Physics 2B
Quiz 4
Summer Session I 2009
Name:
PID:
Version 0
Quiz ID:
READ FIRST: Make sure to write/fill in your scantron form your name,
quiz ID and the test version.
e ≈ 1.6 × 10−19 C , k ≈ 8.9 × 109 N m2 /C2 , 0 ≈ 8.9 × 10−12 C2 /N m2 , µ0 = 4π × 10−7 N/A2 .
Z
dx
x
= √
2
2
3/2
(x + a )
a2 a2 + x 2
All coordinate system are right-handed.
1. A particle with charge -3.00 C initially moves at ~v = 1.00x̂ + 7.00ŷ m/s. If it encounters
~ = 10.00ẑ T, find the foce on the particle.
a magnetic field B
Solution: Let algebra takes care of the right-hand-rule:
~ = q~v × B
~ = −3 × (1x̂ + 7ŷ) × 10ẑ = 30ŷ − 210x̂
F
A. −210x̂ + 30ŷ N
B. −210x̂ − 30ŷ N
C. 210x̂ + 30ŷ N
D. 210x̂ − 30ŷ N
2. What is the kinetic energy of a 10-g particle with a charge of 4.0 C moving with a radius
of 9.0 m in a constant magnetic field with strength 1.0 T?
Solution: From the cyclotron radius, we find the velocity:
v=
qBr
m
And the kinetic energy is by definition:
1
q 2 B 2 r2
(4.0 C · 1.0 T · 9 m)2
K = m v2 =
=
≈ 64800J
2
2m
2 · .010 kg
A. 64,800 J
B. 255 J
C. 1800 J
D. 0.101 J
3. A wire carries a 2.0-A current along the x-axis through a magnetic field B = 4.0x̂ + 7.0ŷ
T. If the wire experiences a force of 22.0ẑ N as a result, how long is the wire?
Physics 2B
Quiz 4
Summer Session I 2009
Version 0
Solution: Compare both side of the equation:
~ = I ~l × B
~
F
22.0ẑ N = 2 A · l · x̂ × (4x̂ + 7ŷ) T
22.0 N
≈ 1.57 m
l=
2A × 7T
A. 1.57 m
B. 1.36 m
C. 2.8 m
D. 1.00 m
4. A single loop equilateral triangle conductor with 3-cm sides has a 2-A current in the
direction as indicated by the arrow. What is its magnetic dipole moment?
y
I
x
3 cm
Solution: By the right hand rule, the direction of√the magnetic dipole moment is
in the −ẑ direction. The height of the triangle is 3 3/2 cm. The magnitude is:
√ !
(3 cm)2 3
≈ 7.79 A · cm2
µ = N I A = 2A ·
4
A. -7.8 ẑ A · cm2
B. 7.8 ẑ A · cm2
C. 18 ẑ A · cm2
D. -18 ẑ A · cm2
5. A segment of a wire lies on the x-axis between x = −1 and x = 1. A current I is running
from x = 1 to x = −1. Which of the following is the correct expression of the magnetic
field due to this segment at distance y away from the wire on the y-axis?
Solution: Cut the segment into infinitesimal pieces as usual. Taken into account
the direction of the current, the segment at x is Id~l = −Idx~x. By the right hand
Page 2
Physics 2B
Quiz 4
Summer Session I 2009
Version 0
rule, the direction the magnetic field due to each segment is in the −ẑ direction. The
magnitude is given by:
Z1
Z1
µ
I
y dx
sin
θ
dx
µ
I
0
0
~ =
B
(−ẑ) = −
ẑ
2
2
2
4π x + y
4π
(x + y 2 )3/2
−1
−1
1
µ0 I
µ0 I
x ẑ
ẑ
p
p
=−
=−
2
2
4π y x + y 2π y 1 + y 2
−1
y
(0, y)
−1
1
x
Note: Use the right hand rule and the given direction of the current, one find that
the magnetic field is into the page (the −ẑ direction) when y > 0 and out of the
page (the ẑ direction) when y < 0. The only choice that have this property is
µ0 I
ẑ
p
−
.
2π y 1 + y 2
A. −
µ0 I
ẑ
p
2π y 1 + y 2
B.
µ0 I
ẑ
p
2π y 1 + y 2
C. −
µ0 I
ẑ
p
4π 1 + y 2
D.
µ0 I
ẑ
p
4π 1 + y 2
6. A wire is formed into a circle with radius 10.00 mm. A current flow through the wire
and causes a magnetic field B at the center of the loop. If the wire is heated and the
radius expands by 9.00%, what does the magnetic field become at the center of the loop.
Solution: The magnetic field in the center of a circular loop, as shown in class and
textbook example is inversely proportional to the radius of the circle. Expanding
the radius by 9% means the new area is 1.09 times of old area. The new magnetic
field is thus B/1.09.
A. B/1.09
B. 1.09B
C. B/(2π1.09)
D. B
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Physics 2B
Quiz 4
Summer Session I 2009
Version 0
7. As shown in the following figure, a 1-m long metal rod of mass 10 g is free to slide up
and down two guide wires. The whole setup is in a perpendicular magnetic field and
gravity is acting down in the −~y direction. If the rod is suspended and not moving,
what is the magnetic field?
g
y
1111111
0000000
x
20Ω
z
10V
+
Solution: The current through the rod is 0.5 A in the x̂ direction. The magnetic
force to be upward (ŷ) in order to cancel gravity. So, the magnetic field is in the
(−ẑ) direction. To find the magnitude, set the magnetic force and gravitational force
equal:
mg
0.01 kg · 9.8 m/s2
mg = lIB
B=
=
= 0.196 T
lI
1 m · 0.5 A
A. -0.196ẑ T
B. 0.196ẑ T
C. 2ẑ T
D. -2ẑ T
8. A solenoid 3.0-cm long consists of 6075 loops of wire. If the magnetic field inside the
solenoid is 2.0 T, what is the magnitude of the current that flow through it?
Solution: In the equation B = µ0 nI, n is number of loop per length:
n = 6075/.03 m−1 ≈ 202500 m−1 .
I=
A. 8 A
B. 0.13 A
2.0T
B
≈ 7.86 A
=
−7
µ0 n
4π × 10 N/A2 × 202500 m−1
C. 100 A
D. 3.0 A
9. Four long, parallel wires are located at the corners of a square 15 cm on a side. Each
carries a current of 2500 A with the top two into the page and the bottom two out of
the page. Find the magnetic field at the center of the square.
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Physics 2B
Quiz 4
Summer Session I 2009
Version 0
Solution: Use the principle of super
√ position. The distances between each wire to
the center are the same: r = 0.15/ 2 m. The total contribution to the y-component
by the four wires cancel. The x-components add:
µ0 I
π
4
4
π · 10−7 N/A2 · 2.5 A
B =4×
cos = √ ×
≈ 13.3 × 10−6 T
√
2πr
4 2
2
π · 0.15/ 2 m
A. 13.3 mT
B. 3.3 mT
C. 18.8 mT
D. 4.7 mT
~ = −y 2 x̂ + x2 ŷ. How much
10. The magnetic field in the x-y plane of some system is B
current flows through the square with two diagonal corners at (0, 0) and (1, 1).
Solution: Use Ampère’s law and perform line integral along each of the four segments of the square:
I
Z
Z
Z
Z
~
~
~ · d~l
µ0 Iinside = B · dl =
+
+ +
B
bott
Z1
=
(x2 ŷ) · x̂ dx +
0
righ
Z1
top
left
(−y 2 x̂ + 1ŷ) · ŷ dy
0
Z0
+
(−1x̂ + x2 ŷ) · x̂ dx +
1
Z0
(−y 2 x̂) · ŷ dy
1
=2
Alternative: An alternaltive way to state Amère’s law is that the curl of the magnetic field is the current density:
1
∂B
∂B
1
1
y
x
~ =
~J = ∇
~ ×B
−
ẑ = (2x + 2y)ẑ
µ0
µ0 ∂x
∂y
µ0
The current is simpliy the area integral over the desired area:
Iinside
1
=
µ0
Z
~J · dA
~ = 1
µ0
Z1 Z1
2(x + y)dxdy =
0
A. 2/µ0
B. 0
C. 4/µ0
D. 1/µ0
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0
2
µ0