2.4 × 10−5 F ,
Vmax = 149 V , and
ω = 100 rad/s .
Kapoor (mk9499) – oldmidterm 03 – Turner – (60230)
Thus
The capacitive reactance is
flow
fhigh
73.1 Hz
= (29.7 Ω)
11500 Hz
XC(high) = XC(low)
1
XC =
ωC
1
(100 rad/s) (2.4 × 10−5 F)
= 416.667 Ω .
=
The inductive reactance is
XL = ω L
= (100 rad/s) (0.01 H)
= 1 Ω.
The maximum current is
Imax =
Vmax
Z
=p
=p
8
= 0.188789 Ω .
015 10.0 points
A 72 mH inductor is connected to a outlet
where the rms voltage is 131 V and the frequency is 18 Hz.
Determine the energy stored in the inductor
at t = 4.5 ms, assuming that this energy is
zero at t = 0.
Correct answer: 4.42394 J.
Explanation:
Vmax
R2 + (XL − XC )2
149 V
(18 Ω)2 + (1 Ω − 416.667 Ω)2
= 0.358125 A .
Let : t = 4.5 ms = 0.0045 s ,
f = 18 Hz ,
L = 72 mH = 0.072 H ,
Vrms = 131 V .
and
The inductive reactance is
XL = ω L = 2 π f L
keywords:
014 10.0 points
A certain capacitor in a circuit has a capacitive reactance of 29.7 Ω when the frequency is
73.1 Hz.
What capacitive reactance does the capacitor have at a frequency of 11500 Hz?
Correct answer: 0.188789 Ω.
Explanation:
Let : XC = 29.7 Ω ,
flow = 73.1 Hz , and
fhigh = 11500 Hz .
The capacitive reactance is XC =
C
, so
2πf
XC(high)
f
2 π flow C
= low .
=
XC(low)
2 π fhigh C
fhigh
and the maximum current is
√
√
2 Vrms
2 Vrms
Imax =
=
XL
2πf L
The current at time t is
I = Imax sin(ω t)
√
2 Vrms
sin(2 π f t)
=
2πf L
√
2 (131 V)
=
2 π (18 Hz) (0.072 H)
× sin[(2 π (18 Hz) (0.0045 s)]
= 11.0855 A ,
so the potential energy stored in the inductor
is
1
U = L I2
2
1
= (0.072 H) (11.0855 A)2
2
= 4.42394 J .
Kapoor (mk9499) – oldmidterm 03 – Turner – (60230)
43 Ω
8 turns
20 turns
110 Vrms
016 10.0 points
A ideal transformer shown in the figure
below having a primary with 20 turns and
secondary with 8 turns.
The load resistor is 43 Ω.
The source voltage is 110 Vrms .
What is the rms electric potential across
the 43 Ω load resistor?
Correct answer: 44 Vrms .
Explanation:
Let : N1 = 20 turns ,
N2 = 8 turns , and
V1 = 110 Vrms .
The rms voltage across the transformer’s
secondary is
N2
V1
N1
8 turns
=
(110 Vrms )
20 turns
= 44 Vrms ,
V2 =
which is the same as the electric potential
across the load resistor.
9
Version 118/ABDBC – midterm 03 – Turner – (60230)
This print-out should have 18 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
A magnetic dipole is falling in a conducting
metallic tube. Consider the induced current
in an imaginary current loop when the magnet
is moving away from the upper loop and when
the magnet is moving toward the lower loop.
1
5. no current flow
Explanation:
When the falling magnet is below the up→
per loop, −
µ ind must be up to attract the
falling magnet and slow it down; i.e., counterclockwise as viewed from above.
→
Before reaching the lower loop, −
µ ind must
be down to oppose the falling magnet; i.e.,
clockwise as viewed from above.
Iabove
?
Iabove
Npole
x
z
v
N
y
S
Npole
x
v
z
dipole
magnet
N
dipole
magnet
S
Spole
y
Spole
Ibelow
?
Ibelow
Determine the directions of the induced
currents Iabove and Ibelow in an imaginary loop
shown in the figure, as viewed from above,
when the loop is above the falling magnet and
when the loop is below the falling magnet.
1. Iabove = clockwise and
Ibelow = clockwise
2. Iabove = clockwise and
Ibelow = counter-clockwise
3. Iabove = counter-clockwise and
Ibelow = counter-clockwise
4. Iabove = counter-clockwise and
Ibelow = clockwise correct
002 10.0 points
A conducting bar moves as shown near a long
wire carrying a constant I = 100 A current.
v
A
B
L
a
I
If a = 12 mm, L = 78 cm, and v = 9.8 m/s,
what is the potential difference, ∆V ≡ VA −
VB ?
1. 85.8701
2. 12.74
3. 25.6286
4. 45.2308
Version 118/ABDBC – midterm 03 – Turner – (60230)
5. 90.5589
6. 37.4026
7. 26.4
8. 113.554
9. 102.358
10. 43.7349
2
µ0 I
r
µ0 I
2. B =
correct
4r
µ0 I
3. B =
πr
µ0 I
4. B =
4πr
µ0 I
5. B =
2πr
µ0 I
6. B =
3πr
µ0 I
7. B =
3r
µ0 I
8. B =
2r
Explanation:
By the Biot-Savart Law,
Z
µ
I
d~s × r̂
0
~ =
B
.
4π
r2
1. B =
Correct answer: 12.74 mV.
Explanation:
Given;
dΦB
dt
d
= − (B ℓ x)
dt
dx
= −B ℓ
dt
= −B ℓ v .
E =−
From Ampere’s law, the strength of the
magnetic field created by the long currentcarrying wire at a distance a from the wire
is
µ0 I
B=
.
2πa
Hence the potential difference is
∆V = B
Lv µ0 I
(L v)
=
2πa
= 12.74 mV .
003 10.0 points
The wire is carrying a current I.
y
180◦
| d~s × r̂| = ds
I
I
r
O
Consider the left straight part of the wire.
The line element d~s at this part, if we come
in from ∞, points towards O, i.e., in the xdirection. We need to find d~s × r̂ to use the
Biot-Savart Law. However, in this part of the
wire, r̂ is pointing towards O as well, so d~s
and r̂ are parallel meaning d~s × r̂ = 0 for this
part of the wire. It is now easy to see that the
right part, having a d~s antiparallel to r̂, also
~ at O.
gives no contribution to B
Let us go through the semicircle C. The
element d~s, which is along the wire, will now
be perpendicular to r̂, which is pointing along
the radius towards O. Therefore
x
I
~
Find the magnitude of the magnetic field B
at O due to a current-carrying wire shown in
the figure, where the semicircle has radius r,
and the straight parts to the left and to the
right extend to infinity.
using the fact that r̂ is a unit vector. So the
Biot-Savart Law gives for the magnitude B of
the magnetic field at O:
Z
ds
µ0 I
.
B=
4 π C r2
Since the distance r to the element d~s is constant everywhere on the semicircle C, we will
Version 118/ABDBC – midterm 03 – Turner – (60230)
be able to pull it out of the integral. The
integral is
Z
Z
ds
1
1
ds = 2 LC
= 2
2
r C
r
C r
where LC = π r is the length of the semicircle.
Thus the magnitude of the magnetic field is
B=
µ0 I 1
µ0 I
.
πr =
2
4π r
4r
004 10.0 points
Given: µ0 = 1.25664 × 10−6 N/A2 .
A 3 m long large coil with a radius of
19.4 cm and 170 turns surrounds a 5.5 m
long solenoid with a radius of 9.4 cm and 6100
turns, see figure below.
The current in the solenoid changes as
I = I0 sin(2 π f t) ,
3
Explanation:
Let :
R = 32 Ω ,
r1 = 19.4 cm = 0.194 m ,
r2 = 9.4 cm = 0.094 m ,
A1 = π r12 = 0.118237 m2 ,
A2 = π r22 = 0.0277591 m2 ,
N1 = 170 ,
N2 = 6100 ,
ℓ1 = 3 m ,
ℓ2 = 5.5 m ,
N1
n1 =
= 56.6667 turns/meter,
ℓ1
N2
n2 =
= 1109.09 turns/meter,
ℓ2
I0 = 30 A , and
ω = 2 π f = 376.991 rad/s .
where I0 = 30 A and f = 60 Hz.
5.5 m
3m
ℓ2
ℓ1
19.4 cm
A1
9.4 cm
32 Ω
E = E0 sin ω t
Find the maximum induced current Imax in
the large coil.
1. 1.86693
2. 0.19212
3. 1.79372
4. 1.6525
5. 3.63875
6. 0.203105
7. 4.92693
8. 3.30593
9. 2.32452
10. 5.36093
Correct answer: 2.32452 A.
A2
E = E0 sin ω t
R
Basic Concepts: Faraday’s law
E = −N
dΦ
dt
and Ohm’s law
E = I R.
Solution: The angular velocity is
ω = 2 π f = 2 π (60 Hz) = 376.991 rad/s .
Version 118/ABDBC – midterm 03 – Turner – (60230)
The solenoid carries a current I =
I0 sin ω t . The maximum magnetic field in the
solenoid is
N2 I0
ℓ2
(6100) (30 A)
= µ0
(5.5 m)
= 0.0418117 T ,
B2max = µ0
where B2 = B2max sin ω t .
The flux Φ12 through coil 1 due to coil 2
(the solenoid) is Φ12 = B2 A2 , so the mutual
inductance is
N1 Φ12
I
N1 B2 A2
=
I
N1 N2 A2
= µ0
ℓ2
(170) (6100) (0.0277591 m2 )
= µ0
(5.5 m)
= 0.00657706 H ,
M12 =
2
2
where A2 = π (0.094 m) = 0.0277591 m .
The induced emf is
E1 = −M12
dI
dt
d
sin ω t
= −M12 I0
dt
= M12 I0 ω cos ω t
= (0.00657706 H) (30 A)
× (376.991 rad/s) cos ω t
= (74.3847 V) cos ω t ,
where the maximum emf
is E1max =
74.3847 V .
We can then determine the maximum current in the resistor using Ohm’s law.
Imax
005
E1max
=
R
74.3847 V
=
32 Ω
= 2.32452 A
10.0 points
4
In a certain series RLC circuit, the rms current is 6.26 A , the rms voltage is 149 V and
the current leads the voltage by 37 ◦ .
What is the total resistance of the circuit?
1. 9.47634
2. 12.5703
3. 11.5495
4. 19.0091
5. 8.55232
6. 25.3514
7. 16.1081
8. 13.0571
9. 7.22575
10. 13.3246
Correct answer: 19.0091 Ω.
Explanation:
Let : Irms = 6.26 A ,
Vrms = 149 V , and
φ = −37 ◦ .
The average power delivered by the generator
is dissipated as heat in the resistor. The
average power delivered by the generator is
given by
Pav = Irms Vrms cos φ
= (6.26 A) (149 V) cos(−37 ◦ )
= 744.919 W .
Note: In the above expression, the phase angle is −37 ◦ , since the voltage lags the current.
The power dissipated in the resistor is given
by
2
Pav = Irms
R.
So, the resistance is
R=
744.919 W
Pav
=
= 19.0091 Ω .
2
Irms
(6.26 A)2
keywords:
006 10.0 points
Consider a rotating conducting rod with
length a. It is fixed at the end O and is
rotating clockwise with angular frequency ω
Version 118/ABDBC – midterm 03 – Turner – (60230)
(see figure). The rotation occurs in a region in
which there is a constant, uniform magnetic
field perpendicular to the plane of rotation
(pointed up from the page). The magnitude
of this field is B.
5
Consider the figure shown below. The switch
is initially set at position b. There is no charge
nor current in the top loop while at position
b. At t = t0 the switch is set to position a.
C
A
a
O
B
Correct answer: 21.2814 V.
Explanation:
The E induced between the two ends of the
rod is given by the work done by the magnetic
force in bringing a unit charge from A to O,
Z O~
Fmag · d~r
Eind =
q
ZAa
=
r ω B dr
0
a2 ω B
2
(3.71 m)2 (1.07 rad/s) (2.89 T)
=
2
= 21.2814 V .
007
10.0 points
a
ω
If B = 2.89 T, ω = 1.07 rad/s and
a = 3.71 m, what is the magnitude of the
E induced between O and A?
1. 19.5006
2. 27.3898
3. 21.2814
4. 10.12
5. 16.1518
6. 54.7779
7. 9.44317
8. 13.0959
9. 22.8241
10. 43.5656
=
L
b S
E
d
c
R
After a long time at position a, the switch
is set back to position b. Denote this time as
t = 0.
What is the maximum energy stored in the
capacitor, expressed in terms of L, E, and R.
(Hint: Use conservation of energy.)
1. Umax = L
2. Umax
3. Umax
4. Umax
5. Umax
6. Umax
7. Umax
8. Umax
9. Umax
10. Umax
E
R
2
1 2 E
= L
2
R
2
E
=L
R
2
E
= L2
R
E
1
= L 2
2 R
2
E
1
correct
= L
2
R
1
E
= L
2 R
E
=L 2
R
E2
=L
R
1 E2
= L
2
R
Explanation:
By conservation of energy
Umax
1 Q2max
1
2
.
= L Imax =
2
2 C
Version 118/ABDBC – midterm 03 – Turner – (60230)
Since
E
Imax = ,
R
the maximum stored energy in the capacitor
is
2
E
1
Umax = L
.
2
R
008 10.0 points
In a series RLC AC circuit, the resistance is
8 Ω, the inductance is 29 mH, and the capacitance is 12 µF. The maximum potential is
154 V, and the angular frequency is 100 rad/s.
Calculate the maximum current in the circuit.
1. 0.569734
2. 0.16316
3. 0.0869858
4. 0.185437
5. 0.139386
6. 0.327402
7. 0.211382
8. 0.33389
9. 0.284374
10. 0.0483527
Correct answer: 0.185437 A.
Explanation:
Let : R = 8 Ω ,
L = 29 mH = 0.029 H ,
C = 12 µF = 1.2 × 10−5 F ,
Vmax = 154 V , and
ω = 100 rad/s .
6
The maximum current is
Imax =
Vmax
Z
Vmax
=p
2
R + (XL − XC )2
154 V
=p
(8 Ω)2 + (2.9 Ω − 833.333 Ω)2
= 0.185437 A .
keywords:
009 10.0 points
A small rectangular coil composed of 26 turns
of wire has an area of 21 cm2 and carries a
current of 1.4 A. When the plane of the coil
makes an angle of 34◦ with a uniform magnetic
field, the torque on the coil is 0.16 N m.
What is the magnitude of the magnetic
field?
1. 0.538357
2. 2.52479
3. 0.241903
4. 0.218239
5. 0.436031
6. 0.550452
7. 0.706435
8. 0.643048
9. 0.579532
10. 0.610031
Correct answer: 2.52479 T.
Explanation:
The capacitive reactance is
XC =
1
ωC
1
(100 rad/s) (1.2 × 10−5 F)
= 833.333 Ω .
=
The inductive reactance is
XL = ω L
= (100 rad/s) (0.029 H)
= 2.9 Ω .
Let : N = 26 turns ,
I = 1.4 A ,
θ = 34◦ ,
A = 21 cm2 = 0.0021 m2 ,
τ = 0.16 N m .
The magnetic force on the current is
~ = I ~ℓ × B
~
F
and
Version 118/ABDBC – midterm 03 – Turner – (60230)
7
and the torque is
6. upward.
so the torque on the loop due to the magnetic
field is
τ = 2 F r cos θ
= (N I ℓ B) w cos θ
= N I B (ℓ w) cos θ
= N I B A cos θ ,
where A is the area of the loop and θ is the
angle between the plane of the loop and the
magnetic field.
The magnetic field from above is
B=
=
τ
N I A cos θ
0.16 N m
(26 turns) (1.4 A) (0.0021 m2 ) cos(34◦ )
= 2.52479 T .
010 10.0 points
The switch S has been closed for a long time
and a constant current is flowing through a
solenoid, creating a magnetic field.
iron core
S
The force which the magnetic field exerts
on a conducting ring positioned as shown is
1. downward.
2. to the right.
3. There is no force, only a torque.
4. to the left.
5. There is neither a force nor a torque.
correct
Explanation:
The magnetic field is constant and the ring
is not moving. Therefore, the flux through
the ring is constant, and there is no induced
current in the ring. Thus, neither a force nor
a torque will be created.
011 10.0 points
Assume: The mobile charge carriers are either electrons or holes.
The holes have
the same magnitude of charge as the electrons. The number of mobile charge carriers for this particular material is n =
8.49 × 1028 electrons/m3 .
Note: In the figure, the point at the upper
edge P1 and at the lower edge P2 have the
same x coordinate.
A constant magnetic field of magnitude
B = 1.2 T points out of the paper. There
is a steady flow of a horizontal current flowing
from left to right in the x direction.
y
x
7
cm
B = 1.2 T
P1
~
B
7.6 A
4.4 cm
~,
~τ = ~r × F
V
P2
2.2 m
The charge on the electron is 1.6021 ×
10−19 C.
What is the magnitude of the electric field
between the upper and lower surfaces?
1. 2.7774e-07
2. 2.81212e-07
3. 2.06237e-07
4. 1.59825e-07
5. 1.64005e-07
6. 4.19657e-07
7. 2.10624e-07
8. 4.34724e-07
Version 118/ABDBC – midterm 03 – Turner – (60230)
where the area
9. 2.17694e-07
10. 1.8905e-07
A= a·b
= (0.07 m) (0.044 m)
= 0.00308 m2 .
Correct answer: 2.17694 × 10−7 N/C.
Explanation:
Let :
8
a = 7 cm = 0.07 m ,
b = 4.4 cm = 0.044 m ,
B = 1.2 T ,
n = 8.49 × 1028 electrons/m3 ,
q = 1.6021 × 10−19 C ,
I = 7.6 A , and
L = 2.2 m .
y
B
B
~
B
v
FA
a
x
012 10.0 points
A rectangular loop of wire is pulled through
a magnetic field B (out of the page). Shown
below are four different stages of development
of this procedure, labeled A, B, C, and D.
The rectangular loop has a constant speed
as it is pulled through the loop.
P1
B
~
B
B
I
b
B
L
v
FB
V
P2
B
B
B
For the Hall effect the magnetic force balances the electric force which means
q vd B = q E ,
B
B
or
E = vd B .
v
FC
Also we know
I = n q vd A
I
vd =
,
nqA
B
or
B
so that the magnitude of the electric field is
E=
=
IB
nqA
(7.6 A) (1.2 T)
n (1.6021 × 10−19 C) (0.00308 m2 )
= 2.17694 × 10−7 N/C ,
FD
B
B
B
B
v
Version 118/ABDBC – midterm 03 – Turner – (60230)
AL
= 4.
AR
i2
i3
dΦ
E =−
dt
D
i1
Figure: The force vector is not
to scale. The velocity vector is
to scale; (i.e., constant speed).
Select the correct rank ordering of the magnitudes of the force pulling the rectangular
loop (at constant speed) in the four stages
shown above.
9
B
Y
R
1. FA = FC > FB = FD
B
AL
X
R
AR
B
B
2. FB > FD = FA > FC
3. FA > FC = FB > FD
4. FA = FD > FB = FC
5. FB > FC > FA > FD
6. FB = FC > FA = FD
7. FA > FC > FB > FD
8. FB > FC = FA > FD
9. FB = FD > FA = FC correct
10. FA > FD = FB > FC
Explanation:
FA is zero because there is no field. FC is
zero because there is no change in flux within
the loop; i.e., no current in the loop. FB and
FD both have a changing flux in the loop,
producing a current flow in the loop. The
right-hand-rule produces a force which will
tend to hold the loop stationary, (FB = FD ) .
013 10.0 points
A solenoid (with magnetic field B) produces a steadily increasing uniform magnetic
flux through its circular cross section. A
octagonal circuit surrounds the solenoid as
shown in the figure. The wires connecting
in the circuit are ideal, having no resistance.
Two identical resistors with resistance R (labeled X and Y ) are in the circuit. A wire
connects points C and D. The ratio of the
solenoid’s area AL left of the wire CD and
the solenoid’s area AR right of the wire CD is
C
Figure: Let i1 , i2 , and i3 be defined
as positive if the currents flow in the
same direction as shown by the arrows in
the figure (otherwise the currents i1 , i2 ,
and i3 have negative values). Also, let
the induced emf be defined as positive,
E > 0.
What are the equations for the (right) loop
CXDC and the (left) loop CDY C, respectively?
Assume the induced emf for the closed loop
octagonal CXDY C is E.
4E
E
+ i1 R = 0 and
− i2 R = 0 .
5
5
E
4E
− i2 R = 0 .
2. + i1 R = 0 and
5
5
E
4E
3. + i1 R = 0 and
+ i2 R = 0 . cor5
5
rect
1.
4.
5.
6.
7.
8.
E
− i1 R = 0 and
5
4E
+ i1 R = 0 and
5
4E
− i1 R = 0 and
5
E
− i1 R = 0 and
5
4E
− i1 R = 0 and
5
4E
5
E
5
E
5
4E
5
E
5
− i2 R = 0 .
+ i2 R = 0 .
− i2 R = 0 .
+ i2 R = 0 .
+ i2 R = 0 .
Explanation:
By definition, the areas of the left and right
Version 118/ABDBC – midterm 03 – Turner – (60230)
loops are related by
A = AL + AR .
AL
= 4, we can solve for AL and AR in
AR
terms of A.
Since
AL =
4A
5
AR =
A
.
5
Then we can compute the magnitude of the
induced emf around the right and left loops.
dB
A dB
1
E R = AR
=
= E
dt
5 dt
5
4A dB
4
dB
=
= E.
E L = AL
dt
5 dt
5
1. 51.0
2. 34.0
3. 92.0
4. 69.0
5. 84.0
6. 81.0
7. 72.0
8. 50.0
9. 48.0
10. 28.0
Correct answer: 81 Vrms .
Explanation:
Let :
The induced emf and the changing magnetic flux are related by
E =−
dΦ
dB
= −A
.
dt
dt
left loop :
1
E + i1 R = 0
5
(1)
4
E + i2 R = 0 .
5
(2)
21 Ω
24 turns
40 turns
014 10.0 points
A ideal transformer shown in the figure
below having a primary with 40 turns and
secondary with 24 turns.
The load resistor is 21 Ω.
The source voltage is 135 Vrms .
135 Vrms
N1 = 40 turns ,
N2 = 24 turns , and
V1 = 135 Vrms .
The rms voltage across the transformer’s
secondary is
Since the magnetic flux is increasing, the induced emf is in the clockwise direction and the
direction of the current is counter-clockwise,
as shown in the figure.
From Kirchoff’s laws, the loop equations for
the right and left loops respectively are
right loop :
10
What is the rms electric potential across
the 21 Ω load resistor?
N2
V1
N1
24 turns
(135 Vrms )
=
40 turns
= 81 Vrms ,
V2 =
which is the same as the electric potential
across the load resistor.
015 10.0 points
An ac power generator produces 23 A rms
at 2556 V. The voltage is stepped up to
1.6 × 105 V by an ideal transformer, and the
energy is transmitted through a long-distance
power line that has a resistance of 47 Ω.
What percentage of the power delivered
by the generator is dissipated as heat in the
power line?
1. 0.121522
2. 0.149486
3. 0.190898
4. 0.243199
5. 0.063281
6. 0.30959
7. 0.0148407
8. 0.0554721
9. 0.315798
Version 118/ABDBC – midterm 03 – Turner – (60230)
10. 0.0107931
Which expression gives the magnitude
B(r3 ) at D of the magnetic field in the region b < r3 < a?
Correct answer: 0.0107931.
Explanation:
Let :
11
µ0 i r3
2 π a2
µ0 i (a2 − r32 )
2. B(r3 ) =
correct
2 π r3 (a2 − b2 )
µ0 i (a2 − b2 )
3. B(r3 ) =
2 π r3 (r32 − b2 )
1. B(r3 ) =
I1 = 23 A ,
V1 = 2556 V ,
V2 = 1.6 × 105 V ,
R = 47 Ω .
and
4. B(r3 ) = 0
The current after being stepped up is
I1 V1
V2
(23 A) (2556 V)
=
1.6 × 105 V
= 0.367425 A .
5. B(r3 ) =
I2 =
6. B(r3 ) =
7. B(r3 ) =
Then,
Plost = I22 R = (0.367425 A)2 (47 Ω) = 6.34505 W . 8. B(r3 ) =
The original power is
9. B(r3 ) =
P = I1 V1 = (23 A) (2556 V) = 58788 W ,
10. B(r3 ) =
so the percentage of the power lost is
µ0 i r3
2 π c2
µ0 i (a2 + r32 − 2 b2 )
2 π r3 (a2 − b2 )
µ0 i
π r3
µ0 i r3
2 π b2
µ0 i (r32 − b2 )
2 π r3 (a2 − b2 )
µ0 i
2 π r3
6.34505 W
Plost
=
×100% = 0.0107931% . Explanation:
P
58788 W
Ampere’s
Law states that the line inteI
~ · d~ℓ around any closed path equals
gral
B
016 10.0 points
µ0 I, where I is the total steady current passThe figure below shows a cylindrical coaxial
ing through any surface bounded by the closed
cable of radii a, b, and c in which equal, unipath.
formly distributed, but antiparallel currents i
Considering the symmetry of this problem,
exist in the two conductors.
a
we choose a circular path, so Ampere’s Law
simplifies to
b
iout ⊙
c
η=
B (2 π r3 ) = µ0 Ien ,
iin ⊗
O
F
E
D
C
r1
r2
r3
r4
where r3 is the radius of the circle and Ien is
the current enclosed.
Aen
π (r32 − b2 )
Since
, when b <
=
Acylinder
π (a2 − b2 )
r3 < a for the cylinder,
B=
µ0 Ien
2 π r3
Version 118/ABDBC – midterm 03 – Turner – (60230)
π (r32 − b2 )
R1
µ0 i − i
π (a2 − b2 )
=
2 π r3
2
R2
L
a − r32
µ0 i
a2 − b2
=
E
2 π r3
=
12
S b
a
µ0 i (a2 − r32 )
.
2 π r3 (a2 − b2 )
Let : R2 = 12 Ω and
E = 12.2 V .
017 10.0 points
One application of an RL circuit is the generation of time-varying high-voltage from a
low-voltage source, as shown in the figure.
1286 Ω
12 Ω
3H
12.2 V
S b
a
The switch is initially set to position a. After a long time at position ”a” the switch is
quickly thrown to ”b”. Compute the voltage across the inductor immediately after the
switch is thrown from ”a” to ”b”.
1. 942.293
2. 1053.07
3. 1216.68
4. 976.8
5. 1284.22
6. 1375.5
7. 1393.66
8. 1319.63
9. 905.22
10. 865.567
Correct answer: 1319.63 V.
Explanation:
When the switch is at “a”, the circuit comprises the battery, the inductor L, and the
resistor R2 . A long time after the switch has
been in position “a”, the current is steady.
This means that the inductor has no response
to the current. In other words, the circuit can
be considered as consisting of E and R2 only,
with the inductor reduced to a wire. In this
case, the current is simply found by Ohm’s
Law
I0 =
E
12.2 V
=
= 1.01667 A .
R2
12 Ω
Let : R1 = 1286 Ω and
I0 = 1.01667 A .
When the switch is thrown from “a” to “b”,
the current in the circuit is the current passing
through R2 , which was found in Part 1 to be
1.01667 A . From Kirchhoff’s Loop Law, the
initial voltage across the inductor is equal to
the initial voltage across R1 and R2 . So, we
have
VL = VR1 + VR2
= I0 R 1 + I0 R 2
= (1.01667 A) (1286 Ω) + (1.01667 A) (12 Ω)
= 1319.63 V .
018 10.0 points
A 4.6 Ω square loop, whose dimensions are
2.6 m × 2.6 m, is placed in a uniform 0.068 T
magnetic field that is directed perpendicular
to the plane of the loop (see figure).
Version 118/ABDBC – midterm 03 – Turner – (60230)
The loop, which is hinged at each vertex, is
pulled as shown until the separation between
points C and D is d = 1.6 m. The process
takes 0.04 s.
C
2.
6
m
m
6
.
2
F
F
d
D
What is the average current generated in
the loop?
1. 0.0604337
2. 0.455443
3. 0.62083
4. 0.686499
5. 0.677514
6. 1.03545
7. 0.0948682
8. 2.3187
9. 0.53644
10. 0.484514
Correct answer: 1.03545 A.
Explanation:
Basic Concept: Faraday’s Law
E =−
d ΦB
.
dt
Ohm’s Law
I=
V
.
R
Let : R = 4.6 Ω ,
a = 2.6 m ,
d = 1.6 m ,
∆t = 0.04 s .
∆ΦB
∆t
∆A
.
= −B
∆t
The change in area for the square with sides
a is
∆A = Af − Ai ,
where
Ai = a2
= 6.76 m2 .
To find Af , we note that the final shape is
made up of four right triangles of hypotenuse
d
a with one leg being . The area of each of
2
these triangles is
s
2
d
1
1 d
2
a −
Af =
.
4
2 2
2
Thus
Af = d
s
a2
2
d
−
2
s
= (1.6 m)
(2.6
= 3.95818 m2 .
m)2
−
1.6 m
2
2
We then find the average emf to be
B [Ai − Af ]
∆t
(0.068 T) [(6.76 m2 ) − (3.95818 m2 )]
=
0.04 s
= 4.76309 V .
Eavg =
Applying Ohm’s law, the average current is
Eavg
R
(4.76309 V)
=
(4.6 Ω)
= 1.03545 A .
Iavg =
and
Solution: From Faraday’s law, the average
induced emf is
Eavg = −
13
Kapoor (mk9499) – oldmidterm 04 – Turner – (60230)
This print-out should have 29 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
A light ray passes through substances 1, 2,
and 3, as shown.
The indices of refraction for these three
substances are n1 , n2 , and n3 , respectively.
Ray segments in 1 and 3 are parallel.
1
n1
n2
2
3
θ2
d
x
θ1
d
0.52 m
If the fountain is 0.520 m deep, find the
minimum diameter, d of the piece of wood
that would prevent the diamond from being
seen from outside the water.
Correct answer: 1.17992 m.
Explanation:
Basic Concept:
n3
One can conclude that
1
sin θc =
nr
ni
Given:
ni = 1.333
nr = 1.00
∆y = 0.520 m
1. n1 must be equal to 1.00.
2. n2 must be less than n1 .
Solution:
3. all three indices must be the same.
4. n2 must be less than n3 .
5. n3 must be the same as n1 . correct
Explanation:
n1 sin θ1 = n2 sin θ2 = n3 sin θ3 ,
Since θ1 = θ3 , sin θ1 = sin θ2 and n1 = n3 .
Since sin θ1 = sin θ3 > sin θ2 , n2 > n1 = n3 .
So, we can only conclude that n3 must be
the same as n1 .
002 10.0 points
A jewel thief decides to hide a stolen diamond
by placing it at the bottom of a crystal-clear
fountain. He places a circular piece of wood
on the surface of the water and anchors it
directly above the diamond at the bottom of
the fountain.
nr
θc = sin
n
i 1
= sin−1
1.333
◦
= 48.6066
−1
tan θc =
d
2
∆y
d
=
2∆y
d = 2∆y(tan θc )
= 2(0.52 m)(tan 48.6066◦ )
= 1.17992 m
003 (part 1 of 2) 10.0 points
Hint: Construct a ray diagram.
7
Given: A real object is located at “p1 = f ”
4
to the left of a convergent lens with a focal
length f as shown in the figure below.
Kapoor (mk9499) – oldmidterm 04 – Turner – (60230)
f
f
7
0
4
The image distance q1 to the right of the
lens is
(× f )
1. q1 =
2. q1 =
3. q1 =
4. q1 =
5. q1 =
6. q1 =
7. q1 =
8. q1 =
9. q1 =
10. q1 =
9
f.
4
12
f.
5
13
f.
6
15
f.
7
15
f.
8
10
f.
3
8
f.
3
11
f.
6
7
f . correct
3
13
f.
7
f
7
0
4
Basic Concepts:
1 1
1
+ = .
p q
f
Solution:
1
1
1
=− +
q1
p1 f
4
1
=−
+
7f
f
7−4
=
7f
3
=
7f
7
q1 = f .
3
f < q < ∞ 0 > m > −∞
−∞ < q < 0 ∞ > m > 1
Convergent (concave) lenses f > 0 have
real images q > 0 when the object ∞ > p > f .
7
3
004 (part 2 of 2) 10.0 points
A second convergent lens with the same focal
length f is placed behind the first lens, as
shown in the figure below (the first lens has
the lighter image).
f
∞ >p> f
f >p> 0
f
(× f )
Explanation:
Basic Concepts:
1 1
1
h′
q
+ =
m=
=−
p q
f
h
p
Converging Lens
f >0
2
f
f
f
7
2
0
4
3
The second image location q2′ , measured
with respect to the second lens, is
(× f )
5
f . correct
8
8
f.
2. q2′ =
13
1
3. q2′ = f .
2
1. q2′ =
Kapoor (mk9499) – oldmidterm 04 – Turner – (60230)
1
1
+
2
7
f
f− f
3
3
5
3
+
=
5f
5f
3+5
=
5f
8
=
5f
5
q2 = f .
8
The final image due to the second lens is real
and to the right of the second lens. As can
be seen from the figure, the final image due to
the second lens relative to the first lens is
2
5
q2′ = f + f
3
8
31
=
f .
24
=
4
4. = f .
7
9
5. q2′ =
f.
14
3
6. q2′ = f .
4
2
7. q2′ = f .
3
3
8. q2′ = f .
5
9
9. q2′ =
f.
16
8
10. q2′ =
f.
11
Explanation:
q2′
5
f
3
f
f
keywords:
/* If you use any of these, fix the comment
symbols.
31
f
24
(× f )
3
7
4
0
2 5
3 8
7
3
2
Given: Distance between lenses is d = f ,
3
7
and from Part 2, we have q1 = f .
3
Solution: q1 > d and for the second lens
p2 = −q1 + d
7 2
= − +
f
3 3
5
=− f.
3
Note: Since the object for the second lens
is behind the lens it is virtual and therefore
p2 < 0.
Using the lens equation for the second lens,
we have
1
1
1
=− +
q2
p2 f
1
1
=
+
q1 − d f
005 (part 1 of 2) 10.0 points
A person looks at a gem using a converging
lens with a focal length of 14.8 cm. The lens
forms a virtual image 34.0 cm from the lens.
a) Find the magnification.
Correct answer: 3.2973.
Explanation:
Basic Concept:
1
1 1
= +
f
p q
Given:
q = −34.0 cm
f = 14.8 cm
Solution:
1 1
1
= −
p
f
q
−1
1 1
−
p=
f
q
−1
1
1
=
−
14.8 cm −34 cm
= 10.3115 cm
Kapoor (mk9499) – oldmidterm 04 – Turner – (60230)
−q
p
−(−34 cm)
=
10.3115 cm
= 3.2973
M=
006 (part 2 of 2) 10.0 points
b) Describe the image.
1. virtual, inverted, larger
4
and the magnification of the image is
M =−
q
p
R
2p − R
(−2.52 cm)
=−
2(8.5 cm) − (−2.52 cm)
=−
= 0.129098 .
2. real, upright, larger
6. real, upright, smaller
008 (part 1 of 2) 10.0 points
A dentist wants a small mirror that produces
an upright image with a magnification of 3.5
when the mirror is located 2.1 cm from a
tooth.
What should be the radius of curvature of
the mirror?
Correct answer: 5.88 cm.
7. real, inverted, larger
Explanation:
3. virtual, inverted, smaller
4. virtual, upright, larger correct
5. virtual, upright, smaller
8. real, inverted, smaller
Explanation:
q < 0 so the image is virtual. M > 0 so
the image is upright. |M | > 1 so the image is
magnified.
007 10.0 points
A spherical Christmas tree ornament is
5.04 cm in diameter.
What is the magnification of the image of
an object placed 8.5 cm away from the ornament?
Correct answer: 0.129098.
Explanation:
p = 8.5 cm and
−5.04 cm
= −2.52 cm .
R=
2
Using the mirror equation,
1
2
1 1
=
= +
f
R
p q
2
1
2p − R
1
= − =
q
R p
Rp
Rp
q=
,
2P − R
m = 3.5 , and
s = 2.1 cm .
The lateral magnification of the mirror is
given by
s′
m=− ,
s
so
s′ = −m s .
From the mirror equation, we have
1
2
1
1
s′ + s
= = + ′ = ′ ,
f
r
s s
s s
Let :
r=
=
=
=
=
2 s s′
s + s′
2 s (−m s)
s + (−ms)
−2 m s
1−m
−2 (3.5) (2.1 cm)
1 − 3.5
5.88 cm .
so
Kapoor (mk9499) – oldmidterm 04 – Turner – (60230)
009 (part 2 of 2) 10.0 points
Should the mirror be concave or convex?
1. Concave correct
2. Convex
Explanation:
The mirror must be concave because a convex mirror always produces a diminished virtual image.
010 10.0 points
In a particular medium the speed of light
is six-sevenths that of the speed of light in
vacuum and the electric field amplitude has a
maximum value of 6.7 mV/m.
The speed of light is 2.99792 × 108 m/s .
Calculate the maximum value of the magnetic field.
Correct answer: 2.60736 × 10−11 T.
Explanation:
5
011 10.0 points
Consider a monochromatic electromagnetic
plane wave propagating left to right (as shown
below). At a particular point in space, the
magnitude of the electric field has an instantaneous value of 16.7 V/m.
The permeability of free space is 4π ×
10−7 N/A2 , the permittivity of free space is
8.85419 × 10−12 C2 /N/m2 and the speed of
light is 2.99792 × 108 m/s.
propagation direction
E
B
What is the instantaneous magnitude of the
Poynting vector at the same point and time?
Correct answer: 0.740291 W/m2 .
Explanation:
Let : E = 6.7 mV/m and
c = 2.99792 × 108 m/s .
c
The index of refraction is n ≡ , so the index
v
of refraction of light in this particular medium
is seven-sixths.
According to Maxwell’s laws, for an electromagnetic wave, the relation between the
amplitude of the electric field and the magnitude of the magnetic field is given by
E=
c
B = vB,
n
where n is the index of refraction and v is
the velocity of light in that medium. For
7
6
our problem n =
since v = c , so the
6
7
maximum value of the magnetic field is
7E
6 c
7
V
6.7 mV/m
=
· 3
8
6 2.99792 × 10 m/s 10 mV
B=
= 2.60736 × 10−11 T .
Let :
E = 16.7 V/m ,
c = 2.99792 × 108 m/s ,
µ0 = 4π × 10−7 N/A2 .
and
Basic Concepts:
E
=c
B
b ×B
b =S
b.
E
~ is given by
The Poynting vector S
~ ×B
~.
~ = 1 E
S
µ0
Solution: For a plane, electromagnetic
~ and B
~ are always perpendicular to
wave, E
each other and to the direction of propagation
of the wave.
The magnetic field is
16.7 V/m
2.99792 × 108 m/s
= 5.57052 × 10−8 T .
B=
Kapoor (mk9499) – oldmidterm 04 – Turner – (60230)
In this case, the Poynting vector is in the
direction of propagation and has magnitude
Top View
blackened
0.8 cm
light
source
5 cm
EB
S=
µ0
(16.7 V/m) (5.57052 × 10−8 T)
=
4π × 10−7 N/A2
= 0.740291 W/m2 .
6
0.8 cm
mirrored
2m
012 10.0 points
Two plates, one black (perfectly absorbing)
and one mirrored (perfectly reflecting), are
affixed one at each end of a 5 cm rod, which
is then suspended at its center from a torsion
balance (see figure below).
torsion
balance
0.8 cm
black
surface
mirrored surface
The plates are squares whose sides are of
length 0.8 cm. A point light source is then
placed at a distance 2 m away from the rod,
as shown below. The point light source radiates an average power of 28 kW. The torsion
balance keeps the rod from turning by counteracting any torque on the rod.
Assume: The direction of the EM waves are
nearly perpendicular to each plate’s surface
(i.e., a distance 2 m from each plate), the
light is radiated isotropically from the source,
and uniformly distributed over the surface of
each plate.
Figure: Not drawn to scale.
What is the magnitude of the torque k~τ k
exerted on the torsion balance by the rod?
Correct answer: 2.97295 × 10−12 N m2 .
Explanation:
Let : c = 2.99792 × 108 m/s ,
ℓ = 5 cm ,
a = 0.8 cm ,
d = 2 m , and
Pav = 28 kW = 28000 W .
1 ~
~ deThe Poynting vector ~S =
E×B
µ0
scribes the rate of flow of energy in an EM
wave. Time average of S is equal to the intensity. In the present case, since ℓ ≪ d, intensity
at each of the mirrors is approximately equal
to the intensity at a distance d from the point
source. Hence
P
4 π d2
28000 W
=
4 π (2 m)2
= 557.042 kg/s3 .
S=I=
Thus the pressure p on the black plate is
p=
S
c
557.042 kg/s3
2.99792 × 108 m/s
= 1.8581 × 10−6 kg/m/s2 .
=
Kapoor (mk9499) – oldmidterm 04 – Turner – (60230)
The magnitude of the force is then
~ b k = (Pressure) (Area)
kF
= p a2
= (1.8581 × 10−6 kg/m/s2 )
× (0.008 m)2
= 1.18918 × 10−10 N .
pulse of 240 ns duration at a small 6 mg pellet
at rest. The pulse hits the mass squarely in
the center of its bottom side.
The speed of light is 3 × 108 m/s and the
acceleration of gravity is 9.8 m/s2 .
h
y
b
For both plates, the distance from the center
of the plate to the point of rotation is
ℓ
r= .
2
~ are perpendicuAlso, for both plates ~r and F
lar. For the black plate
k~τb k = r Fb .
For the mirrored plate
b
b
b
b b b
b b
b
b
The definition of torque is
~.
~τ = ~r × F
7
b
b
b
T is the time
to reach its
b
maximum
b
height
b
b
h
b
v0
b
b
b
b
b
6 mg
t
T
1000 MW
240 ns
If the radiation is completely absorbed
without other effects, what is the maximum
height the mass reaches?
Correct answer: 907.029 µm.
Explanation:
k~τm k = r Fm = 2 r Fb .
Using the right hand rule, we see that the
~ b will be pointing up, and the
torque due to F
~ m will be pointing down. So the
one due to F
total torque is
k~τ k = k~τb k − k~τm k
= r Fb − r Fm
= r Fb − 2 r Fb
= −r Fb
5 cm
1
=−
2 100cm/m
× (1.18918 × 10−10 N)
= −2.97295 × 10−12 N m
Let : P = 1000 MW = 1 × 109 W ,
∆t = 240 ns = 2.4 × 10−7 s ,
m = 6 mg = 6 × 10−6 kg ,
L = 5 cm = 5 × 106 m ,
c = 3 × 108 m/s , and
g = 9.8 m/s2 .
Applying conservation of energy, we obtain
Kf − Ki + Uf − Ui = 0 .
p2i
Since Ui = Kf = 0 and Ki =
, we have
2m
k~τ k = 2.97295 × 10−12 N m2 .
The “−” sign indicating the torque points
down, which means the system rotates clockwise, and the direction of the torque is −k̂
(vertically downward).
013 10.0 points
A vertically pulsed laser fires a 1000 MW
−
−
p2i
+ Uf = 0
2m
p2i
+ mgh = 0.
2m
(1)
Using conservation of momentum,
pi = pem =
P ∆t
U
=
.
c
c
(2)
Kapoor (mk9499) – oldmidterm 04 – Turner – (60230)
Substituting pi from Eq. 2 into Eq. 1, we
obtain
2
P ∆t
c
−
+ mgh = 0
2m
P 2 (∆t)2
−
+ mgh = 0.
(3)
2 m c2
Solving Eq. 3 for the maximum height of the
pellet’s trajectory gives us
P 2 (∆t)2 1
(4)
2 m2 g c2
(1 × 109 W)2 (2.4 × 10−7 s)2
=
2 (6 × 10−6 kg)2 (9.8 m/s2 )
1
×
(3 × 108 m/s)2
= (0.000907029 m) (1 × 106 µm/m)
h=
= 907.029 µm .
014 10.0 points
A plane electromagnetic wave is generated
due to the initiation of current along the x
direction in a current sheet in the zx plane
at y = 0. A steady flow current is switched
on at t = 0. This current sheet generates a
traveling “front” of the electric and magnetic
fields.
z
infinite
current
sheet in
z, x plane
Py
y
x
Current is in
−x direction, as
shown by arrows.
Consider a new situation, where the current
sheet has the sinusoidal oscillation with an
angular frequency ω. The fields at any point
x and time t are given by:
E = Emax cos(k y − ω t)
B = Bmax cos(k y − ω t)
Choose the selection for all possible expressions for the intensity I of the EM waves at
Py .
1. I =
8
1
2
ǫ0 c Emax
only
2
2
2. I = ǫ0 c Emax
only
1
1
2
2
3. I =
ǫ0 Emax +
c Bmax only
4
4 µ0
1
2
4. I =
c Bmax
only
µ0
1
2
5. I =
c Bmax
only
2 µ0
1
2
2
c Bmax
,
6. I = ǫ0 c Emax
,
µ0
1
1
2
2
and
ǫ0 Emax +
c Bmax
2
2 µ0
1
1
2
2
7. I = ǫ0 c Emax
,
c Bmax
,
2
2
µ
0
1
1
2
2
and
c Bmax
ǫ0 Emax +
4
4 µ0
correct
1
1
2
2
8. I =
ǫ0 Emax +
c Bmax only
2
2 µ0
Explanation:
The time-average of the energy density u at
any point is
1
1
ǫ0 E 2 +
B2 ,
2
2 µ0
= ǫ0 E 2 , or
1 2
B , and
=
µ0
S
I
≡ = .
c
c
u=
For the traveling wave form given, u is independent of x .
I = u c , and
u = uE + uB = 2 uE = 2 uB , so
1
uE = ǫ0 [ Emax cos(k x − ω t)]2
2
1
2
= ǫ0 Emax
.
4
Kapoor (mk9499) – oldmidterm 04 – Turner – (60230)
Similarly, uB =
we have
1
2
Bmax
. Since uE = uB ,
4 µ0
I = 2 uE c =
= 2 uB c =
Explanation:
When a light ray goes from glass into air, it
refracts away from the normal.
1
2
ǫ0 c Emax
,
2
1
2
c Bmax
,
2 µ0
9
016 (part 2 of 2) 10.0 points
and
air
= (uE + uB ) c
glass
1
1
2
2
=
ǫ0 c Emax
+
c Bmax
.
4
4 µ0
What is the approximate refracted ray?
015 (part 1 of 2) 10.0 points
Given: A ray approaching an interface.
air
air
1.
glass
glass
What is the approximate refracted ray?
correct
2. None of these
1.
3.
air
air
glass
glass
2. None of these
4.
3.
air
air
glass
glass
air
4.
correct
glass
Explanation:
When a light ray goes from air into glass, it
refracts toward the normal.
017 (part 1 of 2) 10.0 points
The index of refraction of a transparent liquid
(similar to water but with a different index of
refraction) is 1.22. A flashlight held under the
Kapoor (mk9499) – oldmidterm 04 – Turner – (60230)
transparent liquid shines out of the transparent liquid in a swimming pool. This beam of
light exiting the surface of the transparent liquid makes an angle of θa = 33 ◦ with respect
to the vertical.
flashlight
ray
θa
air
water
θw
At what angle (with respect to the vertical)
is the flashlight being held under transparent
liquid?
Correct answer: 26.5146 ◦.
Explanation:
By Snell’s Law
10
018 (part 2 of 2) 10.0 points
The flashlight is slowly turned away from the
vertical direction.
At what angle will the beam no longer be
visible above the surface of the pool?
Correct answer: 55.052 ◦.
Explanation:
This is solved in the same fashion as Part 1.
When the light ceases to be visible outside
the transparent liquid, then θa ≥ 90◦ . The
sin 90◦ = 1. Hence (from above),
na
sin θw =
nw
1
θw = arcsin
1.22
θw = 55.052 ◦ .
019 10.0 points
A coin is at the bottom of a beaker. The
beaker is filled with 2.1 cm of water (n1 =
1.33) covered by 2.2 cm of liquid (n2 = 1.4)
floating on the water.
na sin θa = nw sin θw ,
where na and nw are the indices of refraction
for each substance and θa and θw are the incident angles to the boundary in each medium,
respectively.
Assume that the surface of the transparent
liquid is a level horizontal plane, thus each
angle with respect to the vertical represents
the incident angle in each medium.
The index of refraction of air is (nearly)
na = 1.0 while the index of refraction of transparent liquid is given as nw = 1.22. The incident angle in the air is given to be θa = 33 ◦ .
Hence
sin θw
na
=
sin θa
nw
sin θw
1
=
sin 33 ◦
1.22
0.544639
sin θw =
1.22
θw = arcsin(0.446426)
θw = 26.5146 ◦ .
How deep does the coin appear to be from
the upper surface of the liquid (near the top
of the beaker)?
Correct answer: 3.15038 cm.
Explanation:
For small angles
x
.
ℓ
The appearance of the width of the coin x
remains the same. Applying Snell’s Law,
sin θ ≈ tan θ =
n1 sin θ1 = n2 sin θ2
x
x
ni = nf
ℓi
ℓf
ℓi
ℓf =
ni
Kapoor (mk9499) – oldmidterm 04 – Turner – (60230)
since nf ≈ 1 for air and the apparent distance
d = ℓf . Thus
11
Rays for all other small angles with the vertical appear to diverge from the image of the
coin at this depth.
ℓ2
ℓ1
+
n2 n1
2.2 cm 2.1 cm
+
=
1.4
1.33
= (1.57143 cm) + (1.57895 cm)
d=
= 3.15038 cm .
The coin appears to be closer to the surface
by
(2.1 cm) + (2.2 cm) − (3.15038 cm)
= 1.14962 cm .
Alternate Solution:
Light coming
straight up from the coin falls on each interface at 0◦ and continues straight up. Consider light making a 1 ◦ angle (therefore we
can use the small angle approximation) with
the vertical in the water. It enters the liquid
a distance
keywords:
020 10.0 points
Hint: The convergent mirror in this problem
is a part of a lens/mirror system so the object
in this problem may be either real or virtual.
Construct a ray diagram.
Given: A virtual object is located to the
right of a divergent mirror. The object’s distance from the mirror and its focal length are
shown in the figure below.
(2.1 cm) tan 1◦ ≃ (2.1 cm) × θ1 u
= 0.0366519 cm
from the vertical ray. In the liquid its angle
with the vertical is given by
f
1.33 sin 1◦ = 1.4 sin θ2
θ2 = 0.950001◦
This same ray reaches air at distance
(0.0366519 cm) + (2.2 cm) tan(0.950001◦ )
= 0.0731293 cm ,
19
f
10
Which diagram correctly shows the image?
0
from the vertical ray, and the angle of refraction is found from
1.4 sin(0.950001◦ ) = 1 sin θ3
θ3 = 1.33◦
1.
f
Your brain automatically finds the intersection of this ray with the vertical ray, at an
apparent depth of
0.0731293 cm
0.0731293 cm
≃
◦
tan(1.33 )
0.0232129 rad
= 3.15038 cm .
−19
f
9
0
19
f
10
Kapoor (mk9499) – oldmidterm 04 – Turner – (60230)
12
q
f
2.
p
f
19
−19
f f
109
0
0
−19
−19
ff
109
Basic Concepts:
1
1 1
+ = ,
p q
f
3.
where f < 0 for a divergent mirror.
Solution:
f
19
−19
f f
109
0
correct
4.
1
1
1
=− −
q1
p1 f
(−10) 1
−
=−
19 f
f
−(19) − (−10)
=
19 f
−9
=
19 f
−19
q1 =
f.
9
The magnification m of this mirror is
f
q1
p1
19
f
= − −9
19
f
−10
−10
=−
−9
−10
=
9
m=−
−19
f
9
Explanation:
0
19
f
10
021 10.0 points
Hint: The convergent mirror in this problem
is a part of a lens/mirror system so the object
in this problem may be either real or virtual.
Construct a ray diagram.
Kapoor (mk9499) – oldmidterm 04 – Turner – (60230)
Given: A real object is located to the left
of a convergent mirror. The object’s distance
from the mirror and its focal length are shown
in the figure below.
13
3.
f
f
5
f
2
5
f
0
2
Which diagram correctly shows the image?
1.
5
5
f
f
2
3
5
f
3
4.
f
5
5
f
f
2
3
Explanation:
f
correct
0
q
0
0
f
p
5
5
f
f
0
2
3
Basic Concepts:
2.
f
1
1 1
+ = ,
p q
f
5
f
2
0
5
f
3
where 0 < f for a convergent mirror.
Solution:
1 1
1
=− +
q
p f
(2) 1
=−
+
5f
f
Kapoor (mk9499) – oldmidterm 04 – Turner – (60230)
−(2) + (5)
5f
3
=
5f
5
q= f.
3
Solution: Substituting these values into the
lens equation
=
q=
The magnification m of this mirror is
=
q1
m=−
p1
5
f
=− 3
5
f
2
2
=−
3
−2
.
=
3
1
1 1
−
f
p
1
1
1
−
(10.6 cm) (33 cm)
= 15.6161 cm .
023 (part 2 of 2) 10.0 points
Find the magnification.
Correct answer: −0.473214.
Explanation:
022 (part 1 of 2) 10.0 points
A convergent lens has a focal length of
10.6 cm . The object distance is 33 cm .
q
p
(15.6161 cm)
=−
(33 cm)
M =−
= −0.473214 .
h
q
p
f
14
f
h′
024 (part 1 of 2) 10.0 points
A divergent lens has a focal length of 17.6 cm .
The object distance is 5.9 cm .
Scale: 10 cm =
Find the distance of the image from the
center of the lens.
Correct answer: 15.6161 cm.
1 1
1
h′
q
+ =
M=
=−
p q
f
h
p
Convergent Lens
f >0
f < q < ∞ 0 > M > −∞
Note: The focal length for a convergent
lens is positive, f = 10.6 cm.
h′
pq
f
Explanation:
∞ >p> f
h
f
Scale: 10 cm =
Find the distance of the image from the
center of the lens.
Correct answer: 4.41872 cm.
Explanation:
Kapoor (mk9499) – oldmidterm 04 – Turner – (60230)
∞ >p> 0
θB
f <q< 0 0 <M < 1
Note: The focal length for a divergent lens
is negative, f = −17.6 cm.
Solution: Substituting these values into the
lens equation
q=
mirror 2
B
1 1
1
h′
q
+ =
M=
=−
p q
f
h
p
Divergent Lens
f <0
15
θA
90◦
mirror 1
1
1 1
−
f
p
1
=
A
1
1
−
(−17.6 cm) (5.9 cm)
= −4.41872 cm
|q| = 4.41872 cm .
x
What is the angle of the reflection of ray B
(with respect to the normal) on mirror 2?
Correct answer: 46◦ .
Explanation:
If the angle of the incident ray A is θA , the
angle of reflection must also be θA . Since the
mirrors are perpendicular to each other, angle
θB is equal to 90◦ − θA
θB = 90◦ − θA
= 90◦ − 44◦
= 46◦ .
025 (part 2 of 2) 10.0 points
Find the magnification.
Correct answer: 0.748936.
Explanation:
q
p
(−4.41872 cm)
=−
(5.9 cm)
M =−
= 0.748936 .
027 (part 2 of 2) 10.0 points
Determine the angle between the rays A and
B.
1. 44◦
2. 180◦ correct
3. 0◦
026 (part 1 of 2) 10.0 points
Consider the case in which light ray A is incident on mirror 1, as shown in the figure.
The reflected ray is incident on mirror 2 and
subsequently reflected as ray B. Let the angle of incidence (with respect to the normal)
on mirror 1 be θA = 44 ◦ and the point of
incidence be located 20 cm from the edge of
contact between the two mirrors.
4. 46◦
5. 92◦
6. 88◦
7. 90◦
Explanation:
Kapoor (mk9499) – oldmidterm 04 – Turner – (60230)
Refer to the figure below. Consider the
angle each ray makes with mirror 1. Ray A
makes an angle of φA with mirror 1. This is
given by
φA = 90◦ − θA = θB .
The normal to mirror 2 is parallel to mirror
1. Thus the angle ray B makes with mirror
1 (assumed to be extended) is θB . Both rays
make the same angle with respect to mirror
1. They are parallel, but point in the opposite
directions. Therefore the angle between them
is 180◦ .
mirror 2
B
θB
φA
A
θA
φA
mirror 1
90◦
x
028 (part 1 of 2) 10.0 points
A biology student uses a simple magnifier to
examine the structural features of the wing of
an insect. The wing is held 4.75 cm in front
of the lens, and the image is formed 49.3 cm
from the eye.
What is the focal length of the lens?
Correct answer: 5.25645 cm.
Explanation:
Given :
p = 4.75 cm and
q = −49.3 cm .
From the thin lens equation,
pq
f=
p+q
(4.75 cm) (−49.3 cm)
=
4.75 cm + (−49.3 cm)
= 5.25645 cm .
16
029 (part 2 of 2) 10.0 points
What angular magnification is achieved?
Correct answer: 10.3789.
Explanation:
With the image at the normal near point,
the angular magnification is
49.3 cm
f
49.3 cm
=1+
5.25645 cm
= 10.3789 .
m=1+
Version 145/ACBAB – midterm 04 – Turner – (60230)
This print-out should have 19 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
A jewel thief decides to hide a stolen diamond
by placing it at the bottom of a crystal-clear
fountain. He places a circular piece of wood
on the surface of the water and anchors it
directly above the diamond at the bottom of
the fountain.
1
Solution:
nr
θc = sin
n
i 1
= sin−1
1.333
◦
= 48.6066
−1
tan θc =
d
2
∆y
d
=
2∆y
d = 2∆y(tan θc )
= 2(0.7 m)(tan 48.6066◦ )
= 1.58836 m
d
0.7 m
If the fountain is 0.700 m deep, find the
minimum diameter, d of the piece of wood
that would prevent the diamond from being
seen from outside the water.
1. 3.40363
2. 1.58836
3. 1.83796
4. 2.7229
5. 3.85744
6. 4.08435
7. 2.26908
8. 3.17672
9. 2.49599
10. 1.9741
Correct answer: 1.58836 m.
Explanation:
Basic Concept:
sin θc =
nr
ni
Given:
002 10.0 points
A thin lens has a crescent shape, which is
defined by two spherical surfaces A and B.
The radius of the surface A is |RA | = 3 a ,
and of surface B is |RB | = a and the index of
refraction of the lens is 2.0.
CA
CB
O
B
What is the focal length using the small
angle approximation?
3
1. f = − a
4
1
2. f = − a
3
3
3. f = a correct
2
1
4. f = − a
4
5. f = 3 a
ni = 1.333
nr = 1.00
∆y = 0.700 m
3a
a
A
6. f = 4 a
7. f = −a
Version 145/ACBAB – midterm 04 – Turner – (60230)
2
A coin 1.05 cm in diameter is embedded in
a solid glass ball of radius 38.3 cm(see the
figure).
1
8. f = − a
2
5
9. f = a
2
n1 > n2
10. f = 2 a
Explanation:
R1 is the x-coordinate of the center of the
spherical surface A ; i.e., R1 = −|RA | = −3 a.
Similarly, R2 = −a.
The lens maker’s formula gives
1
1
1
−
= (n − 1)
f
R1 R2
1
1
−
= (2 − 1)
−3 a −a
2
=
.
3a
Hence, f =
3
a .
2
003 10.0 points
Which travels most slowly in glass?
1. Red light
2. Green light
3. Orange light
d
n1
n2
p
The index of refraction of the ball is 1.64,
and the coin is p = 19.8 cm from the surface.
Find the distance the image’s position is
from the surface of the glass ball.
1. 12.884
2. 13.8881
3. 13.1187
4. 16.6466
5. 12.5449
6. 14.7349
7. 15.1245
8. 11.3205
9. 14.4782
10. 12.1357
Correct answer: 15.1245 cm.
4. Brown light
5. Yellow light
6. Violet light correct
7. Blue light
Explanation:
c
v
Since violet light is refracted more than
the light of other frequencies its index of refraction is the highest, making its speed the
smallest. In other words, violet light travels
slowest in glass.
Explanation:
Because n1 > n2 , where n1 = 1.64 is the
index of refraction of the ball and n2 = 1.00 is
the index of refraction of the air, the rays are
refracted away from the normal at the surface
and diverge outward. Hence, the image is
formed in the glass and is virtual. Applying
the equation
n2 − n1
n1 n2
+
=
,
p
q
R
n=
004
10.0 points
with d = |q| and R = −38.3 cm , we get
−1
n2 − n1 n1
−
q = n2
R
p
−1
(1.64)
(1) − (1.64)
= (1)
−
(−38.3 cm)
(19.8 cm)
Version 145/ACBAB – midterm 04 – Turner – (60230)
= −15.1245 cm
d = |q| = 15.1245 cm .
The negative sign of q indicates that the image
is in the same medium as the object (the side
of incident light).
005 10.0 points
A certain kind of glass has an index of refraction of 1.654 for blue light of wavelength 430
nm and an index of 1.619 for red light of wavelength 680 nm. A beam containing these two
colors is incident at an angle of 16.29 ◦ on a
piece of this glass.
What is the angle between the two beams
inside the glass?
1. 0.612007
2. 0.397618
3. 0.174793
4. 0.162088
5. 0.56236
6. 0.501478
7. 0.205985
8. 0.213215
9. 0.468476
10. 0.439717
Correct answer: 0.213215◦.
Explanation:
Given : nb
λb
nr
λr
θi
= 1.654 ,
= 430 nm ,
= 1.619 ,
= 680 nm , and
= 9.97711◦ .
Due to different refraction indices of glass
for red and blue light, they will have different
refraction angles, so the mixture-beam will
be separated inside the glass. According to
Snell’s law,
nr sin θr = ni sin θi
sin θi
sin θr =
nr
−1 sin θi
.
θr = sin
nr
3
For red light
sin 16.29◦
θr = arcsin
1.619
◦
= 9.97711 ,
and for blue light
sin 16.29◦
θb = arcsin
1.654
◦
= 9.76389 .
Thus the angle between the beams is
∆θ = θr − θb
= 9.97711◦ − 9.76389◦
= 0.213215◦ .
006 10.0 points
Assume:
Refraction index for diamond
ndiamond = 2.33 .
The smallness of the critical angle θc for diamond means that light is easily “trapped”
within a diamond and eventually emerges
from the many cut faces. This makes a diamond more brilliant than stones with smaller
n and larger θc . Traveling inside a diamond, a
light ray is incident on the interface between
diamond and air.
What is the critical angle for total internal
reflection?
1. 29.5123
2. 27.1693
3. 20.8469
4. 29.1964
5. 25.4158
6. 22.4378
7. 21.3237
8. 24.7343
9. 26.7726
10. 22.9934
Correct answer: 25.4158 ◦.
Explanation:
Basic Concept: Critical angle θc for total
internal reflection
n2
sin θc =
.
n1
Version 145/ACBAB – midterm 04 – Turner – (60230)
Solution: For diamond, the critical angle
sin θc =
1
.
2.33
θc = 25.4158 ◦ .
007 10.0 points
Consider a light ray which enters from air to
a liquid, where the index
√ of refraction of the
liquid is given by n = 2.
light ray
n=1
n=
√
Air
Liquid
2
v′
What is the ratio
(where v ′ is defined in
c
the liquid)
1.
2.
3.
4.
v′
c
v′
c
v′
c
v′
c
v′
c
=
=
√
2
1
2
A thin lens of focal length 21 cm forms a real
image 5.2 times as high as the object.
How far apart are the object and image?
1. 155.238
2. 510.069
3. 578.875
4. 564.713
5. 293.148
6. 185.834
7. 418.476
8. 206.25
9. 314.659
10. 472.582
Correct answer: 155.238 cm.
Explanation:
The mirror equation is
1 1
1
= + ,
f
p q
where f is the focal length, p is the image
distance and q is the object distance. We
want p + q, which is the distance between the
image and object. The magnification M is
related to these distances by
M=
=2
q
,
p
so we may solve to get
=1
q = M p.
1
= √ correct
2
Explanation:
The frequency of an electromagnetic wave
is independent of the media in which it is
present; that is, f = f ′ .
A ray with a frequency f has a wavelength
c
λ = in the vacuum. In a medium with an
f
index of refraction n, and from the definition
c
of the index of refraction, n ≡ ′ .
v
√
So for n = 2
5.
4
v′
1
1
≡ =√ .
c
n
2
008
10.0 points
On substituting this into the mirror equation,
1
1 1
= 1+
.
f
M p
results, whence
1
p=f 1+
M
1
= (21 cm) 1 +
5.2
= 25.0385 cm .
From this,
q=Mp
= (5.2) (25.0385 cm)
= 130.2 cm .
Version 145/ACBAB – midterm 04 – Turner – (60230)
Thus the distance between them is
p + q = (25.0385 cm) + (130.2 cm)
= 155.238 cm .
009 10.0 points
The real image produced by a concave mirror
is found to be 0.692 times the size of the
object.
If the distance from the mirror to the screen
on which the image appears is 1.17 m, what is
the focal length of the mirror?
1. 0.718008
2. 0.529066
3. 0.645849
4. 0.836551
5. 0.54678
6. 0.691489
7. 0.623072
8. 0.639098
9. 0.944363
10. 0.670769
Correct answer: 0.691489 m.
Explanation:
5
according to E = (0.04 N/C) sin(2000 s−1 ) t,
where t is in seconds.
The permittivity of free space is 8.85 ×
10−12 C2 /N · m2 .
Find the maximum displacement current
~
through a 1.5 m2 area perpendicular to E.
1. 6.8145e-10
2. 1.062e-09
3. 2.3364e-09
4. 8.496e-10
5. 2.9205e-10
6. 1.2744e-09
7. 1.8585e-09
8. 3.186e-10
9. 2.2302e-09
10. 1.3275e-09
Correct answer: 1.062 × 10−9 A.
Explanation:
Let : E0 = 0.04 N/C ,
ω = 2000 s−1 ,
A = 1.5 m2 , and
ǫ0 = 8.85 × 10−12 C2 /N · m2 .
The displacement current is
1 1
1
2
h′
q
+ = =
m=
=−
p q
f
R
h
p
Concave Mirror
f >0
∞ >p> f
f >p> 0
f < q < ∞ 0 > m > −∞
−∞ < q < 0 ∞ > m > 1
Let : q = 1.17 m and
M = −0.692 .
For a concave mirror, where
q
M = − = −0.692 ,
p
1 1
1
+ = , and
p q
f
f=
1.17 m
q
=
= 0.691489 m .
1+M
1 + 0.692
010 10.0 points
In a region of space, the electric field varies
d φe
dt
d
= ǫ0 (E A)
dt
d
= ǫ0 E0 A (sin ω t)
dt
= ǫ0 E0 A ω cos ω t ,
Id = ǫ0
so the maximum displacement current is
Id,max = ǫ0 E0 A ω
= (8.85 × 10−12 C2 /N · m2 ) (0.04 N/C)
× (1.5 m2 ) (2000 s−1 )
= 1.062 × 10−9 A .
011 10.0 points
Consider a polarizer-analyzer arrangement as
shown in the figure.
Version 145/ACBAB – midterm 04 – Turner – (60230)
Unpolarized
light
percent ( % ), we obtain
Polarizer
E0
Analyzer
θ
E 0 cos θ
Polarized
light
Transmission
axis
At what angle is the axis of the analyzer
to the axis of the polarizer if, after passing
through both sheets, the beam intensity is
reduced by 82.5 percent?
1. 52.8917
2. 35.4263
3. 50.5348
4. 46.6051
5. 39.582
6. 52.1796
7. 46.0316
8. 53.7288
9. 40.048
10. 31.564
Correct answer: 53.7288◦.
Explanation:
The unpolarized light I0 = Iunpolarized is
reduced by 50% when the light passes through
the polarizer.
Ipolarized =
I0
2
Itransmitted =
Ipolarized
I0
=
cos2 θ ,
2
2
2r
100%
r
2 (82.5%)
= arccos 2 −
100%
= 0.937744 rad
= 53.7288◦ .
θ = arccos
2−
012 10.0 points
An object is 16 cm from the surface of a
reflective spherical Christmas-tree ornament
9.08 cm in radius.
What is the magnification of the image?
1. 0.103563
2. 0.148855
3. 0.0982439
4. 0.24177
5. 0.134385
6. 0.183588
7. 0.143711
8. 0.171107
9. 0.0800319
10. 0.221032
Correct answer: 0.221032.
1 1
1
2
h′
q
+ = =
M=
=−
p q
f
R
h
p
Convex Mirror 0 > f
∞ >p> 0 f <q< 0
and the fact that
Itransmitted = I0 − Iabsorbed .
Iabsorbed = I0 − Itransmitted = I0 −
r
Explanation:
We use the formula for the intensity of the
transmitted (polarized) light
I0
cos2 θ .
2
These relations yield
s
2 (I0 − Iabsorbed )
.
θ = arccos
I0
Iabsorbed
is the ratio of the intensiI0
ties of the absorbed and the incident light in
So, if r =
6
0 <M < 1
Let : R = 9.08 cm ,
p = 16 cm .
and
p is positive since it is in front of the mirror
and R is negative since it is behind the mirror. A spherical Christmas-tree ornament is a
convex mirror, so
2
1 1
+ =−
.
p q
|R|
Version 145/ACBAB – midterm 04 – Turner – (60230)
We are given the object distance, p and |R|.
Inserting these values into the mirror equation
and solving for q, we find:
q=−
θa
air
glass
1
1
2
+
R p
water
θw
1
=−
7
2
1
+
(9.08 cm) (16 cm)
= −3.53651 cm .
The magnification is given by
q
p
−3.53651 cm
=−
16 cm
= 0.221032 .
M =−
013 10.0 points
A diver shines light up to the surface of a
flat glass-bottomed boat at an angle of 36 ◦
relative to the normal.
If the indices of refraction of air, water,
and glass are 1.0, 1.33, and 1.6 respectively,
at what angle does the light leave the glass
(relative to its normal)?
1. 41.6823
2. 60.7573
3. 37.1431
4. 27.0577
5. 51.4215
6. 58.7495
7. 49.7167
8. 40.1504
9. 31.3103
10. 54.9678
Correct answer: 51.4215 ◦.
Explanation:
n1 sin θ1 = n2 sin θ2
Since
na sin θa = ng sin θg = nw sin θw ,
we do not need to consider the index of refraction of glass in the solution to the question.
The angle is
θair
nwater sin θwater
= sin
nair
◦
−1 1.33 × sin 36
= sin
1.0
−1
= 51.4215 ◦ .
014 10.0 points
A screen is placed between two lamps so that
they illuminate the screen equally. The first
lamp emits a luminous flux of 1064 lm and is
5.3 m from the screen.
What is the distance of the second lamp
from the screen if the luminous flux is
2411 lm?
1. 8.79882
2. 10.6874
3. 7.15113
4. 12.3253
5. 7.27282
6. 7.01002
7. 6.70638
8. 9.23351
9. 7.97818
10. 9.41648
Correct answer: 7.97818 m.
Explanation:
Basic Concepts
The illumination is
E=
P
4 π d2
The screen is illuminated equally by both
lamps so
P1
P2
=
4 π d22
4 π d21
P2
P1
= 2
2
d2
d1
Version 145/ACBAB – midterm 04 – Turner – (60230)
s
d22
d21
263 W/m2
=
×
P2
P1
2.99792 × 108 m/s
2P
d
2
d22 = 1
= 1.48487 × 10−6 T .
P1
s
d21 P2
d2 =
016 10.0 points
P1
015 10.0 points
Consider an electromagnetic plane wave with
a time averaged intensity 263 W/m2 .
The speed of light is 2.99792 × 108 m/s
and the permeability of free space is 4 π ×
10−7 T·N/A .
What is the maximum magnetic field?
1. 1.0756e-06
2. 1.33629e-06
3. 1.15454e-06
4. 1.10254e-06
5. 1.53484e-06
6. 1.48487e-06
7. 1.7872e-06
8. 1.26208e-06
9. 1.69078e-06
10. 9.55923e-07
Correct answer: 1.48487 × 10−6 T.
Q
D
F
S
1. Q, Z, and F
2. Q, Z, and K
3. D and K
5. Z and K
Let : I = 263 W/m2 ,
µ0 = 4 π × 10−7 T·m/A ,
c = 2.99792 × 108 m/s .
6. Q, D, K, and S
and
The average intensity of an electromagnetic
wave is the magnitude of its average Poynting
vector
EB
Emax Bmax
=
µ0
2µ0
7. D and F correct
8. Q, D, Z, and S
9. Q, D, and F
10. Q, F, and S
Explanation:
Since E = c B,
2
Bmax
c
2µ
r 0
2 µ0 I
=
c
q
= 2 (4 π × 10−7 T·N/A)
I=
Bmax
Z
Which of the glass lenses above, when
placed in air, will cause rays of light (parallel to the central axis) to diverge?
4. Q, K, and S
Explanation:
I = Save =
K
8
convergent
divergent
Use the lens makers’ equation
1
1
1
,
= (n − 1)
−
f
R1 R2
Version 145/ACBAB – midterm 04 – Turner – (60230)
where R1 and R2 are + (or −) if the center of
curvature is behind (or in front of) the lens.
To cause parallel rays of light to converge,
f must be positive, (n > 1).
Q: R1 > 0 and R2 = ∞, =⇒ f > 0 , therefore
convergent.
D: R1 = ∞ and R2 > 0, =⇒ f < 0 , therefore
divergent.
Z: R1 > 0 and R2 > 0, but |R1 | > |R2 |,
=⇒ f > 0 , therefore convergent.
F: R1 < 0 and R2 < 0, but |R1 | < |R2 |,
=⇒ f < 0 , therefore divergent.
K: R1 < 0 and R2 < 0, but |R1 | > |R2 |,
=⇒ f > 0 , therefore convergent.
S: R1 > 0 and R2 > 0, but |R1 | = |R2 |,
=⇒ f = ∞ , therefore neither convergent
or divergent.
Thus, lenses Q, Z, and K are convergent
lenses and Lenses D and F are divergent.
Alternate (Elegant) Solution:
Converging Lens: The glass is thicker on
the axis than at the edge.
Q, Z, and K satisfy these conditions for a
converging lens (f > 0).
Diverging Lens: The glass is thinner on
the axis than at the edge.
D and F satisfy these conditions for a diverging lens (f < 0).
Neutral Lens: The glass has a constant
thickness. Rays of light are parallel to the
central axis on both sides of the lens.
S satisfy this conditions for a non-focusing
lens (f = ∞).
Digression based on Huygen’s principle:
Consider the passage of a wave front of
a plane wave through a lens. If the center
part of the lens is thicker, the center of the
exit vertical plane has a larger phase change
compared to that in the region surrounding
the center. So the surrounding region needs
to travel farther to acquire the same phase
change. Analogously, if the lens at the center
is thinner, at the exit side the center part
needs to travel farther to acquire the same
phase as that in the peripheral region. This
leads to a divergent wave front at the exit side
of the lens.
9
017 10.0 points
Consider a light ray traveling from medium 1
to medium 2 with medium 2 extending downward indefinitely. Let the index of refraction
of medium 1 be 1.0.
interface A
θ1
medium1
medium2
interface B
θ2
If the index of refraction of medium 2 is
1.58 and θ1 is nearly 90◦ , what is θ2 ?
1. 57.2796
2. 61.1125
3. 50.7348
4. 50.1317
5. 60.1644
6. 56.4623
7. 58.6118
8. 55.3765
9. 62.1415
10. 54.6876
Correct answer: 50.7348 ◦.
Explanation:
interface A
θ1
θR
medium1
medium2
interface B
θ2
The angle of refraction at interface A (see
the above figure) is found from Snell’s law:
n2 sin θR = n1 sin θ1
1.00
sin 90◦
sin θR =
1.58
= 0.632911
θR = 39.2652 ◦
Version 145/ACBAB – midterm 04 – Turner – (60230)
where θR is angle of refraction in medium 2.
θ2 is related to θR by observing that
θ2 + θR = 90◦ . Thus
θ2 = 90◦ − θR
= 90◦ − 39.2652 ◦
= 50.7348 ◦
018 10.0 points
A telescope is constructed with two lenses
separated by a distance of 54.3 cm. The focal
length of the objective is 48 cm. The focal
length of the eyepiece is 6.3 cm.
Calculate the magnitude of the angular
magnification of the telescope.
1. 8.33333
2. 9.0
3. 6.0274
4. 7.61905
5. 3.28767
6. 6.96203
7. 14.5161
8. 4.28571
9. 11.1111
10. 1.15152
Correct answer: 7.61905.
Explanation:
The angular magnification of the telescope
is
fo
fe
48 cm
=
6.3 cm
= 7.61905.
m=
019 10.0 points
At what distance from a 70 W electromagnetic wave point source (like a light bulb) is
the amplitude of the electric field 6 V/m?
µo c = 376.991 Ω .
1. 1.46059
2. 2.76504
3. 6.0208
4. 3.99446
5. 10.0623
6. 1.99489
10
7. 10.8012
8. 0.489898
9. 2.46256
10. 24.7656
Correct answer: 10.8012 m.
Explanation:
Let : P = 70 W ,
Emax = 6 V/m , and
µ0 c = 376.991 Ω .
The wave intensity (power per unit area) is
given by
I = Sav
2
Emax
=
2 (µo c)
(6 V/m)2
=
2 (376.991 Ω)
= 0.0477465 W/m2 .
The constant µo c (impedance of free space)
turns out to be 120 π Ω.
You should verify that (V/m)2 Ω = W/m2 .
With this we can find the intensity, which
multiplied by the area of a circle of radius R
must gives us the source’s power P , therefore
4 π R2 I = P = 70 W .
Solving for the radius we get,
s
r
P
70 W
R=
=
4π I
4 π (0.0477465 W/m2 )
= 10.8012 m .
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
This print-out should have 33 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
Three identical point charges, each of mass
130 g and charge +q, hang from three strings,
as in the figure.
The acceleration of gravity is 9.8 m/s2 ,
and the Coulomb constant is 8.98755 ×
109 N · m2 /C2 .
1
F acting on this charge, keeping it balanced
against the force of gravity m g. The electrostatic force is due to the other two charges
and is therefore horizontal.
In the x-direction
F − T sin θ = 0 .
In the y-direction
T cos θ − m g = 0 .
These can be rewritten as
2
.7
41◦
cm
10
F = T sin θ
.7
cm
10
9.8 m/s
130 g
130 g
130 g
+q
+q
+q
If the lengths of the left and right strings
are each 10.7 cm, and each forms an angle of
41◦ with the vertical, determine the value of
q.
Correct answer: 0.696975 µC.
and
m g = T cos θ .
Dividing the former by the latter, we find
F
= tan θ ,
mg
or
Explanation:
Let : θ = 41◦ ,
m = 130 g = 0.13 kg ,
L = 10.7 cm = 0.107 m ,
g = 9.8 m/s2 , and
ke = 8.98755 × 109 N · m2 /C2 .
Newton’s 2nd law:
X
F = m g tan θ
= (0.13 kg) (9.8 m/s2 ) tan 41◦
= 1.10747 N .
The distance between the right charge and
the middle charge is L sin θ, and the distance
to the left one is twice that. Since all charges
are of the same sign, both forces on the right
charge are repulsive (pointing to the right).
We can add the magnitudes
F = ma.
Electrostatic force between point charges q1
and q2 separated by a distance r
q1 q2
F = ke 2 .
r
All three charges are in equilibrium, so for
each holds
X
F = 0.
Consider the forces acting on the charge on
the right. There must be an electrostatic force
(1)
F = ke
qq
qq
+
k
e
(L sin θ)2
(2 L sin θ)2
or
F =
5 ke q 2
.
4 L2 sin2 θ
(2)
We have already found F , and the other quantities are given, so we solve for the squared
charge q 2
q2 =
4 F L2 sin2 θ
5 ke
(3)
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
or, after taking the square root (we know
q > 0) and substituting F from Eq. 1 into Eq.
3 and solving for q, we have
r
m g tan θ
q = 2 L sin θ
5 ke
= 2 (0.107 m) sin 41◦
s
(0.13 kg) (9.8 m/s2 ) tan 41◦
×
5 (8.98755 × 109 N · m2 /C2 )
= 6.96975 × 10−7 C = 0.696975 µC .
002 10.0 points
Two identical conducting spheres, A and B,
carry equal charge. They are stationary and
separated by a distance much larger than
their diameters. A third identical conducting sphere, C, is uncharged. Sphere C is first
touched to A, then to B, and finally removed
(to a far away distance).
As a result the electrostatic force Ff between A and B, which was originally Fi , becomes
1. Ff = Fi .
2. Ff =
1
Fi .
8
3. Ff = 0 Fi .
1
Fi .
16
1
5. Ff = Fi .
2
3
6. Ff = Fi .
4
1
7. Ff = Fi .
4
3
8. Ff =
Fi .
16
3
9. Ff = Fi . correct
8
Explanation:
Since the two conducting spheres are identical (i.e., same radius), when the spheres touch
the charges redistribute themselves equally
between the two spheres. Let spheres A and
4. Ff =
2
B have an initial charge Q. When an identical uncharged sphere C comes in contact with
sphere A and is removed, then by conservation of charge, each sphere will carry charge
QC1 = QA =
1
Q.
2
When sphere C touches sphere B, then each
sphere will carry charge
QC2
1
Q+Q
QC1 + QB
3
= QB =
= 2
= Q.
2
2
4
Hence if the initial force is given by
Fi = ke
then the final force is
Ff = ke
=
Q2
,
d2
3
Q
4
1
Q
2
d2
3
Fi .
8
003 (part 1 of 2) 10.0 points
A deuteron (a nucleus that consists of one proton and one neutron) is accelerated through a
2 kV potential difference.
How much kinetic energy does it gain?
Correct answer: 3.20435 × 10−16 J.
Explanation:
Let : ∆V = 2 kV = 2000 V ,
q = 1.60218 × 10−19 C ,
mproton = 1.67262 × 10−27 kg ,
mneutron = 1.67493 × 10
−27
and
kg .
Note that the neutron has no charge.
By conservation of energy
∆K = q ∆V
= 1.60218 × 10−19 C (2000 V)
= 3.20435 × 10−16 J .
004 (part 2 of 2) 10.0 points
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
How fast is it going if it starts from rest?
Correct answer: 4.37544 × 105 m/s.
Explanation:
The speed of the deuteron may be determined from
1
(mproton + mneutron ) v 2 = ∆K
2
Solving for v,
s
2 (∆K)
v=
mproton + mneutron
s
2 (3.20435 × 10−16 J)
=
1.67262 × 10−27 kg + 1.67493 × 10−27 kg
= 4.37544 × 105 m/s .
005 (part 1 of 2) 10.0 points
A long coaxial cable consists of an inner cylindrical conductor with radius R1 and an outer
cylindrical conductor shell with inner radius
R2 and outer radius R3 as shown. The cable extends out perpendicular to the plane
shown. The charge on the inner conductor
per unit length along the cable is λ and the
corresponding charge on the outer conductor
per unit length is −λ (same in magnitudes
but with opposite signs) and λ > 0.
−Q
R2
Q
b
R3
Find the magnitude of the electric field at
the point a distance r1 from the axis of the
inner conductor, where R1 < r1 < R2 .
λ R1
2. E =
4 π ǫ0 r1 2
2λ
3. E = √
3 π ǫ0 r1
λ R1
3 π ǫ0 r1 2
λ2 R1
5. E =
4 π ǫ0 r1 2
λ
6. E = √
3 π ǫ0 r1
4. E =
7. None of these.
λ
2 π ǫ0 R1
λ
9. E = √
2 π ǫ0 r1
λ
10. E =
correct
2 π ǫ0 r1
Explanation:
Pick a cylindrical Gaussian surface with the
radius r1 and apply the Gauss’s law; we obtain
8. E =
E · ℓ · 2 π r1 =
E=
Q
ǫ0
λ
2 π ǫ0 r1
006 (part 2 of 2) 10.0 points
For a 100 m length of coaxial cable with
inner radius 0.471383 mm and outer radius
1.42122 mm.
Find the capacitance C of the cable.
Correct answer: 5.041 nF.
Explanation:
R1
1. E = 0
3
Let :
ℓ = 100 m ,
R1 = 0.471383 mm ,
R2 = 1.42122 mm .
and
We calculate the potential across the capacitor by integrating −E · ds. We may choose
a path of integration along a radius; i.e.,
−E · ds = −Edr.
Z
1 q R1 dr
V =−
2 π ǫ0 l R2 r
R
1
1 q
ln r
=−
2 π ǫ0 l
R2
R2
q
ln
.
=
2 π ǫ0 l
R1
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
Since C =
q
, we obtain the capacitance
V
2 π ǫ0 l
R2
ln
R1
2 π (8.85419 × 10−12 c2 /N · m2 )
=
1.42122 mm
ln
0.471383 mm
× (100 m)
C=
= 5.041 nF .
c
µF
12
a
20
4
µF
53 V
37
S2
4
µF
b
F
6µ
S1
d
Find the charge on the 12 µF upper-left
capacitor between points a and c.
Correct answer: 480.245 µC.
Explanation:
007 10.0 points
Two identical parallel-plate capacitors are
connected in series.
Which of the following is true of the equivalent capacitance?
1. It is the same as that of each capacitor.
Let : C1
C2
C3
C4
EB
2. It depends on the charge on each capacitor.
3. It is smaller than that of each capacitor.
correct
4. It depends on the potential difference
across both capacitors.
C1
a
C2
EB
= 12 µF ,
= 20 µF ,
= 37 µF ,
= 46 µF ,
= 53 V .
c
S2
and
C3
b
C4
S1
d
5. It is larger than that of each capacitor.
Explanation:
The equivalent capacitance for two capacitors in series is
Redrawing the figure, we have
C1
C3
c
a
1
1
1
C2 + C1
=
+
=
Ceq
C1 C2
C1 C2
Ceq =
C1 C2
C1 + C2
which is always less than C1 or C2 individually. (Convince yourself!)
008 (part 1 of 2) 10.0 points
In the figure below consider the case where
switch S1 is closed and switch S2 is open.
EB
C2
C4
d
C1 and C3 are in series, so
−1
1
1
+
C13 =
C1 C3
C1 C3
=
C1 + C3
(12 µF) (37 µF)
=
12 µF + 37 µF
= 9.06122 µF .
b
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
C2 and C4 are in series, so
C2 and C4 are in series, so
C24
−1
1
1
=
+
C2 C4
C2 C4
=
C2 + C4
(20 µF) (46 µF)
=
20 µF + 46 µF
= 13.9394 µF .
Q4 = Q2 = 738.788 µC .
009 (part 2 of 2) 10.0 points
Now consider the case where switch S2 is also
closed.
c
37
F
µ
µF
2
1
a
Simplifying the circuit, we have
C13
a
C24
S2
µF
4
b
F
6µ
S1
d
Find the charge on the 12 µF upper-left
capacitor between points a and c.
Correct answer: 459.026 µC.
C13 and C24 are parallel, so
Cab = C13 + C24
= 9.06122 µF + 13.9394 µF
= 23.0006 µF .
Cab
20
53 V
b
EB
a
5
Explanation:
A good rule of thumb is to eliminate junctions connected by zero capacitance.
Switch S2 is closed.
C1
C3
a
b
EB
C2
c
d
b
C4
EB
C1 and C3 are in series, so
Q1 = Q3 = Q13
= C13 EB
= (9.06122 µF) (53 V)
When S2 is closed, C1 and C2 are parallel,
so
C12 = C1 + C2
= 12 µF + 20 µF
= 32 µF .
= 480.245 µC .
C2 and C4 are in series, so
Q2 = Q4 = Ql
= C24 EB
= (13.9394 µF) (53 V)
= 738.788 µC .
C1 and C3 are in series, so
Q3 = Q1 = 480.245 µC .
C3 and C4 are parallel, so
C34 = C3 + C4
= 37 µF + 46 µF
= 83 µF .
a
EB
C12
C34
b
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
C12 and C34 are in series, so
−1
1
1
Cab =
+
C12 C34
C12 C34
=
C12 + C34
(32 µF) (83 µF)
=
32 µF + 83 µF
= 23.0957 µF .
Q34 = Q12 = Qab
= Cab EB
= (23.0957 µF) (53 V)
= 1224.07 µC .
a
Cab
b
EB
Since C1 and C2 are parallel,
V1 = V2 = V12
Q12
=
C12
1224.07 µC
=
32 µF
= 38.2522 V .
and
Q1 = V1 C1
= (38.2522 V) (12 µF)
= 459.026 µC .
Q2 = V2 C2
= (38.2522 V) (20 µF)
= 765.043 µC .
V3 = V4 = V34
Q34
=
C34
1224.07 µC
=
83 µF
= 14.7478 V .
6
Q3 = V3 C3
= (14.7478 V) (37 µF)
= 545.67 µC .
Q4 = V4 C4
= (14.7478 V) (46 µF)
= 678.4 µC .
010 (part 1 of 2) 10.0 points
Consider a air-filled parallel-plate capacitor
with plates of length 9.7 cm, width 6.15 cm,
spaced a distance 1.57 cm apart.
Now imagine that a dielectric slab with dielectric constant 7.7 is inserted a length 5.9 cm
into the capacitor. The slab has the same
width as the plates. The capacitor is completely filled with dielectric material down to
a length of 5.9 cm.
A battery is connected to the plates so that
they are at a constant potential 0.15 V while
the dielectric is inserted.
dielectric
constant 7.7
top
5.9 cm
dielectric
none - air 1.0
bottom
9.7 cm
1.57 cm
What is the ratio of the new potential energy to the potential energy before the insertion of the dielectric? Edge effects can be
neglected.
Correct answer: 5.07526.
Explanation:
Let : w = 6.15 cm ,
d = 1.57 cm ,
L = 9.7 cm ,
ℓ = 5.9 cm ,
κ = 7.7 , and
V = 0.15 V .
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
The given capacitor can be considered as
two capacitors in parallel, as shown below.
Therefore, we have
U′
C′
=
U
C
n
oǫ w
0
L + ℓ [κ − 1]
d
=
ǫ0 w
L
d
h
i
ℓ
=1+
κ−1
L
i
(5.9 cm) h
(7.7) − 1
=1+
(9.7 cm)
C1
C2
The potential energy in a capacitor is
(6)
= 5.07526 .
1
U = CV2,
2
When the dielectric is fully inserted ℓ = L,
so
so the ratio is
U′
U
=
C ′ V ′2
CV2
i
Lh
U′
κ − 1 = κ.
=1+
U
L
.
Since the electric potential difference (voltage) across the capacitor is constant,
U′
C′
=
.
U
C
(1)
The capacitance of a parallel plate capacitor is
C=
7
ǫ0 w L
ǫ0 · area
=
.
separation
d
(2)
(3)
The total capacitance of the new system is
C ′ = C1′ + C2
= [L − ℓ + κ ℓ] C
= [L + ℓ (κ − 1)]
(4)
ǫ0 w
.
d
011 (part 2 of 2) 10.0 points
From this, we can deduce that if a person
inserts the slab part way into the conductor,
and then releases it, the capacitor will exert a
force on the dielectric in what direction?
1. there will be no force exerted on the slab
2. pulling it into the capacitor correct
3. toward the positively charged plate
After the insertion of the dielectric, the capacitor can be considered in two parts: the
part filled with dielectric, and the part with
no dielectric.
Since the dielectric is inserted with a length
ℓ,
C ′ = C1′ + C2 , where
ǫ0 w κ ℓ
C1′ =
d
ǫ0 w (L − ℓ)
.
C2 =
d
(7)
(5)
4. toward the negatively charged plate
5. pushing it out of the capacitor
Explanation:
Force is the negative gradient of potential;
i.e., since potential energy U ′ is gained as ℓ
increases, you must pull the dielectric out of
the capacitor.
We found earlier (from Eq. 5) that the potential energy stored in the capacitor is given
by
1 ′ 2
C V
2
1
= [L + ℓ (κ − 1)] C V 2 ,
2
1
= [L + ℓ (κ − 1)] Q V .
2
′
=
Ucap
or
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
The energy drained from the battery is
′
Ubat
= ∆Q V
= (L − ℓ) Q + ℓ Q′ V
= [(L − ℓ) Q + ℓ κ Q] V
= [L + ℓ (κ − 1)] Q V ,
so the combined energy (Ucap − Ubat ) is
1
ℓ
′
U =−
1 + (κ − 1) Q V ,
2
L
since the energy drained from the batter is
twice the energy gained in the capacitor.
Note: κ is always greater than one in a
dielectric, so the work to remove the dielectric
has to increase with increasing ℓ.
The capacitor exerts a force in the direction
of decreasing potential energy. So as the slab
is released, the capacitor will pull the slab into
its center.
IF BATTERY IS DISCONNECTED
If the battery is disconnected before the dielectric is inserted, the force on the dielectric
will be in the opposite direction; i.e., “pulling
it further into the capacitor”.
The potential energy in a capacitor is
U=
1 Q2
,
2 C
so the ratio is
Q′ 2 C
U′
= ′
.
U
C Q2
Since the charge on the capacitor is constant,
U′
C
= ′,
(7)
U
C
which is the inverse of Eq. 1. Therefore (using
Eq. 2 and 7), we have
ǫ0 w
L
U′
C
d o
= ′ =n
ǫ0 w
U
C
L + ℓ [κ − 1]
d
L
h
i
(8)
=
L+ℓ κ−1
=
=
(9.7 cm)
h
i
(9.7 cm) + (5.9 cm) (7.7) − 1
1
= 0.197034 (see Part 1).
5.07526
8
When the dielectric is fully inserted ℓ = L ,
U′
=
U
1
1
= ,
h
i
L
κ
1+
κ−1
L
which is also the inverse of Eq. 6.
012 (part 1 of 2) 10.0 points
A parallel-plate capacitor of dimensions
2.8 cm × 3.16 cm is separated by a 0.69 mm
thickness of paper.
Find the capacitance of this device. The
dielectric constant κ for paper is 3.7.
Correct answer: 42.0094 pF.
Explanation:
Let : κ = 3.7 ,
d = 0.69 mm = 0.00069 m , and
A = 2.8 cm × 3.16 cm = 0.0008848 m2 .
We apply the equation for the capacitance of
a parallel-plate capacitor and find
A
d
= (3.7) (8.85419 × 10−12 C2 /N · m2 )
0.0008848 m2 1 × 1012 pF
×
0.00069 m
1F
C = κ ǫ0
= 42.0094 pF .
013 (part 2 of 2) 10.0 points
What is the maximum charge that can be
placed on the capacitor? The electric strength
of paper is 1.6 × 107 V/m.
Correct answer: 0.463784 µc.
Explanation:
Let : Emax = 1.6 × 107 V/m .
Since the thickness of the paper is 0.00069 m,
the maximum voltage that can be applied
before breakdown is
Vmax = Emax d .
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
9
Hence, the maximum charge is
Qmax = C Vmax = C Emax d
= (42.0094 pF)(11040 V)
1 × 10−12 F 1 × 106 µC
·
·
1 pF
1C
6. iA decreases, iD decreases, ibattery decreases
7. iA remains the same, iD remains the same,
ibattery remains the same
= 0.463784 µc .
014 (part 1 of 2) 10.0 points
All the bulbs in the figure below have the
same resistance R. The switch S is initially
closed.
8. iA increases, iD remains the same, ibattery
increases
9. iA increases, iD increases, ibattery increases
10. iA increases, iD decreases, ibattery decreases
iD
S
iB
iA
V
5. iA increases, iD increases, ibattery decreases
iC
i0
If bulb B is removed from the circuit, i.e.,
the switch S is opened, what happens to the
current through
1) the battery,
2) bulb A, and
3) bulb D;
i.e., the brightness of bulb A, and bulb D?
Hint: You may find it helpful to work out the
currents through bulb A and bulb D, and the
battery for both cases by using V = 1 volt
and R = 1 Ω.
1. iA decreases, iD decreases, ibattery remains
the same
2. iA increases, iD increases, ibattery remains
the same
3. iA remains the same, iD increases, ibattery
increases
4. iA decreases, iD remains the same, ibattery
decreases correct
Explanation:
Denote the equivalent resistance of the left
branch by
The potential across the left branch and
bulb D te the equivalent resistance of the left
branch by RL and the entire circuit by Req .
The potential across the left branch and bulb
D is V , the potential across the battery.
With bulb B:
1
RBC
RBC
1
1
2
+ =
R R
R
R
=
2
=
R
3R
=
2
2
1
1
2
1
5
1
=
+ =
+ =
Req
RL R
3R R
3R
3R
Req =
5
RL = RA + RBC = R +
Therefore
V
2V
=
3R
3R
2
V
iD =
R
5V
ibattery =
3R
iA =
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
10
Without bulb B:
RL = 2 R
1
1
1
3
=
+ =
Req
2R R
2R
2R
Req =
3
Therefore
V
iA =
2R
V
iD =
R
3V
ibattery =
2R
Qualitatively, if bulb B is removed, this increases the equivalent resistance of the left
branch, so iA decreases. The voltage drop
across bulb D remains the same since it is
in parallel with the battery; so iD remains
the same. Since the equivalent resistance of
the left branch increases, the equivalent resistance of the entire circuit also increases.
Hence ibattery decreases.
015 (part 2 of 2) 10.0 points
3. iA remains the same, iD increases, ibattery
increases
4. iA decreases, iD decreases, ibattery decreases
5. iA increases, iD increases, ibattery increases
6. iA remains the same, iD remains the same,
ibattery remains the same
7. iA increases, iD remains the same, ibattery
increases correct
8. iA decreases, iD remains the same, ibattery
decreases
9. iA increases, iD increases, ibattery remains
the same
Explanation:
When the wire is added to the circuit, no
current flows through bulbs B and C so the
equivalent circuit is bulbs A and D parallel
with the battery. Req = R/2. Hence
iD
iA =
S
iC
iA
V
V
,
R
iD =
V
,
R
ibattery = 2
V
.
R
Comparing with the answers from part 1,
iA and ibattery increase while iD remains the
same.
iB
i0
If a wire is added to the circuit, what happens to the current through
1) the battery,
2) bulb A, and
3) bulb D;
i.e., the brightness of bulb A, and bulb D?
1. iA increases, iD decreases, ibattery decreases
2. iA increases, iD increases, ibattery decreases
016 (part 1 of 2) 10.0 points
The switch S has been in position b for a
long period of time.
R3
C
R2
R1
S b
E
a
When the switch is moved to position “a”,
find the characteristic time constant.
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
1. τ = √
1
R1 R2 C
Let : E
R1
R2
R3
C
2. τ = (R1 + R2 ) C correct
R1 + R2
C
2
2
4. τ =
(R1 + R2 ) C
1
5. τ =
R1 C
1
6. τ =
R2 C
3. τ =
R2 +R3
Udissip
=
1
C E2
2
. Since R2 and R3 are in series, the energy
R2
of
dissipated by R2 is only a fraction
R2 + R3
the total energy:
1
R2
R2
2
CE
Udissip =
R2 + R3
2
8. τ = R1 C
1
(R1 + R2 ) C
p
10. τ = R1 R2 C
9. τ =
Explanation:
In charging an R C circuit, the characteristic time constant is given by
τ = RC ,
where in this problem R is the equivalent
resistance, or
R = R1 + R2 .
017 (part 2 of 2) 10.0 points
4 MΩ
1.1 µF
1 MΩ
= 1.1 V ,
= 1 MΩ ,
= 3 MΩ ,
= 4 MΩ , and
= 1.1 µF .
The total energy dissipated
7. τ = R2 C
3 MΩ
11
S b
1.1 V
a
We observe that power consumption consideration provides an independent check on the
fraction used. Since the two resistors are in
series they share a common current, I. The
corresponding power consumptions by R2 and
R3 are respectively P2 = I 2 R2 and P3 = I 2 R3 .
This shows the correctness of the fraction, i.e.
P2 /(P2 + P3 ) = R2 /(R2 + R3 ). Alternate
solution:
More formally, noting that the initial current
E
is I0 =
, the total energy dissipated
R2 + R3
by R2 is
Z ∞
R2
U =
I(t)2 R2 dt
Z0 ∞
=
I02 R2 e−2t/[C(R2 +R3 )] dt
0
2
C (R2 + R3 )
E
=
R2 −
R2 + R3
2
∞
−2t/[C(R2 +R3 )]
×e
S has been left at position “a” for a long
time. It is then switched from “a” to “b” at
t = 0.
Determine the energy dissipated through
the resistor R2 alone from t = 0 to t = ∞.
Correct answer: 0.285214 µJ.
R2
1
=
C E2
R2 + R3 2
1
(3 MΩ)
(1.1 µF) (1.1 V)2
=
(3 MΩ) + (4 MΩ) 2
Explanation:
= 0.285214 µJ .
0
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
018 10.0 points
An electron enters a region of uniform magnetic field and is observed to execute circular motion, completing 142 revolutions in
∆T = 1.24 µs .
The mass of the electron is 9.10939 ×
10−31 kg and its charge is 1.60218 × 10−19 C.
Calculate the magnitude of the magnetic
field in this region.
Correct answer: 4.09096 mT.
Explanation:
Give the variation of the strength of the
magnetic field B as a function of distance z
from the current sheet.
1. Bz ∝ constant correct
2. Bz ∝ ln z
3. Bz ∝
∆T
N
me
qe
= 1.24 µs = 1.24 × 10−6 s ,
= 142 ,
= 9.10939 × 10−31 kg , and
= 1.60218 × 10−19 C .
The observed frequency is
ν=
N
.
∆T
Identifying this frequency with the cyclotron
qB
, the magnetic field is
frequency ν =
2πm
1
z2
4. Bz ∝ z −3/2
5. Bz ∝
Let :
12
6. Bz ∝
1
z
r
1
z
Explanation:
Ampére’s law can be used to determine the
magnitude of theI magnetic field. To evaluate
~ · d~ℓ in Ampére’s law, let
the line integral B
us take a rectangular path through the sheet.
The dimensions of the rectangle are ℓ in the x
direction and w in the z direction.
2 π me N
qe ∆T
2 π (9.10939 × 10−31 kg)
142
=
×
−19
1.60218 × 10
C
1.24 × 10−6 s
B=
B
B
= 4.09096 mT .
019 (part 1 of 2) 10.0 points
Consider an infinite sheet of current, perpendicular to the z axis, located at z = 0 .
The linear current density λ−y flows in the
−y direction.
x
infinite
current
sheet in
x, y plane
z
y
Current is in
−y direction, as
shown by arrows.
x
ℓ B
B
z
w
Note: The top and bottom sides of length
~ = 0), and B
~
w do not contribute (since ~ω · B
is equal and opposite on opposite sides of the
xy plane.
The contributions from the left and right
sides of length ℓ are equal and have the same
magnitude. The amount of current enclosed
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
is λ−y ℓ, where λ−y is a current per unit length
in the direction shown in the figure and ℓ is
the length perpendicular to the λ−y direction
in the plane of the current sheet. The units of
λ−y is [A/m].
Using Ampére’s law, we have
I
~ · d~s = µ0 I = µ0 λ−y ℓ
B
2 B ℓ = µo λ−y ℓ .
Thus the magnetic field is constant as we
vary z . The constant magnetic field in the z
direction is
λ−y
Bz = µ0
.
2
020 (part 2 of 2) 10.0 points
What is the direction of the magnetic field at
a point on the positive z side of the sheet (on
the right-hand side of the sheet)?
b z is in the +x direction
1. B
b z is in the −x direction correct
2. B
During the time the field is changing its
direction, how much charge flows through the
coil if the resistance is 2.3 Ω?
Correct answer: 2.47727 C.
Explanation:
From Faraday’s Law for Solenoids
E = −N
dΦB
dt
and Ohm’s Law
V
,
R
I=
the current through R is
V
R
E
=
R
N d ΦB
=
R dt
dBA 1
=N
dt R
dB A
.
=N
dt R
I=
b z is in the −z direction
4. B
b z is in the +z direction
5. B
b z is in the +y direction
6. B
021 10.0 points
A circular coil enclosing an area of 119 cm2
is made of 228 turns of copper wire as shown
schematically in the figure. Initially, a 1.05 T
uniform magnetic field points perpendicularly
left-to-right through the plane of the coil. The
direction of the field then reverses to right-toleft. The field reversal takes 1.0 ms.
Magnetic
Field B(t)
R
b z is in the −y direction
3. B
Explanation:
The direction of the magnetic field can be
determined from Ampére’s law as well, or by
placing your thumb in the direction of the
current flow, the curl of your fingers gives the
direction of the magnetic field. For all positive
b z is in the −x direction.
values of z , B
13
Integrating both sides of the equation above
yields
Z
t
I dt =
Z
t
N
t0
B
t0
=
Z
−B
A dB
dt
R dt
N
A
dB
R
A
∆B
R
A
= N 2B.
R
=N
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
The left hand side of the above equation is
just the charge flowing through the R during
this period of time! So,
Z t
I dt
Q=
t0
A
2B
R
= 2.47727 C .
=N
022 (part 1 of 2) 10.0 points
In the arrangement shown in the figure,
the resistor is 9 Ω and a 8 T magnetic field
is directed into the paper. The separation
between the rails is 1 m . Neglect the mass of
the bar.
An applied force moves the bar to the left
at a constant speed of 9 m/s . Assume the
bar and rails have negligible resistance and
friction.
9 m/s
m≪1 g
9Ω
1m
I
8T
14
From Ohm’s law, the current flowing through
the resistor is
I=
E
72 V
=
= 8 A.
R
9Ω
Thus, the magnitude of the force exerted on
the bar due to the magnetic field is
FB = I ℓ B
= (8 A)(1 m)(8 T)
= 64 N .
To maintain the motion of the bar, a force
must be applied on the bar to balance the
magnetic force
F = FB = 64 N .
023 (part 2 of 2) 10.0 points
At what rate is energy dissipated in the resistor?
Correct answer: 576 W.
Explanation:
The power dissipated in the resistor is
P = I2 R
= (8 A)2 (9 Ω)
8T
= 576 W .
Calculate the applied force required to
move the bar to the left at a constant speed
of 9 m/s.
Correct answer: 64 N.
Explanation:
Motional emf is
E = Bℓv.
Magnetic force on current is
~ = I ~ℓ × B
~ .
F
Ohm’s Law is
024 10.0 points
A coil is suspended around an axis which is
co-linear with the axis of a bar magnet. The
coil is connected to a resistor with ends labeled a and b. The bar magnet moves from
left to right with North and South poles labeled in the figure.
Use Lenz’s law to answer the following question concerning the direction of induced currents and magnetic fields.
V
.
R
The motional emf induced in the circuit is
I=
E = Bℓv
= (8 T) (1 m) (9 m/s)
= 72 V .
N
a
R
b
S
v
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
What is the direction of the induced magentic field in the coil and the direction of the
induced current in the resistor R when the
bar magnet is moving left to right?
1. the induced field is zero tesla and current
flows from b through R to a (←− I)
2. Induced magnetic field is right to left
(⇐= Binduced ) and current flows the induced
current is zero amperes
magnet winds around the solenoid from terminal b counter-clockwise.
Since the induced field is right to left
(⇐= Binduced ), the induced current must flow
clockwise and therefore it goes from b through
R to a (←− I).
025 10.0 points
Consider the following circuit. After leaving
the switch at the position “a” for a long time,
move the switch from “a” to “b”. There will
be current oscillations.
3. Induced magnetic field is left to right
(Binduced =⇒) and current flows from a
through R to b (I −→)
L
C
4. Induced magnetic field is left to right
(Binduced =⇒) and current flows from b
through R to a (←− I)
5. Induced magnetic field is right to left (⇐=
Binduced ) and current flows from a through R
to b (I −→)
6. the induced field is zero tesla and current
flows from a through R to b (I −→)
7. the induced field is zero tesla and current
flows the induced current is zero amperes
8. Induced magnetic field is left to right
(Binduced =⇒) and current flows the induced
current is zero amperes
9. Induced magnetic field is right to left (⇐=
Binduced ) and current flows from b through R
to a (←− I) correct
Explanation:
The induced magnetic field depends on
whether the flux is increasing or decreasing.
The magnetic flux through the coil is from
right to left. When the magnet moves from
left to right, the magnetic flux through the
coils decreases.
The induced current in the coil must produce an induced magnetic field from right
to left (⇐= Binduced ) to resist any change of
magnetic flux in the coil (Lenz’s Law).
The helical coil when viewed from the bar
15
E
S b
a
R
The maximum current will be given by
r
L
1. Imax = E
C
r
1
2. Imax = E
LC
√
E
LC
3. Imax =
R
E
4. Imax =
R
r
C
correct
5. Imax = E
L
r
E
C
6. Imax =
R L
E
7. Imax = √
RL
r
L
E
8. Imax =
R C
√
9. Imax = E L C
EC
10. Imax = √
RL
Explanation:
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
The energy stored in a RCL circuits is
Find the distance of the focal point from
the center of the lens.
Correct answer: 23.7333 cm.
1
1 2
2
L Imax
=
q
2
2 C max
1
= C E2
2
r
C
⇒ Imax =
E.
L
Explanation:
026 10.0 points
Light goes from flint glass into ethanol. The
angle of refraction in the ethanol is 15 ◦ , the
index of refraction for flint glass is 1.61, and
the index of refraction for ethanol is 1.36.
What is the angle of incidence in the glass?
Correct answer: 12.6286 ◦.
1
h′
q
1 1
+ =
M=
=−
p q
f
h
p
Convergent Lens
f >0
f >p> 0
n2 sin β
α = arcsin
n1
1.36 sin 15◦
= arcsin
1.61
f=
027 10.0 points
A convergent lens forms a virtual image 1.78
times the size of the object. The object distance is 10.4 cm .
h′
h
Scale: 10 cm =
(1)
(2)
Substituting these values into the mirror
equation
= 12.6286◦ .
p
q
= 1.78 .
p
q = −M p
= − (1.78) (10.4 cm)
= −18.512 cm .
n2 sin β
n1
q
M =−
Solving for q, we have
n1 sin α = n2 sin β
f
−∞ < q < 0 ∞ > M > 1
Note: The focal length for a convergent
lens is positive.
Explanation:
By Snell’s Law for an angle of incidence α
and an angle of refraction β
sin α =
16
f
=
1
(3)
1 1
+
p q
1
1
1
+
10.4 cm −18.512 cm
= 23.7333 cm .
028 10.0 points
Three polarizing disks whose planes are
parallel are centered on a common axis. The
directions of the transmission axes relative
to the common vertical direction are shown.
A linearly polarized beam of light with the
plane of polarization parallel to the vertical
reference direction is incident from the left
on the first disk with intensity Ii = 8.2 units
(arbitrary).
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
26◦
◦
52.8
79.8◦
Ii
I1
I2
If
Calculate the transmitted intensity If when
θ1 = 26◦ , θ2 = 52.8◦ , and θ3 = 79.8◦ .
Correct answer: 4.18983.
Explanation:
Given : Ii
θ1
θ2
θ3
= 8.2 ,
= 26◦ ,
= 52.8◦ ,
= 79.8◦ .
and
17
029 10.0 points
A binary star system in the constellation
Orion has an angular separation between the
two stars of 9.53 × 10−6 rad.
If the wavelength is 715 nm, what is the
smallest diameter a telescope can have and
just resolve the two stars?
Correct answer: 9.1532 cm.
Explanation:
From the formula
λ
,
D
where D is the diameter of the telescope, we
infer that the smallest diameter the telescope
needs to resolve the two stars is
λ
D = 1.22
θ
715 nm
= 1.22
9.53 × 10−6 rad
θ = 1.22
= 0.091532 m = 9.1532 cm .
We use the formula
I = Ii cos2 θ ,
for the intensity of the transmitted (polarized)
light three times. After passing through the
first polarizer, the intensity of the transmitted
light is
I1 = Ii cos2 θ1
= (8.2) cos2 26◦
= 6.62421 .
After the second polarizer, the intensity becomes
I2 = I1 cos2 (θ2 − θ1 )
= (6.62421) cos2 (52.8◦ − 26◦ )
= 5.27757 .
and finally, the transmitted intensity If will
be
2
If = I2 cos (θ3 − θ2 )
= (5.27757) cos2 (79.8◦ − 52.8◦ )
= 4.18983 .
030 (part 1 of 2) 10.0 points
In a series RLC ac circuit, the resistance is
15.7 Ω, the inductance is 17 mH, and the
capacitance is 20.9 µF. The maximum potential is 203 V, and the angular frequency is
636.62 rad/s.
Calculate the maximum current in the circuit.
Correct answer: 3.06539 A.
Explanation:
Let : ω = 636.62 rad/s ,
C = 20.9 µF = 2.09 × 10−5 F ,
L = 17 mH = 0.017 H ,
R = 15.7 Ω , and
Vmax = 203 V .
The capacitive reactance is
XC =
1
ωC
1
(636.62 rad/s) (2.09 × 10−5 F)
= 75.1577 Ω ,
=
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
and the inductive reactance is
18
is located at the first single-slit diffraction
minimum.
XL = ω L
= (636.62 rad/s) (0.017 H)
= 10.8225 Ω .
The maximum current is
R2 + (XL − XC )2
203 V
(15.7 Ω)2 + (10.8225 Ω − 75.1577 Ω)2
= 3.06539 A .
031 (part 2 of 2) 10.0 points
What is the power factor for the circuit?
Correct answer: 0.237077.
Explanation:
The phase angle between the voltage and
the current in the circuit is
−1 XL − XC
.
φ = tan
R
Since
S2
L
d
Determine the ratio ; i.e., the slit separaa
tion d compared to the slit width a.
1.
2.
3.
4.
XL − XC
10.8225 Ω − 75.1577 Ω
=
R
15.7 Ω
= −4.09778 ,
the power factor is
−1 XL − XC
cos φ = cos tan
R
−1
= cos tan (−4.09778)
= 0.237077 .
032 10.0 points
Hint: Use a small angle approximation; e.g.,
sin θ ≈ tan θ ≈ θ and cos θ ≈ 1 .
Consider the setup of double-slit experiment
in the schematic drawing below.
Note: As can be seen in the figure below,
one of the double-slit interference minima
θ
viewing screen
=p
S1
Vmax
d
=p
y
Vmax
Z
a
Imax =
5.
6.
7.
8.
9.
10.
d
a
d
a
d
a
d
a
d
a
d
a
d
a
d
a
d
a
d
a
=3
=
7
correct
2
=5
=
5
2
=4
=2
13
2
9
=
2
=
=6
=
11
2
Explanation:
At y there is a minimum for single-slit
diffraction and a minima for double-slit interference, as noted in the question.
The first minimum for single-slit diffraction
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
19
occurs when
λ
,
(1)
a
and the minima for double-slit interference
occur when
1 λ
sin θ = m +
.
(2)
2 d
The first diffraction minimum for single-slit
diffraction and the third double-slit interference minimum (m = 3) occur at the same
position y, as seen in the figure below.
sin θ =
033 10.0 points
The emf E drives the circuit shown below
at an angular velocity ω.
E
R
When does the light bulb (with resistance
R) glow most brightly?
1. higher frequencies, √
3
L
C
1
<ω<∞
LC
2. steady DC voltage, ω = 0 correct
1
LC
1
4. the resonant frequency, ω = √
LC
1
5. both higher frequecies, √
< ω < ∞,
LC
1
or lower frequencies, 0 < ω < √
LC
3. lower frequencies, 0 < ω < √
0
m
Figure: The dashed curve on the
left of the screen is due to single
slit interference. The dashed curve
on the right of the screen is due to
double slit interference. The screen
position is zero amplitude and the
positive direction is reflected on either side of the screen.
Since the single-slit diffraction minimum
masks the third double-slit interference minima, one must estimate where the third
double-slit minima occurs using the spacing
between the double-slit interference pattern
shown on the right-hand side of the viewing
screen, as seen in the figure above.
Using Eq. 1 and 2, we have
sin θ
1
1 1
= = m+
λ
a
2 d
1
d
= 3+
a
2
7
= .
2
Explanation:
The voltage across Cq
is E = IC ZC , where
the impedance is ZC = XC2 .
1
Since XC =
, the current through the
ωC
capacitor is
IC = s E
1
ωC
2 .
As the angular velocity ω decreases, the current IC will also decrease.
The voltage across R and L is E q
= IRL ZRL ,
where the impedance is ZRL = R2 + XL2 .
Since XL = ω L , the current through the
inductor is
E
IRL = q
.
2
2
R + (ω L)
Kapoor (mk9499) – oldfinal0 6 − −T urner − −(60230)
Ilight bulb
As the angular velocity ω decreases, the current IRL will increase. The limiting case is
constant voltage.
Since the light bulb is in series with the
inductor, it will glow most brightly at a constant voltage; i.e., steady DC voltage, ω = 0.
ω0 =
ω0
R
L
ω
20