oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
−3.7 C
Question 1, chap 22, sect 5.
part 1 of 1
10 points
q
What will happen if the charge is moved a
little away from the middle?
1. The charge will remain stationary.
2. The charge will move farther away from
the middle. correct
̂
ı̂
4m
Imagine a charge in the middle between
two parallel plate conductors. There is no net
charge on the plates, and the plates are not
connected to ground.
1
60◦
P
−3.7 C
−3.7 C
Find the magnitude of the electric field vec~ at P .
tor kEk
Correct answer: 2.77115×109 N/C (tolerance
± 1 %).
Explanation:
Let : a = 4 m ,
q = −3.7 C , and
k = 8.9875 × 109 N · m2 /C2 .
q
3. All of these can happen, depending on the
size of the charge.
̂
ı̂
a
4. The charge will return to the middle.
5. There is not enough information to tell.
q
P
q
Electric field vectors due to bottom two
charges cancel out each other. The magnitude
of the field vector due to charge at to top of
the triangle, which gives
Explanation:
There will be an image-charge attracting it
toward each metal surface. The charge will
move toward the plate closest to it because
the closest image-charge will be the strongest.
Any charge (free to move) will move toward
the closest conductor it can find.
Question 2, chap 23, sect 1.
part 1 of 2
10 points
Three point charges are placed at the vertices of an equilateral triangle.
The value of the Coulomb constant is
8.9875 × 109 N · m2 /C2 .
~ =
kEk
=
4 kq
kq
√ !2 = 3 a2
3
a
2
4 (8.9875 × 109 N · m2 /C2 ) (−3.7 C)
3
(4 m)2
= 2.77115 × 109 N/C ,
√
3
a is the height of
where h = a cos(30 ) =
2
the triangle.
◦
Question 3, chap 23, sect 1.
part 2 of 2
10 points
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
2
~ at
Find the direction of the field vector E
P.
1
1. √ (ı̂ + ̂)
2
2. −̂
(c)
1
3. − √ (ı̂ − ̂)
2
Which electrostatic field patterns are physically possible?
4. −ı̂
1. (a) and (c)
5. ̂ correct
2. (b) and (c)
6. ı̂
3. (a) only
1
7. − √ (ı̂ + ̂)
2
1
8. √ (ı̂ − ̂)
2
4. (b) only correct
5. (a) and (b)
6. (c) only
Explanation:
~ at P due to q = −3.7 C is
By inspection, E
along ̂ direction.
Question 4, chap 23, sect 3.
part 1 of 1
10 points
The diagrams below depict three electric
field patterns. Some of these patterns are
physically impossible. Assume these electric
field patterns are due to static electric charges
outside the regions shown.
Explanation:
Electrostatic lines of force do not intersect
one another. Neither do they form a closed
circuit (unless there is a changing magnetic
field present).
Question 5, chap 23, sect 2.
part 1 of 3
10 points
In an early model for the atom the nucleus
of charge q was surrounded by a circular ring
of radius a of negative charge −q.
Find the magnitude of the force on the
nuclear
√ charge if it is displaced the distance
x = 2 a along the axis of the ring.
1. F =
(a)
(b)
1 ke q 2
3 a2
2. None of these
1 ke q 2
√
3. F =
2 a2
ke q 2
4. F = 2
a
1 ke q 2
5. F =
2 a2
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
3
√
2 ke q 2
6. F = √
correct
1. restore the nucleus to the center of the
3 3 a2
ring. correct
7. F = 0
√
2. repel the nucleus from the center of the
3 ke q 2
ring.
8. F = √
2 2 a2
1 ke q 2
3. be undetermined in direction.
9. F = √
2
a
3
Explanation:
2 ke q 2
The force between the negatively charged
10. F = √
3 3 a2
ring and the positively charged nucleus is attractive.
Explanation:
By a simple integration, we know that the
Question 7, chap 23, sect 2.
electric field from the ring in the yz plane with
part 3 of 3
10 points
radius a is expressed as
dE = ke
dq
r2
Find the minimum energy required to
move the nucleus from the equilibrium position to infinity for this model for hydrogen with charge 1.602 × 10−19 C and radius
0.11 nm? The value of the Coulomb constant
is 8.98755 × 109 N · m2 /C2 . 1 electron Volt =
1.602 × 10−19 C .
Correct answer: 13.0891 eV (tolerance ± 1
%).
dEy = 0
dEx = dE cos θ
dq x
= ke 2
r r
x
dq .
= ke
3/2
(x2 + a2 )
Therefore
E = Ex =
Z
= ke
= ke
ke
Explanation:
x
3/2
(x2 + a2 )
Z
x
(x2 + a2 )
x
(x2
+
3/2
3/2
a2 )
dq
dq
a = 0.11 nm = 1.1 × 10−10 m ,
9
ke x q 2
F =− 2
(x + a2 )3/2
√
ke 2 a q 2
=− √
[( 2 a)2 + a2 ]3/2
√
2 ke q 2
,
=− √
3 3 a2
√
where x = 2 a, as given in the problem.
Question 6, chap 23, sect 2.
part 2 of 3
10 points
2
and
2
ke = 8.98755 × 10 N · m /C .
q.
Therefore, the force on the nucleus is
This force will
Let : q = 1.602 × 10−19 C ,
The required minimum energy is
−
Z
0
∞
F dx = −
Z
∞
0
q2
ke x q 2
− 2
dx
(x + a2 )3/2
ke
a
= 8.98755 × 109 N · m2 /C2
2
1.602 × 10−19 C
×
1.1 × 10−10 m
=
= 2.09688 × 10−18 J .
Question 8, chap 24, sect 3.
part 1 of 1
10 points
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
Consider oppositely charged concentric
spherical shells as shown. The charges are uniformly distributed throughout the two shells.
Denote the electric fields in regions I, II and
III by EI , EII and EIII .
I
II
4
Two parallel wires carry opposite current
as
shown.
i2
i1
+Q
Find the direction of the magnetic force on
i2 due to the magnetic field of i1 .
−Q
III
What is true about the fields?
1. to the left and upward
2. into the paper
1. EI 6= 0, EII 6= 0, EIII = 0
3. to the right and downward
2. EI = 0, EII 6= 0, EIII = 0 correct
4. to the right correct
3. EI = 0, EII 6= 0, EIII 6= 0
5. out of the paper
4. EI 6= 0, EII = 0, EIII = 0
6. to the right and upward
5. EI = 0, EII = 0,EIII 6= 0
7. to the left and downward
6. EI 6= 0, EII = 0, EIII 6= 0
8. to the left
Explanation:
By Gauss’ law,
~ · dA
~ = Q.
E
ǫ0
S
For spherical charge distribution we have
ΦS = E (4 π r 2 ), so
ΦS =
I
Q
.
4 π r 2 ǫ0
In region I, qI = 0, so E = 0.
Similarly, since qII = −Q 6= 0 in region II ,
EII 6= 0 .
In region III,
Explanation:
i1
i2
E=
qIII = qI + qII
= (−Q) + (+Q)
= 0 , so
EIII = 0 .
Question 9, chap 29, sect 1.
part 1 of 1
10 points
B1
F2
Wire 1, which carries a current i1 , sets up a
~ 1 , which points into the paper
magnetic field B
~1
at the position of wire 2. The direction of B
is perpendicular to the wire, as illustrated
below. The magnetic force on a length l
~ 2 = i2 ~l × B
~ 1 . Since i2 flows
of wire 2 is F
~ 1 is to the right.
downward, ~l × B
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
Question 10, chap 30, sect 1.
part 1 of 1
10 points
A proton is moving in a circle in a magnetic
field of magnitude 6.6 T.
The proton has a charge of 1.60218 ×
10−19 C and a mass of 1.67262 × 10−27 kg.
Calculate the proton’s angular frequency.
Correct answer: 6.32203 × 108 rad/s (tolerance ± 1 %).
Explanation:
5
µ0 I
4
µ0 I
2. B =
2
1. B =
3. B = µ0 n I
4. B = 2 µ0 I
5. B =
µ0 n I
4
6. B = 2 µ0 n I
Let : B = 6.6 T ,
m = 1.67262 × 10−27 kg ,
|q| = 1.60218 × 10−19 C .
7. B = 4 µ0 I
and
The centripetal force is
m ω2 r = q B v .
v = ω r, so the cyclotron frequency is
qB
m
(1.60218 × 10−19 C) (6.6 T)
=
1.67262 × 10−27 kg
ω=
= 6.32203 × 108 rad/s .
Question 11, chap 29, sect 3.
part 1 of 1
10 points
A conductor consists of an infinite number
of adjacent wires, each infinitely long and
carrying a current I (whose direction is out-ofthe-page), thus forming a conducting plane.
A
C
If there are n wires per unit length, what is
~
the magnitude of B?
8. B = 4 µ0 n I
9. B =
µ0 n I
correct
2
10. B = µ0 I
Explanation:
B
A
l
W
C
B
By symmetry the magnetic fields are equal
and opposite through point A and C and horizontally oriented. Following the dashed curve
in
I a counter-clockwise direction, we calculate
~ · d~s, which by Ampere’s law is proporB
tional to the current through the dashed loop
coming out of the plane of the paper. In
this problem this is a positive current. Hence
~ along the horizontal legs points in the diB
rection in which we follow the dashed curve.
Ampere’s Law is
I
~ · d~s = µ0 I .
B
To evaluate this line integral, we use the rectangular path shown in the figure. The rectangle has dimensions l and w. The net current
through the loop is n I l. Note that since there
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
~ in the direction of w, we
is no component of B
are only interested in the contributions along
sides l
I
~ · d~s = 2 B l = µ0 n l I
B
µ0 n I
B=
.
2
Question 12, chap 29, sect 4.
part 1 of 1
10 points
A solenoid 111.9 cm long has 480 turns and
a radius of 1.12 cm.
If it carries a current of 2.93 A, find the
magnetic field along the axis at its center.
Correct answer: 0.00157939 T (tolerance ± 1
%).
Explanation:
Let : l = 111.9 cm = 1.119 m ,
N = 480 turns ,
I = 2.93 A , and
µ0 = 1.25664 × 10−6 N/A2 .
Since l >> R, the magnetic field at the center
is approximately
B ≈ µ0 n I
N
= µ0 I
l
= (1.25664 × 10−6 N/A2 )
480 turns
×
(2.93 A)
1.119 m
= 0.00157939 T .
Question 13, chap 31, sect 2.
part 1 of 1
10 points
A plane loop of wire of area A is placed in a
region where the magnetic field is perpendicular to the plane. The magnitude of B varies in
time according to the expression B = B0 e−at .
That is, at t = 0 the field is B0 , and for t > 0,
the field decreases exponentially in time.
Find the induced emf, E, in the loop as a
function of time.
6
1. E = a A B0 e−at correct
2. E = A B0 e−at
3. E = a A B0
4. E = a B0 e−at
5. E = a B0 t
6. E = a A B0 e−2at
Explanation:
Basic Concepts: Faraday’s Law:
I
d ΦB
E ≡ E · ds = −
dt
Solution: Since B is perpendicular to the
plane of the loop, the magnetic flux through
the loop at time t > 0 is
ΦB = B A
= A B0 e−a t
Also, since the coefficient AB0 and the parameter a are constants, and Faraday’s Law
says
d ΦB
E=−
dt
the induced emf can be calculated the from
Equation above:
d ΦB
E =−
dt
d
= −A B0 e−a t
dt
= a A B0 e−a t
That is, the induced emf decays exponentially
in time.
Note: The maximum emf occurs at t = 0 ,
where E = a A B0.
B0
B = B0 e−at
0
~t
0
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
The plot of E versus t is similar to the B
versus t curve shown in the figure above.
Question 14, chap 31, sect 3.
part 1 of 1
10 points
A bar of negligible resistance and mass m in
the figure below is pulled horizontally across
frictionless parallel rails, also of negligible resistance, by a massless string that passes over
an ideal pulley and is attached to a suspended
mass M .
The uniform magnetic field has a magnitude B, and the distance between the rails is
ℓ. The rails are connected at one end by a
load resistor R.
The acceleration of gravity is g.
m
ℓ
B
B
R
M
B
B
B
B
B
M g R2
ℓ B2
M gR
10. v∞ =
ℓ B2
Explanation:
Basic Concepts:
9. v∞ =
~ g = M ~g
F
~ net
F
1. v∞ =
d ΦB
dt
~
~
ΦB = B · A
E = Bℓv
Solution: It follows from Lenz’s law that the
magnetic force opposes the motion of the bar.
When the wire acquires steady-state speed,
the gravitational force Fg is counter-balanced
by the magnetic force Fm (see figure below):
m
a
2. v∞ =
3. v∞ =
4. v∞ =
5. v∞ =
6. v∞ =
7. v∞ =
8. v∞ =
T
ℓ
Fm
R
T
B
a
M 2 g 2 R2
ℓ2 B 2
M gR
ℓ2 B
M 2 g2 R
ℓB
M gR
ℓB
M g R2
ℓB
M gR
correct
ℓ2 B 2
M g R2
ℓ2 B 2
M g R2
ℓ2 B
~ m = I ~ℓ × B
~
F
~g − F
~m
= (M + m) ~a = F
E =IR=−
a
What is the magnitude of the terminal velocity (i.e., the eventual steady-state speed
v∞ ) reached by the bar?
7
M
Fg
B
B
B
Fg = M g = Fm = ℓ I B
B
(1)
Mg
.
(2)
ℓB
To find the induced current, we use Ohm’s law
dΦ
and substitute in the induced emf, E = −
dt
I=
I=
|E|
1 dΦ
=
.
R
R dt
(3)
Note: We have ignored the minus sign from
the induced emf E because we will eventually evaluate the magnitude of the terminal
velocity. The flux is Φ = B A . So
|E| =
dA
dΦ
=B
= Bℓv,
dt
dt
and
(4)
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
Bℓv
.
(5)
R
Using Eqs. 2 and 5 and noting that v is the
terminal velocity v∞
I=
B ℓ v∞
Mg
=
.
ℓB
R
4.
5.
(6)
6.
Solving for the magnitude of the terminal
velocity v∞
7.
v∞ =
(7)
Question 15, chap 28, sect 4.
part 1 of 1
10 points
Four identical light bulbs are connected either in series (circuit A), or in a parallel-series
combination (circuit B), to a constant voltage
battery with negligible internal resistance, as
shown.
Circuit A
E
=8
=4
=2
=1
1
16
1
9.
=√
8
Explanation:
In circuit A, the equivalent resistance is
RA = 4 R, so the electric current through
each bulb is
V
iA =
4R
and the power of each bulb is
2
V
V2
2
PA = I R =
.
R=
4R
16 R
8.
M gR
.
ℓ2 B 2
PA
PB
PA
PB
PA
PB
PA
PB
PA
PB
PA
PB
8
=
Thus the total power consumed by all four
bulbs in circuit A is
PA,T otal
V2
.
= 4 PA =
4R
In circuit B, the equivalent resistance is
1
1
1
1
=
+
=
RB
2R 2R
R
Circuit B
E
Assuming the battery has no internal resistance and the resistance of the bulbs is
temperature independent, what is the ratio of
the total power consumed bycircuit A to that
PA,T otal
consumed by circuit B; i.e.,
?
PB,T otal
P
1. A = 16
PB
1
P
2. A = correct
PB
4
P
1
3. A =
PB
2
RB = R ,
so the electric current through each bulb is
iB =
V
2R
and the power of each bulb is
2
V
V2
2
PB = I R =
.
R=
2R
4R
Thus the total power consumed by all four
bulbs in circuit B is
PB,T otal
V2
= 4 PB =
R
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
and
C
R1
PA,T otal
1
PA
= .
=
PB,T otal
PB
4
9
V0
Figure (a)
Question 16, chap 28, sect 7.
part 1 of 2
10 points
For a long period of time the switch S is
in position “b”. At t = 0 s, the switch S is
moved from position “b” to position “a”.
4 MΩ
For an RC circuit,
I = I0 e−t/(R C)
V0
e−t/(R1 C) .
=
R1
Because I R1 = VR1 , we have
VR1 = V0 e−t1 /(R1 C)
6
2 µF
2 MΩ
= (20 V) e(−5 s)/[(2×10
S b
Ω)(2×10−6 F)]
= 5.7301 V .
a
20 V
Question 17, chap 28, sect 7.
part 2 of 2
10 points
Find the voltage across the 2 MΩ center-left
resistor at time t1 = 5 s.
Correct answer: 5.7301 V (tolerance ± 1 %).
Explanation:
R2
C
R1
V0
Much later, at some time t′0 = 0 s, the
switch is moved from position “a” to position
“b”.
Find the voltage across the 2 MΩ center-left
resistor at time t′ = 2.3 s.
Correct answer: 5.50388 V (tolerance ± 1
%).
Explanation:
Now the switch moves to position “b”,
thereby excluding the battery from the circuit, see Figure (b).
S b
a
R2
R1
Let : R1
R2
C
V0
= 2 MΩ = 2 × 106 Ω ,
= 4 MΩ = 4 × 106 Ω ,
= 2 µF = 2 × 10−6 F ,
= 20 V .
C
Figure (b)
and
When the switch is closed to position “a”,
the resistor R2 is not part of the circuit and
may be disregarded, see Figure (a).
Note: The equivalent resistance of the circuit is
Req = R1 + R2
= 2 × 106 Ω + 4 × 106 Ω
= 6 × 106 Ω
because R1 and R2 are in series.
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
10
Again, see Figure (b)
Let : R = 3 Ω ,
2R = 6 Ω,
L = 1.1 H ,
E = 31 V .
′
I = I0 e−t /(Req C)
V0 −t′ /(Req C)
=
e
Req
because the capacitor has an initial potential
across it of V0 . Thus,
VR1 = I R1
′
R1
e−t /(Req C)
= V0
Req
Kirchoff’s Laws then give
I1 = I2 + I3 ;
2 × 106 Ω
= (20 V)
6 × 106 Ω
6
× e(−2.3 s)/[(6×10
from the bottom loop
V − I1 R − I2 R = 0 ;
d I3
= 0 . (3)
dt
Using Eq. 1 to eliminate I2 in Eq. 2, we obtain
Ω)(2×10−6 F)]
V − I1 R − I1 R + I3 R = 0
=⇒ I1 (2 R) = I3 R + V .
Question 18, chap 31, sect 6.
part 1 of 1
10 points
Prior to t = 0, the switch in the circuit is
open.
3Ω
6Ω
1.1 H
31 V
3Ω
If the switch is closed at t = 0, find the
current in the inductor at t = 0.176 s.
Correct answer: 1.4442 A (tolerance ± 1 %).
Explanation:
R
I2
L
I3
R
d I3
(I3 R + V ) − I3 (2 R) − L
= 0.
2R
dt
(3′ )
which simplifies to this differential equation:
V −
d I3
+ 5 R I3 = V .
dt
This is the differential equation for a simple
LR circuit with inductance 2 L and resistance
5 R . The the time constant for this circuit is
2L
τ =
, and the current through the induc5R
tor is
V 1 − e−t/τ
Iind = I3 (t) =
5R
i
V h
=
1 − e−5 R t/2 L
5R
(31 V)
1
=
5 (3 Ω)
5 (3 Ω) (0.176 s)
− exp −
2 (1.1 H)
= 1.4442 A .
S
E
I1
(2′ )
Substituting this into Eq. 3 gives
2L
2R
(2)
V − I1 R − I3 (2 R) − L
= 5.50388 V .
S
(1)
and from the outer loop
For the given values
VR1
and
R
Question 19, chap 32, sect 4.
part 1 of 2
10 points
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
In the LC circuit below the capacitor is
charged to its maximum 50 µC while the
switch S is open. The switch is closed at
t = 0.
3 µF
2π T
×
T
4
= −q0 ω
= −0.0166667 A .
Thus
1
L I2
2
1
= L (−q0 ω)2
2
= ULmax
1 2
= UCmax =
q
2C 0
1
(5 × 10−5 C)2
=
−6
2 (3 × 10 F)
106 µJ
×
J
= 416.667 µJ .
UL =
Find the energy stored in the inductor at
T
t = , where T is the period of circuit oscil4
lations.
Correct answer: 416.667 µJ (tolerance ± 1
%).
Explanation:
L
Q
S
Let : C = 3 µF = 3 × 10−6 F ,
L = 3 H , and
Q0 = 50 µC = 5 × 10−5 C .
Basic concept: Oscillation in LC circuit:
q = qmax cos(ω t + δ)
dq
= −ω qmax sin(ω t + δ)
dt
2π
1
, where T is the period
with ω =
=√
T
LC
of oscillation.
Solution: When the switch closes, initially
there is zero current in the circuit. Thus the
solution has the correct phase at t = 0
I=
I=
T
,
4
I = −q0 ω sin
3H
Q
S
C
Now at t =
11
dQ
= −ω qmax sin(ω t) .
dt
T
Observe that at t = , the magnitude of
4
the current flow through the inductor is at a
maximum, and the charge q = qmax cos ω t is
T
π
zero at t =
=
. Thus, at this moment
4
2ω
there is no energy stored in the capacitor and
all the energy is stored in the inductor. At
any other time, one can show that for an
ideal LC circuit where there is no dissipation
from resistance, the total energy stored is the
sum of the energy in the capacitor and the
energy in the inductor, and this energy is
constant in time (although the energy in each
one oscillates).
Question 20, chap 32, sect 4.
part 2 of 2
10 points
Find total stored energy in the circuit at
T
t= .
12
Correct answer: 416.667 µJ (tolerance ± 1
%).
Explanation:
The total energy does not change with respect to time, so
Utotal = ULmax = UCmax = 416.667 µJ .
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
12
whose solution is
Question 21, chap 32, sect 4.
part 1 of 2
10 points
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 = 0 the switch is set to
position a.
C
L
c
E
S b
a
R
What is the current immediately after the
switch is set to position a at t = 0?
1. I0 = ∞
E −t/τL
I(t) =
.
1−e
R
Hence at t = 0 we have I = 0. The inductor
has a magnetic inertia. When S is at position b, i = 0 and there is no magnetic field in
L. Immediately after S is switched to a, the
magnetic field in L continues to be zero temporarily i.e. i continues to be 0 temporarily.
Question 22, chap 32, sect 4.
part 2 of 2
10 points
After being at the position a for a long
time, reset the clock. At a new time t = 0,
the switch is moved from a to b. If T is
the period of the LC circuit, the relationship
between Vb and Vc and the direction of the
current through the inductor, both evaluated
3
at a time t = T , are given by
4
1. Vc > Vb ; no current flow correct
E
2. I0 = √
2R
E
1 − e−1
3. I0 =
R
EL
4. I0 =
1 − e−1
R
2. Vb > Vc ; from c through L to b
3. Vc > Vb ; from b through L to c
4. Vb = Vc ; from b through L to c
5. I0 = 0 correct
5. Vc > Vb ; from c through L to b
E
R
6. Vb = Vc ; from c through L to b
6. I0 =
7. Vb = Vc ; no current flow
7. I0 = E R
EL
R
E
9. I0 = e−1
R
EL −1
10. I0 =
e
R
8. Vb > Vc ; from b through L to c
8. I0 =
9. Vb > Vc ; no current flow
Explanation:
The differential equation for an LC circuit
is given by
E −IR−L
dI
= 0,
dt
Explanation:
Let’s track the simple harmonic oscillation.
At t = 0 the current is at its maximum value
flowing down through the inductor; the capacitoris uncharged. At one quarter of the
1
period i.e., t = T , the right plate is fully
4
1
charged and the current is zero. At t = T
2
the right plate has zero charge and the current
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
flowing up through the inductor is maximum.
3
At t = T , the left plate of the capacitor is
4
fully charged and the current is zero.
3
Hence at t = T , I = 0 and Vc > Vb .
4
Question 23, chap 32, sect 4.
part 1 of 1
10 points
A circuit containing a resistor, a capacitor,
and an inductor is shown below.
a
R
L
C
b
Select the plot commensurate with the
above circuit in which the impedance from
a to b is plotted as a function of angular velocity ω .
5.
6.
Explanation:
ω
Zseries
1
+
R2
1
1
−
XC
XL
1
,
ωC
XL = ω L .
and
======================
a
R
L
C
b
ZLRC = s
R
2
R + XC2
2
+
1
1
XC
−
2
2
XL
R + XC

2






R


= 
2 



1


2

 R + ωC
2
2 −1/2


1





1


ω
C
+

2 −

ω L 
1



R2 +

ωC

ω
ω0 =
1
RC
ω
ω0
======================
a
R
L
C
ω
ω
2 ,
q
= R2 + (XL − XC )2 ,
XC =
correct
ω
1
Impedance
4.
Impedance Impedance
3.
Impedance
2.
Impedance Impedance
1. None of these
Zparallel = s
13
b
ZCRL = s
=
(
R
R2 + XL2
2
R
2
R + (ω L)2
+
2
1
XL
1
− 2
2
XC
R + XL2
2
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
ωL
2
R + (ω L)2
ω0 =
2 )−1/2
a
R
L
b
ω
ω0
======================
a
R
C
ZRCL = R + s
1
Impedance
2
1
1
−
XC
XL
"
2 #−1/2
1
=R+
ωC −
ωL
ω0 = √
1
LC
ω
ω0
======================
a
R
L
C
b
q
R2 + (XL − XC )2
"
2 #1/2
1
= R2 + ω L −
ωC
Impedance
ZRLC =
ω0 = √
ZRLC = s
1
LC
ω
ω0
======================
C
1
2
1
1
1
+
−
R2
XC
XL
"
2 #−1/2
1
1
+ ωC −
=
R2
ωL
L
b
L
R
Impedance
Impedance
+ (ω C)2 −
14
ω0 = √
ω0
1
LC
ω
Question 24, chap 32, sect 4.
part 1 of 1
10 points
Calculate the resonance frequency of a series RLC circuit for which the capacitance is
89 µF , the resistance is 55 kΩ , and the inductance is 190 mH .
Correct answer: 38.7033 Hz (tolerance ± 1
%).
Explanation:
Let :
R = 55 kΩ = 55000 Ω ,
L = 190 mH = 0.19 H , and
C = 89 µF = 8.9 × 10−5 F .
The resonance frequency is the frequency at
which the current becomes maximum, or the
impedance becomes minimum. This occurs
when
XL = XC
1
ωL =
.
ωC
From this condition, the resonance frequency
is given by
1
√
f=
2π LC
1
p
=
2 π (0.19 H) (8.9 × 10−5 F)
= 38.7033 Hz .
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
Question 25, chap 33, sect 3.
part 1 of 1
10 points
An electromagnetic wave in vacuum has an
electric field amplitude of 188 V/m.
The speed of light is 3 × 108 m/s.
Calculate the amplitude of the correspondent magnetic field.
Correct answer: 626.667 nT (tolerance ± 1
%).
Explanation:
15
of the spectrum, which in turn has more energy than a photon in the infrared part of the
spectrum.
Question 27, chap 33, sect 5.
part 1 of 1
10 points
A point light source delivers a power P .
It radiates light isotropically. A mirror is
placed at point B that is at distance r from
the source. The mirror has a cross section
A. Make the approximation that all the light
hitting the mirror comes in perpendicular to
it. The mirror is a totally reflecting surface.
Let : E = 188 V/m and
c = 3 × 108 m/s .
B
r
The relation between the magnitude of the
electric field and that of the magnetic field for
an electromagnetic wave is
E = cB,
so the amplitude of the magnetic field is
B=
E
188 V/m
109 nT
=
·
c
3 × 108 m/s
T
= 626.667 nT .
Question 26, chap 33, sect 5.
part 1 of 1
10 points
Which has the greatest energy, a photon of
infrared light, of visible light, or of ultraviolet
light?
Find the total force on the mirror
1. F =
2. F =
3. F =
4. F =
5. F =
2 π r2 P
c A2
A2 P
2 c π r2
4 π r2 P
cA
AP
correct
c 2 π r2
2AP
c/, r 2
6. F = π r 2 P c
1. visible light
2. ultraviolet light correct
3. a photon of infrared light
4. They all have the same energy.
Explanation:
E∝f
The higher-frequency ultraviolet photon has
more energy than a photon in the visible part
2P
c
2P
8. F =
c π r2
P
9. F =
c
P
10. F =
c2πr
7. F =
Explanation:
Since the mirror is small enough (or far
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
16
enough away) that we may approximate the
incoming light as perpendicular, the force is
simply
F = pA,
The angle must be in degrees.
where p is the radiation pressure. The mirror
is totally reflecting, so
A fiber optic cable (nf iber = 1.5) is submerged in water (nwater = 1.35).
What is the critical angle for light to stay
inside the cable?
Correct answer: 64.1581 ◦ (tolerance ± 1 %).
p=
2I
.
c
Intensity is defined by
Power hitting the mirror
.
Area of mirror A
A
P
P
4πr 2
I=
=
,
A
4 π r2
leading to a force of
I=
F = pA =
2I A
AP
=
c
c 2 π r2
Question 29, chap 34, sect 3.
part 1 of 1
10 points
Explanation:
Basic Concept:
Snell’s law for total internal reflection
n1 sin θc = n2 sin 90◦
where θc is the critical angle for total internal
reflection. For total internal reflection,
n1 sin θc = n2 sin 90◦
For our problem, it is
nf iber sin θc = nwater sin 90◦
on the mirror.
1.5 sin θc = 1.35 sin 90◦
Question 28, chap 34, sect 3.
part 1 of 1
10 points
Light goes from flint glass into ethanol. The
angle of refraction in the ethanol is 12.4◦ . The
index of refraction for flint glass is 1.61. The
index of refraction for ethanol is 1.36.
What is the angle of incidence in the glass?
Correct answer: 10.4508 ◦ (tolerance ± 1 %).
Explanation:
By Snell’s Law for an angle of incidence α
and an angle of refraction β
n1 sin α = n2 sin β
sin α =
n2 sin β
n1
n2 sin β
α = arcsin
n1
1.36 sin(12.4◦ )
= arcsin
1.61
= arcsin (0.181391)
= 10.4508◦ .
So,
θc = 64.1581 ◦ .
Question 30, chap 34, sect 5.
part 1 of 1
10 points
The distance between an object and its upright image is 11.6 cm.
If the magnification is 0.794, what is the
focal length of the lens being used to form the
image?
Correct answer: −217.042 cm (tolerance ± 1
%).
Explanation:
Given :
L = 11.6 cm
M = 0.794 .
The image is upright, so
q
M =−
p
q = −M p
and
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
r1
−1
an
r2 θ = t
S1
d
y
L
O
Q
S2
y
L=p+q
=p−Mp
= p (1 − M )
L
p=
1−M
P
S2 Q S1 ≈ 90◦
6
viewing
screen
The distance between image and object is
17
δ ≈ d sin θ ≈ r2 − r1
L
From the lens equation,
1
1
1
1
1
1−
= −
=
f
p Mq
p
M
1−M M −1
·
=
L
M
2
(1 − M )
=−
LM
(0.794) (11.6 cm)
ML
=−
f =−
2
(1 − M )
(1 − 0.794)2
π
π/2
0
2π
φ
1
3 π/2
2π
or 360◦
0
= −217.042 cm .
−1
Question 31, chap 35, sect 3.
part 1 of 1
10 points
Two narrow parallel slits are illuminated
with light of wavelength 480 nm.
δ
λ = 480 nm
0◦
90◦
180◦
270◦
360◦
δ
d sin θ
d
y
d y
φ
= =
= p
≈
2π
λ
λ
λ L2 + y 2
λ L
θ
15 mm
S1
viewing
screen
0.24 mm
Solution: We obtain
S2
δ = d sin θ
= d sin
(0.015 m)
p
(11 m)2 + (0.015 m)2
= (0.24 mm) sin(0.00136364 rad)
= 327.272 nm ,
= d sin
11 m
What is the phase difference between the
two interfering waves on a screen at a point
15 mm from the central bright fringe?
Correct answer: 245.454 ◦ (tolerance ± 1 %).
Explanation:
Let : L = 11 m
y = 15 mm
d = 0.24 mm .
y
p
2
L + y2
!
φ=
2 πd
y
p
λ
L2 + y 2
2 π (0.00024 m)
(4.8 × 10−7 m)
(0.015 m)
×p
(11 m)2 + (0.015 m)2
= 4.28399 rad
= 245.454 ◦ ,
=
!
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
in agreement with the above diagram.
1. t =
Question 32, chap 35, sect 3.
part 1 of 1
10 points
2. t =
Light falls on a pair of slits 0.00199 cm
apart. The slits are 79.5 cm from the screen.
The first-order bright line is 1.89 cm from the
central bright line.
What is the wavelength of the light?
Correct answer: 473.094 nm (tolerance ± 1
%).
Explanation:
For double slit interference,
4. t =
5. t =
6. t =
x
λ
=
L
d
7. t =
so that
8. t =
xd
λ=
L
Dimensional analysis of λ:
λ
2n
λ
8n
3λ
4n
3nλ
4
nλ
4
nλ
2
nλ
8
λ
correct
4n
9. t = n λ
cm · cm 1 m 109 nm
= nm
cm 102 cm 1 m
10. t =
Question 33, chap 35, sect 1.
part 1 of 1
10 points
A light ray is traveling in air with an index
of refraction of unity and it is reflected at the
boundary of a second medium with an index
of refraction n.
Consider the optical coating on a glass lens
where the index of refraction of the coating
is n, where n is greater than the index of
refraction of the air.
1 2
θ ≈ 0◦ θ
air
t
3. t =
18
n
lens
Assume: The index of refraction of the lens
is greater than that of the coating.
To minimize the reflection of a ray with a
wavelength 500 nm and incident angle θ ≈ 0,
what is the minimum nonzero thickness of the
coating?
λ
n
Explanation:
Destructive interference occurs when the
difference between the phase angle of the incident ray reflected from the outer surface of the
coating (ray 1) and the phase angle of the ray
reflected from the inner surface of the coating
(ray 2) are at π, 3 π, 5 π etc. The phase differences are due to the path difference of the two
rays and the change of their relative phases
due to reflections. Putting them together, it
gives
π, 3 π, 5 π... = (2 t) kn + |π − π| = 2 t kn .
2π
, it implies that the minimum
λn
λ
.
thickness is given by t =
4n
With kn =
Question 34, chap 35, sect 5.
part 1 of 1
10 points
Use a small angle approximation; sin θ =
tan θ .
Consider the setup of a single slit experiment.
×15
θ
S2
L
viewing
screen
a
S1
y6
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
Solution: The first minimum is at β = 2 π,
where β = 2 φ = 2 π, where φ = π is the
phase difference of the two rays for destructive
interference.
The sixth minimum occurs at β = 12 π,
which corresponds to a path difference δ between two end rays
12 π
= 2π
λ
= 6λ
δ
θ=
a
y6
=
L
δ
y6 = L
a
λL
=6
,
a
1. y6 =
where the minima are at
β
= π , 2 π , 3 π , 4 π , 5 π , 6 π , · · · , or
2
β = 2 π , 4 π , 6 π , 8 π , 10 π , 12 π , · · · ,
= 2mπ,
where m is the first, second, third, fourth, · · ·,
minimum in the diffraction pattern.
β
k
δ=
Determine the height y6 , where the sixth
minimum occurs.
9 λL
2 a
λL
2. y6 = 4
a
λL
3. y6 = 6
correct
a
λL
4. y6 = 8
a
15 λ L
5. y6 =
2 a
λL
6. y6 = 7
a
13 λ L
7. y6 =
2 a
λL
8. y6 = 5
a
17 λ L
9. y6 =
2 a
11 λ L
10. y6 =
2 a
Explanation:
Basic Concepts: Light Diffraction


β 2
sin
I

2
=
 ,
β
I0
2
19
where k ≡
2π
.
λ
Question 35, chap 33, sect 3.
part 1 of 1
10 points
An unpolarized light beam with intensity
of
I0
passes
through 2 polarizers shown in the picture.
Unpolarized
light
Polarizer
E0
Analyzer
θ
Transmission
axis
E 0 cos θ
Polarized
lihgt
If θ = 30◦ ,what is the beam intensity after
the second polarizer?
1
1. I2 = I0
8
3
2. I2 = I0 correct
8
3
3. I2 = I0
16
1
4. I2 = I0
2
oldmidterm 05 – JYOTHINDRAN, VISHNU – Due: Dec 11 2007, 11:00 pm
1
5. I2 = I0
4
5
6. I2 = I0
8
5
7. I2 = I0
16
9
8. I2 = I0
16
7
9. I2 = I0
16
1
10. I2 = I0
16
Explanation:
The beam intensity after the first polarizer
is
I0
I1 =
.
2
We use the formula for the intensity of the
transmitted (polarized) light. Thus the beam
intensity after the second polarizer is
I = I1 cos2 θ
I0
cos2 (30◦ )
=
2
3 I0
=
8
Question 36, chap 35, sect 4.
part 1 of 1
10 points
Monochromatic 489.5 nm light is incident
on a diffraction grating containing 4880 lines
per centimeter. Determine the angle of the
first-order maximum.
Correct answer: 13.8202 ◦ (tolerance ± 1 %).
Explanation:
Basic Concepts:
d sin θ = m λ (m = 0, 1, 2, 3, ...)
The condition for intensity maxima in the
interference pattern of a diffraction grating
for normal incidence is
d sin θ = mλ .
The spacing d between adjacent slits is given
by
d = 1/n = 2.04918 × 10−6 m
20
and m = 1 for first-order maxima. Hence
putting everything together and solving for θ,
θ = sin−1
λ
= 13.8202◦ .
d