Version 066 – F30 – Antoniewicz – (57030)
This print-out should have 28 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
1
5. Ia, IIa, IIIa
6. Ib, IIb, IIIc
7. Ic, IIb, IIIb
8. Ic, IIb, IIIc
9. Ic, IIc, IIIb
A long thick wire of radius R carries a
current I1 as shown in the figure above. We
would like to determine the magnetic field B,
inside the wire, a distance r from the center
of the wire, where r > R, near the midpoint
of the wire where the end effect at both ends
are negligible.
We solve this problem based on Ampere’s
Law.
The left hand side of Ampere’s law is given
by:
Ia) 2πrB
Ib) 2πRB
Ic) πr 2B
10. Ib, IIb, IIIb
Explanation:
Along the Amperian loop of radius r and
magnetic field B, the left hand side of Ampere’s Law is given by 2πrB.
Hence the answer is Ia.
The current encircled by the loop is I1 .
Thus, the right hand side of the expression is
given by µ0 I1 .
Hence, the answer is IIb.
The above two answers lead to
µ0 I 1
2πr
Hence, the answer is IIIb.
The right hand side of Ampere’s law is:
IIa) µ0 I1 !
IIb) µ0 I1
IIc) µ0 I1
r2
R2
r
R
002 (part 1 of 2) 4.0 points
"
The field B at r is given by:
IIIa)
µ0 I 1
2πR !
r2
µ0 I 1
R2
IIIb)
2πr
µ0 I 1
IIIc)
2πr
8m
B!
"
1. Ia, IIa, IIIc correct
2. Ia, IIb, IIIa
Consider a changing magnetic field (from
B ! to B) within a circle of radius 8 m. The
magnetic field points into the page ⊗ and its
magnitude is increasing.
B
!
B
!
⇒
8m
B
B
The field is defined by B = b t , where b =
0.7 T/s:
B
"
B
B!
3. Ia, IIa, IIIb
4. Ib, IIc, IIIa
B
0
"t
Version 066 – F30 – Antoniewicz – (57030)
Find the direction of the induced emf Eind .
dB
d
(B A) =
π a2
dt
dt
= (0.7 T/s) π (8 m)2 = 140.743 V .
|Eind | =
1. into the page
2. clockwise
From
3. out of the page
E=
4. counter-clockwise correct
Explanation:
Using brackets [ and ] to denote directions,
Faraday’s Law
#
" ind · d"s = − d ΦB
Eind = E
dt
yields
[Eind ] = −
2
$
%
dΦ
= −⊗ = $ .
dt
The right hand rule then implies that the
direction of Eind is counter-clockwise. This
is consistent with Lenz’s Law: the magnetic
field is increasing into the page, so the induced
emf Eind creates a magnetic field out of the
page to counter this change.
003 (part 2 of 2) 6.0 points
Find the magnitude of the induced electric
field Eind on the perimeter of the circular
magnetic field.
1. 2.8
2. 0.5
3. 2.0
4. 1.2
5. 0.8
6. 2.45
7. 3.6
8. 1.0
9. 2.25
10. 0.3
Correct answer: 2.8 V/m.
Explanation:
Let : a = 8 m and
b = 0.7 T/s .
#
" ind · d"s ,
E
Eind = Eind 2 π a
Eind
dB
π a2
Eind
dt = a d B
=
=
2πa
2πa
2 dt
(8 m)
(0.7 T/s) = 2.8 V/m .
=
2
004 (part 1 of 2) 5.0 points
A circuit consists of an idealized 152 mH inductor and an idealized 11.1 µF capacitor.
What is the natural frequency of this circuit
if the capacitor is first charged to 413 µC
and then allowed to discharge through the
inductor?
1. 68.6855
2. 128.385
3. 58.0475
4. 60.52
5. 122.528
6. 106.703
7. 99.1009
8. 95.9723
9. 90.4523
10. 102.747
Correct answer: 122.528 Hz.
Explanation:
Let : L = 152 mH = 0.152 H , and
C = 11.1 µF = 1.11 × 10−5 F .
The natural frequency is given by
&
1
1
f=
2π LC
'
1
1
=
2 π (0.152 H) (1.11 × 10−5 F)
= 122.528 Hz .
Version 066 – F30 – Antoniewicz – (57030)
Correct answer: 0.284566 A.
Explanation:
Let : Q = 413 µC = 0.000413 C ,
Elost = 80.1% = 0.801 .
and
Q is the initial charge, q the final charge, and
I is the final current. The total energy is
conserved over time, so
Q2
q2
1
=
+ L I2 .
2C
2C 2
Q2
2C
0.46 m
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?
1. 0.114038
2. 0.252973
3. 0.14288
4. 0.0865172
5. 0.504074
6. 0.355488
7. 0.174928
8. 0.17003
9. 0.22743
10. 0.614656
Explanation:
"
=
1
L I2 .
2
Thus the current in the circuit is
&
0.801
I=Q
LC
'
0.801
= (0.000413 C)
(0.152 H) (1.11 × 10−5 F)
= 0.284566 A .
006 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).
← axis of rotation
"
B
y
x
d
#
2C
= 0.801
!
x
0.46 m
" = 0.4 T
B
→ I →
2C
−
q2
y
Correct answer: 0.355488 N m.
According to the problem,
Q2
← axis of rotation
" = 0.4 T
B
→ 4.2 A →
005 (part 2 of 2) 5.0 points
Find the current in the circuit when the capacitor has lost 80.1% of its initial energy.
1. 0.284566
2. 0.15887
3. 0.124784
4. 0.0299622
5. 0.0562862
6. 0.0589615
7. 0.108245
8. 0.206
9. 0.0243764
10. 0.0694731
3
"
B
Let : d = 0.46 m ,
# = 0.46 m ,
B = 0.4 T , and
I = 4.2 A .
Version 066 – F30 – Antoniewicz – (57030)
Only the right side of the loop contributes to
the torque. By definition, the torque is
4
y
++++
L.
"(
τ = ("d × F
= d×I #B
= (0.46 m) (4.2 A) (0.46 m) (0.4 T)
= 0.355488 N m .
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 M and G only
007
10.0 points
2. Configuration S only
y
3. Configurations M and S only
++++
4. Configuration L only correct
S.
x
6. Configurations S and G only
++++
7. Configurations S, P and G only
y
+++++
P.
+
+
+
+
+
+++++
−
−
−
− x
−
+
+
+
−
−
−
−
++
++
++
++
y
++
++
9. Configuration P only
10. Configurations M, S and G only
+
x
G.
8. Configurations M, L and G only
Explanation:
y
M.
5. Configuration M only
−−−−−−
x
∆E =
k∆q
r̂
r2
and
E=
(
∆E .
Symmetry of the configuration will cause
some component of the electric field to be
zero.
Configuration M is anti-symmetric about
the y-axis (opposite sign of charges), so the
electric field has no y-component.
Version 066 – F30 – Antoniewicz – (57030)
y
−
−
−
−
Configuration S is symmetric by a rotation
of 180◦ , so the electric fields generated by
these two pieces have opposite directions and
the total field is zero.
y
++++
S
x
++++
Configuration L is anti-symmetric by rotation of a 180◦ , so the total field has non-zero
components in both x and y directions, just
like the field generated by just one piece of
charge.
y
++++
L
x
−−−−
Configuration P is symmetric about the xaxis, so the y component of the total field
must vanish.
y
+++++
+
−
+
−
+
−
P
+
− x
+
−
+++++
Configuration G is symmetric about the yaxis, so the x component of the total field
must vanish.
G
++
x
++
+
++
+
++
+
++
+
++
y
M
5
x
−−−−−−
008 10.0 points
You are given three parallel conducting
plates that are aligned perpendicular to the
x-axis. They are labeled, from left to right,
as plates 1, 2 and 3, respectively. The corresponding plate charges are Q1 = −2q, Q2 = q
and Q3 = q. The width of the gap between
1 and 2 is d, and the width between plates 2
and 3 is d/2.
Determine the magnitude of the potential
difference across 1 and 3.
5(q/A)d
correct
2%0
2(q/A)d
2.
%0
1.
3. 0
4.
5.
6.
7.
8.
9.
9(q/A)d
2%0
4(q/A)d
%0
5(q/A)d
%0
3(q/A)d
2%0
6(q/A)d
%0
7(q/A)d
2%0
Explanation:
One may regard the 3-plate system as a
composite system which involves two capacitor systems with the 12-capacitor followed by
the 23-capacitor.
The 12-capacitor has charges Q1 and Q2 +
Q3 , i.e charges of −2q and +2q respectively.
The 23-capacitor has charges Q1 + Q2 and
Version 066 – F30 – Antoniewicz – (57030)
V3 − V1 = Egap,12 d + Egap,23 d
V3 − V1 =
(2q/A)d
(q/A)d
+
%0
2%0
V3 − V1 =
5(q/A)d
2%0
009 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
P
Q!!2 , c
Q2
8. EP =
2 ke (Q1 − Q2 )
(a + b)2
9. EP = 0
4 ke Q 2
(a + b)2
Explanation:
Choose as your Gaussian surface the spherical surface S concentric with the centers of
the spheres, which passes through P . Thus
Q1
4 π r 2 EP =
%0
Q1
EP =
4 π %0 r 2
ke Q 1
=
r2
4 ke Q 1
=
.
(a + b)2
10. EP =
010 (part 1 of 2) 5.0 points
A screen is illuminated by monochromatic
light whose wave length is λ, as shown. The
distance from the slits to the screen is L .
Q1
Find the
) magnitude*of the electric field at
" P ( ≡ EP , where the distance
point P (E
a+b
from P to the center is r =
.
2
2 ke Q 2
1. EP =
(a + b)2
2 ke (Q1 + Q2 )
2. EP =
(a + b)2
4 ke (Q1 − Q2 )
3. EP =
(a + b)2
2 ke Q 1
4. EP =
(a + b)2
2 ke Q 1 a
5. EP =
(a + b)3
4 ke (Q1 + Q2 )
6. EP =
(a + b)2
4 ke Q 1
7. EP =
correct
(a + b)2
y
S1
θ
d
S2
L
viewing
screen
Q3 , i.e charges of −q and +q respectively.
The potential difference is
6
At the seventh dark fringe on the screen
(position y on the screen), find δ (the corresponding path length difference) and φ (the
phase angle difference).
1. δ = 6 λ and
φ = 12 π
13
λ and φ = 13 π correct
2
7
3. δ = λ and φ = 7 π
2
2. δ =
4. δ = 7 λ and
5. δ =
5
λ
2
and
φ = 14 π
φ = 5π
Version 066 – F30 – Antoniewicz – (57030)
7. δ =
9
λ
2
φ = 10 π
and
φ = 9π
8. δ = 4 λ and
φ = 8π
9. δ = 2 λ and
φ = 4π
3
λ and φ = 3 π
2
Explanation:
For dark fringes
!
"
1
δ = d sin θ = m +
λ,
2
10. δ =
where m = 0 , ±1 , ±2 , ±3 , so for the seventh dark fringe (m = 6),
"
!
13
1
λ=
λ and
δ = m+
2
2
1 dy
7 L
2 dL
8. λ =
15 y
2 dy
9. λ =
correct
13 L
1 dL
10. λ =
7 y
Explanation:
7. λ =
$
r1
d
S2
2 dy
15 L
1 Ly
2. λ =
7 d
3. None of these
2
15
2
5. λ =
13
2
6. λ =
13
4. λ =
Ly
d
Ly
d
dL
y
) y*
L
Q
y
O
δ ≈ d sin θ ≈ r2 − r1
L
r1
S1
−1
d=
1. λ =
−1
an
r2 θ = t
S1
φ
13
δ
=
=
λ
2π
2
φ = 13 π .
011 (part 2 of 2) 5.0 points
Let L be the distance from the source to the
screen and h be the distance from the source
to the mirror.
Using the small angle approximation (θ =
sin θ = tan θ) , if the distance from the central
bright region to the seventh dark fringe is y,
what is the wavelength of the light?
P
S2 Q S1 ≈ 90◦
viewing
screen
6. δ = 5 λ and
7
θ = t an
h
2
θ
) y*
L
r2
Q
S2
◦
0
≈9
r1
1
S
Q
r2 −
≈
$ S2
nθ
h si
2
δ≈
The angle θ from the slits’ midpoint to the
y position on the screen is
y
tan θ = ≈ sin θ ,
L
so the wavelength of the light for the seventh
dark fringe (m = 6) is
!
"
1
m+
λ = d sin θ
2
13
y
λ=d
2
L
2 dy
.
λ=
13 L
012
10.0 points
Version 066 – F30 – Antoniewicz – (57030)
A convergent lens forms a virtual image 2.12
times the size of the object. The object distance is 14.5 cm .
Substituting these values into the mirror
equation
f=
h!
=
h
q f
p
8
f
1
1 1
+
p q
(3)
1
1
1
+
14.5 cm −30.74 cm
= 27.4464 cm .
013 (part 1 of 2) 5.0 points
Scale: 10 cm =
Find the distance of the focal point from
the center of the lens.
1. 16.0444
2. 18.0
3. 25.873
4. 28.2906
5. 17.5154
6. 27.4464
7. 15.2
8. 27.1599
9. 20.8
10. 17.2128
A coil is suspended around an axis which
is colinear 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 right to left with North and South poles
labeled as in the figure.
Correct answer: 27.4464 cm.
What is the direction of the induced magnetic field in the coil when the bar magnet is
moving from right to left?
Explanation:
−∞ < q < 0
(1)
Solving for q, we have
q = −M p
= − (2.12) (14.5 cm)
= −30.74 cm .
b
v
3. the induced field is zero tesla
∞ >M > 1
q
= 2.12 .
p
R
2. right to left (⇐= Binduced )
Note: The focal length for a convergent
lens is positive.
M =−
a
S
1. left to right (Binduced =⇒) correct
1 1
1
h!
q
+ =
M=
=−
p q
f
h
p
Convergent Lens
f >0
f >p> 0
N
(2)
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
right to left, the magnetic flux through the
coils increases.
The induced current in the coil must produce an induced magnetic field from left to
right (Binduced =⇒) to resist any change of
magnetic flux in the coil (Lenz’s Law).
Version 066 – F30 – Antoniewicz – (57030)
9
Solving for ∆V , we obtain
014 (part 2 of 2) 5.0 points
What is the direction of the induced current
in resistor R when the bar magnet is moving
from right to left?
1. the induced current is zero amperes
∆V =
∆V =
2(9.11 × 10−31
h2
2me eλ2
(6.625 × 10−34 Js)2
kg)(1.6 × 10−19 C)(2.4 × 10−1
∆V = 26.1385 V
2. from å through R to b (I −→) correct
3. from b through R to å (←− I)
016
10.0 points
Explanation:
The helical coil when viewed from the bar
magnet winds around the solenoid from terminal b counter-clockwise.
As the induced field is left to right
(Binduced =⇒), the induced current must flow
counter-clockwise and therefore it goes from
å through R to b (I −→).
015 10.0 points
Roughly, what is the minimum accelerating potential ∆V needed in order that electrons exhibit diffraction effects in a crystal, given that the wavelength observed is
λ = 2.4 × 10−10 m?
Given that:
h = 6.625 × 10−34 Js
e = 1.6 × 10−19 C
me = 9.11 × 10−31 kg.
1. 26.1385
2. 66.9146
3. 150.558
4. 34.1401
5. 185.874
6. 602.231
7. 41.7058
8. 124.428
9. 28.4608
10. 52.0961
Correct answer: 26.1385 V.
Explanation:
We can make use of the formula
λ = √
h
2me e∆V
A water molecule is a permanent dipole
with a known dipole moment p = qs. There
is a water molecule in the air a very short distance x from the midpoint of a long glass
rod of length L carrying a uniformly distributed positive charge Q. The axis of the
dipole is perpendicular to the rod. Note that
s << x << L. You may neglect the small
change in the dipole moment of the water
molecule induced by the rod.
Choose the answer that correctly expresses
the magnitude and direction (along the xaxis) of the electric force on the water
molecule. Your f inal result must be expressed only in terms of k, Q, p, L, s and
x and any constant numerical factors.
1. −k
2Qp
correct
Lx2
Version 066 – F30 – Antoniewicz – (57030)
where the approximation is justified since
s << x.
2Qp
2. k 2
Lx
Qp
3. k 2
Ls
Qp
4. −k 2
Ls
2Qp
5. k 2
Ls
2Qp
6. k 2
xL
2Qp
7. −k 2
xL
Qp
8. k 2
Lx
2Qp
9. −k 2
Ls
Qp
10. −k 2
Lx
017 (part 1 of 2) 5.0 points
Consider the circuit
E1
B
C
E2
r1
i1
r2
A
D
i2
F
R
E
i
What equation does the loop DCFED
yield?
Explanation:
Since we are given the condition that x <<
L, we may use the approximate equation for
the E-field of a long rod
E≈
1 2Qq
4π%0 L
1
x−

1 2Qq 

F =
4π%0 L 
F =
s
2
+
F≈−
1 2Qq
4π%0 L
1
x+

1 
1

−
s
s
x+
x−
2
2
1 2Qq
4π%0 L
−s
x2 −
1 2Qqs
4π%0 Lx2
F ≈ −k
2Qp
Lx2
1. E2 − i2 r2 − iR = 0 correct
2. −E2 − i2 r2 − 2 iR = 0
3. −E2 − i2 r2 + 2iR = 0
4. −E1 + i2 r2 − iR = 0
1 2(Q/L)
4π%0
r
Calculate the force by considering each charge
independently:
F = −
10
s2
4
s
2
5. −E1 − i2 r2 − iR = 0
6. E2 + i2 r2 − iR = 0
7. −E2 + i2 r2 − 2iR = 0
8. E1 − i2 r2 − iR = 0
9. E2 − i2 r2 + iR = 0
10. E2 + i2 r2 + iR = 0
Explanation:
Recall that Kirchhoff’s loop rule states that
the sum of the potential differences across all
the elements around a closed circuit loop is
zero. If a resistor is traversed in the direction
of the current, the change in potential is −IR.
If an emf source is traversed from the − to
+ terminals, the change in potential is +E.
Apply the opposite sign for traversing the
elements in the opposite direction.
Version 066 – F30 – Antoniewicz – (57030)
Thus by inspection,
DCFED : E2 − iR − i2 r2 = 0
018 (part 2 of 2) 5.0 points
Find the current i. Symmetry is applicable
here. Let E1 = E2 = E = 11 V , r1 = r2 = r =
4 Ω , and R = 4.2 Ω .
1. 3.7931
2. 2.47934
3. 1.59091
4. 1.69492
5. 1.04348
6. 1.77419
7. 5.0
8. 2.25
9. 2.42424
10. 4.07407
Correct answer: 1.77419 A.
Explanation:
Let :
E1 = E2 = E = 11 V ,
r1 = r2 = r = 4 Ω , and
R = 4.2 Ω .
E1 = E2 and r1 = r2 . This implies that
i1 = i2 . (Loops DCFED and ABFEA have
identical loop equations.) Hence the junction
rule yields
i1 + i2 = 2i2 = i
i
i2 =
2
Substituting this into the loop equation
DCFED,
i
E2 − iR − r2 = 0
2
i=
11 V
r2 =
4Ω
R+
4.2 Ω +
2
2
E2
= 1.77419 A .
019 10.0 points
Consider a cube with one corner at the origin
11
and with sides of length 11 cm positioned
along the xyz axes. There is an electric field
" = .24, 195 y, 0/ V/m
E
through the region that has a constant x component and a y component that increases linearly with y. Start by drawing a diagram
" for each side of the cube.
showing n̂ and E
How much change is inside the cube? Use
%0 = 8.85 × 10−12
C2
.
N · m2
1. 2.82917e-12
2. 1.416e-12
3. 1.01952e-12
4. 1.29033e-12
5. 1.19356e-12
6. 2.59978e-12
7. 2.17918e-12
8. 2.29697e-12
9. 8.60928e-13
10. 2.00249e-12
Correct answer: 2.29697 × 10−12 C.
Explanation:
We use Gauss’s Law:
#
" · n̂ dA = Qenc .
E
%0
So in order to find the enclosed charge, we
need to evaluate the left hand side of the
above equation.
Geometry is very important in this problem, and can save a lot of mathematical work.
" lies in the xy plane, so that
Note that E
means neither the front nor back of the cube
contributes to the flux through the cube. On
" is to the right so that face
the bottom face, E
doesn’t contribute to the flux. The contributions from the left and right faces add to zero
(contribution of left face is negative and contribution of right face is positive). The only
surviving contribution is from the top face.
We can therefore write
Version 066 – F30 – Antoniewicz – (57030)
#
cube
" · n̂ dA =
E
=
#
#top
" · n̂ dA
E
Ey dA
top
= (195 y V/m)(A)
12
7. 2.40774e-08
8. 6.76862e-08
9. 6.2665e-08
10. 8.51264e-08
Correct answer: 2.32104 × 10−8 C.
Explanation:
2
= (195 V/m )(11 cm)
× (11 cm)2
= 0.259545 V · m .
Let : L = 0.11 m ,
m = 0.03 kg ,
θ = 4◦ .
There must be a net positive charge inside
the box:
= 2.29697 × 10
−12
C .
020 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.
L
Qenc
= 0.259545 V · m
%0
⇒ Qenc = %0 (0.259545 V · m)
"
!
C2
−12
= 8.85 × 10
N · m2
× (0.259545 V · m)
q
m
θ
a
m
q
From the right triangle,
a
sin θ =
L
a = L sin θ = (0.11 m) sin 4◦
= 0.00767321 m .
The separation of the spheres is r = 2 a =
0.0153464 m . The forces acting on one of the
spheres are shown in the figure below.
T
T cos θ
0. 1
Fe
1m
4◦
0.03 kg
and
θ
θ
T sin θ
mg
0.03 kg
Find the magnitude of the charge on
each sphere. The acceleration of gravity is
9.8 m/s2 and the value of Coulomb’s constant
is 8.98755 × 109 N · m2 /C2 .
1. 6.38151e-08
2. 4.25227e-08
3. 3.79806e-08
4. 2.32104e-08
5. 7.91262e-09
6. 3.94836e-08
Because the sphere is in equilibrium, the
resultant of the forces in the horizontal and
vertical directions must separately add up to
zero:
(
Fx = T sin θ − Fe = 0
(
Fy = T cos θ − m g = 0 .
Dividing,
F sin θ
Fe
=
F cos θ
mg
Version 066 – F30 – Antoniewicz – (57030)
Fe = m g tan θ
1
2
= (0.03 kg) 9.8 m/s2 tan 4◦
= 0.0205585 N .
=
= 2.32104 × 10−8 C .
021 (part 1 of 2) 5.0 points
An electron circles at a speed of 7810 m/s in a
radius of 2.69 cm in a solenoid. The magnetic
field of the solenoid is perpendicular to the
plane of the electron’s path.
The charge on an electron is 1.60218 ×
10−19 C and its mass is 9.10939 × 10−31 kg.
Find the strength of the magnetic field inside the solenoid.
1. 3.41698e-06
2. 2.10439e-06
3. 2.16321e-06
4. 2.80278e-06
5. 2.75429e-06
6. 3.977e-06
7. 2.98889e-06
8. 1.65074e-06
9. 3.86296e-06
10. 6.95836e-06
Correct answer: 1.65074 × 10−6 T.
Explanation:
Let : r = 2.69 cm = 0.0269 m ,
v = 7810 m/s ,
q = 1.60218 × 10−19 C , and
m = 9.10939 × 10−31 kg .
The strength of the magnetic field is
mv
B=
qr
(9.10939 × 10−31 kg) (7810 m/s)
(1.60218 × 10−19 C) (0.0269 m)
= 1.65074 × 10−6 T .
From Coulomb’s law, the electric force between the charges has magnitude
|q|2
|Fe | = ke 2
' r
|Fe | r 2
|q| =
ke
'
(0.0205585 N) (0.0153464 m)2
=
(8.98755 × 109 N · m2 /C2 )
13
022 (part 2 of 2) 5.0 points
Find the current in the solenoid if it has
38.4 turns/cm.
1. 0.850251
2. 0.821825
3. 0.342087
4. 2.6459
5. 1.02095
6. 2.22154
7. 0.954931
8. 1.44031
9. 0.613353
10. 0.834266
Correct answer: 0.342087 mA.
Explanation:
Let :
µ0 = 1.25664 × 10−6 N/A2 , and
n = 38.4 turns/cm = 3840 turns/m .
The current in the solenoid is from B = µ0 n I,
we get
I=
=
B
µ0 n
1.65074 × 10−6 T
(1.25664 × 10−6 N/A2 ) (3840 turns/m)
= 0.000342087 A = 0.342087 mA .
023 (part 1 of 3) 3.0 points
Four long, parallel conductors carry equal currents of I. An end view of the conductors is
shown in the figure. Each side of the square
has length of #.
Version 066 – F30 – Antoniewicz – (57030)
y
A
×
C
5.
x
P
B
D
#
Which diagram correctly denotes the directions of the magnetic fields from each conductor at the point P ? The current direction is
out of the page at points A, B, and D indicated by the dots and into the page at point
C indicated by the cross.
BB
14
BA
P
BD
BC
Explanation:
The direction of the magnetic field due to
each wire is given by the right hand rule.
Place the thumb of the right hand along the
direction of the current; your fingers now curl
in the direction of the magnetic field’s circular
path:
A
×
BA
BA
1.
BC
P
P
BD
BC
BB
BB
B
2.
BB
correct
P
BD
3.
P
BA BC
D
024 (part 2 of 3) 3.0 points
At point P , as far as the magnitudes are
concerned, BA = BB = BC = BD ≡ Bi ,
where Bi is introduced to represent any of the
four magnetic fields.
Find Bi .
1. Bi =
BB
2. Bi =
3. Bi =
BB
BD
BC BA
BD
4.
C
BC
BD
P
BA
4. Bi =
5. Bi =
µ0 I
2π#
µ0 I 2
2π#
µ I
√0
correct
2 π#
µ I
√0
2 2 π#
µ0 I #
2π
Explanation:
By Ampere’s law, the line integral around
any closed path is µ0 I, with each path defined
Version 066 – F30 – Antoniewicz – (57030)
15
√
Express your answer in units of nm.
# 2
◦
by a circle with radius r = # cos 45 =
.
Given that:
2
#
h = 4.136 × 10−15 eV.s
" · d"s = µ0 I
c = 3 × 108 m/s.
B
1. 288.558
B i · 2 π r = µ0 I
2. 264.0
µ0 I
µ0 I
3. 282.0
√ .
=
Bi =
2πr
π# 2
4. 217.684
5. 238.615
025 (part 3 of 3) 4.0 points
6. 310.2
Determine the magnitude of the resultant
7. 248.16
magnetic field if the current is 10 A and each
8. 213.931
side of the square has length 0.9 m. The
9. 400.258
permeability of a vacuum is 4 π × 10−7 N/A2 .
10. 229.778
1. 5.22171e-06
2. 6.28539e-06
Correct answer: 288.558 nm.
3. 5.65685e-06
Explanation:
4. 3.046e-06
Apply the Energy Principle. The maxi5. 1.31993e-05
mum wavelength required to eject the elec6. 6.91393e-06
tron means that the photon has the minimum
7. 6.36396e-06
energy possible, which is equal to the energy
8. 4.71404e-06
needed to break the electron free from the
9. 8.17101e-06
metal. This is the work function of the metal.
10. 1.13137e-05
So,
Ephoton = W
Correct answer: 6.28539 × 10−6 T.
Explanation:
Let : I = 10 A ,
# = 0.9 m , and
µ0 = 4 π × 10−7 N/A2 .
λ =
Applying superposition of the vectors,
" =B
"A +B
"B +B
"C +B
"D = 2B
"B,
B
so
√
µ0 I
2 µ0 I
√ =
B = 2 BB = 2
π#
π# 2
√
−7
2
2 (4 π × 10 N/A ) (10 A)
=
π (0.9 m)
= 6.28539 × 10−6 T .
026 10.0 points
If the work function of a metal is W =
4.3 eV, what would be the maximum wavelength of light required to eject an electron
from the metal?
W =
hc
λ
λ =
hc
W
(4.136 × 10−15 eV.s)(3 × 108 m/s)
(4.3 eV)
λ = 288.558 nm
027 10.0 points
A 17 V battery delivers 109 mA of current
when connected to a 77 Ω resistor.
Determine the internal resistance of the
battery.
1. 78.9633
2. 42.1495
3. 125.078
4. 30.5641
5. 96.6667
6. 71.5929
7. 54.1186
Version 066 – F30 – Antoniewicz – (57030)
8. 48.6415
9. 62.6154
10. 83.8785
kq
a2
√ kq
8. EO = 2 2
a
1 kq
9. EO = √
3 2 a2
1 kq
10. EO = √
2 2 a2
Explanation:
The distance between each corner and the
a
center is √ , so the magnitude of each electric
2
field at D is
q
q
E=k !
"2 = 2 k 2
a
a
√
2
The two negative charges yield forces pointing away from them from O and the two positive charges yield forces pointing toward them
from O with the collinear charges adding algebraically:
7. EO =
Correct answer: 78.9633 Ω.
Explanation:
Let :
16
V = 17 V ,
I = 109 mA = 0.109 A ,
R = 77 Ω .
and
The internal resistance is in series with the
given resistor, so
V = I (R + r)
V
17 V
r=
−R=
− 77 Ω
I
0.109 A
= 78.9633 Ω .
028 10.0 points
Consider a square with side a. Four charges
−q, +q, +q, and −q are placed at the corners
A, B, C, and D, respectively
A
B
−
+
O
a
−
+
D
C
What is the magnitude of the electric field
at the center O?
1 kq
1. EO = √
2 a2
√ kq
2. EO = 3 2 2
a
√ kq
3. EO = 2 2 2
a
√ kq
4. EO = 4 2 2 correct
a
1 kq
5. EO = √
4 2 a2
kq
6. EO = 3 2
a
"A + E
" C ( = (E
"B + E
" D( = 2 E = 4 k q .
(E
a2
EA + EC
E
EB + ED
The Cartesian components of the two vectors with the origin at O are
!
"
1
q
1
"A + E
"B = 4k
− √ ı̂ + √ ̂
E
and
a2
2
2
"
!
q
1
1
"B +E
"D = 4k
E
− √ ı̂ − √ ̂ , so
a2
2
2
!
"
q
1
1
" = 4k
E
−√ − √
ı̂
a2
2
2
"
!
1
1
̂
+ √ −√
2
2
√
q
= −4 2 k 2 ı̂ ,
a
√
q
with magnitude −4 2 k 2 .
a