Name: ________________________________________ Code # _________
UCSD Physics 2B
Practice Exam 4
Summer Session 2008
ID: X
Magnetism & Induction
READ ME !!!
Helpful Hints:
Do this exam a few times... Once without the answers,
Once after seeing the solutions & Once again with your formula sheet
This is a good time to add formulas/constants to your sheet.
Remember, this is a timed exam... work swiftly so you have time to check your answers.
1. A proton with charge q = e + is moving with velocity v at an angle of θ = 50° to the direction of a
magnetic field B. The component of the resulting force in the direction of B is
A. evB sin(50º) cos(50º)
B. evB sin(50º)
C. zero
D. evB cos(50º)
2. A charged particle is moving horizontally westward with a velocity of v = 3.5 × 106 m / s in a region
where there is a magnetic field of magnitude B = 5.6 × 10 −5 T directed due North. If the particle
experiences a force F = 7.8 × 10 −16 N upward (out of the page) what is the magnitude and sign of the
charge on the particle?
A. +4.9 × 10 −5 C
B. −4.0 × 10−18 C
C. +1.4 × 10−11 C
3
D. −1.2 × 10−14 C
Name: ________________________
ID: A
3. A long straight wire parallel to the y-axis carries a current of I = 6.3A j8 in a region of uniform
magnetic field B = 1.5T i8. What is the force per unit length on the wire?
A. +6.3
N 8
k
m
B. +9.5
N 8
k
m
C. −6.3
N 8
k
m
D. −9.5
N 8
k
m
4. Two long, parallel wires are spaced 1.0 m apart in air and each carry a current of 1.0 A. What is the
force per unit length that each wire exerts on the other?
A. 2π × 10 −6
N
m
B. 2.0 × 10 −5
N
m
C. 2.0 × 10 −7
N
m
D. 2π × 10 −5
N
m
5. A circular 20 turn coil has a radius r = 5.0 cm. What is the magnitude of the coil's magnetic dipole
moment when it carries a current of I = 2.5 A?
A. 1.5 A•m2
B. 0.15 A•m2
C. 3.9 kA•m2
D. 0.39 A•m2
6. A circular 20 turn coil of radius 5.0 cm is so oriented that its axis makes a 30º angle with a uniform
magnetic field of 0.15 T. What is the torque on the coil when it carries a current of 2.5 A?
A. 1.5 × 10−3 N ⋅ m
B. 9.4 × 10−3 N ⋅ m
C. 2.9 × 10−2 N ⋅ m
D. 5.1 × 10−2 N ⋅ m
2
Name: ________________________
ID: A
7. A circular loop of wire of radius a = 6.0 cm has N = 30 turns and lies in the x-y plane. It carries a
current I = 5.0 A. What is the magnitude of the magnetic field at a point along the axis x = 6.0 cm
from the loop center?
A. 19 μ T
B. 0.56 m T
C. 0.11 μ T
D. 56 μ T
8. Two wires lying in the plane of this page carry equal currents I in opposite directions as shown in
the diagram below. What is the magnetic field at a point midway between the wires?
A.
B.
C.
D.
E.
zero
into the page
out of the page
toward the top or bottom of the page
toward one of the two wires
3
Name: ________________________
ID: A
9. A rectangular loop of wire of height h = 2.2 m is immersed to a depth x = 15 m inside a region of
uniform magnetic field B = 3.2 T which is perpendicular to the loop as shown below. If the loop has
a resistance R = 150 Ohms and is pulled out of the field with velocity v = 3.3 m/s, what will be the
dissipated power?
A. 150 mW
B. 6.5 W
C. 210 mW
4
D. 3.6 W
Name: ________________________
ID: A
10. A long straight wire carries a constant current I as shown below. The magnitude of the magnetic
flux through the illustrated loop of wire is
A. ÊÁË μ0
B. ÊÁË μ0
C. ÁÊË μ0
D. ÊÁË μ0
E. ÊÁË μ0
/ 4π ˆ˜¯ 2 I L
/ 4π ˆ˜¯ 4 I L
/ 4π ˜ˆ¯ 2 I L
/ 4π ˆ˜¯ 2 I L
/ 4π ˆ˜¯ 4 I L
ln(b / a )
ln(b / a )
È
˘
lnÍÍÍÎ (b + a ) / (b − a ) ˙˙˙˚
È
˘
lnÍÍÍÎ (b − a ) / (b + a ) ˙˙˙˚
È
˘
lnÍÍÍÎ (b − a ) / (b + a ) ˙˙˙˚
11. A wire of radius R = 0.35 cm carries a current I = 75 A that is uniformly distributed over its
cross-sectional area (into the page) as shown below. What is the magnitude of the magnetic field at
a distance r = 5.0 cm from the center of the wire?
A. 0.47 mT
B. 1.5 mT
C. 0.30 mT
5
D. 0.56 mT
Name: ________________________
ID: A
12. A tightly wound air-core solenoid is 15 cm long, has 350 turns, and carries a current of 3.0 A. If you
ignore end effects, what is the value of the magnetic field at the center of the solenoid?
A. 6.4 mT
B. 4.4 mT
C. 6.8 μ T
D. 8.8 mT
13. A uniform magnetic field of magnitude B = 0.5 T is parallel to the x-axis. A square coil of side
s = 10 cm has N = 300 turns and its normal makes an angle θ = 60° with the x-axis. What is the
approximate magnetic flux Φ B through the coil?
A. 0.14 Wb
B. 0.75 Wb
C. 1.5 Wb
D. 0.27 Wb
14. A copper ring lies in the y-z plane as shown below. The magnet's long axis lies along the x axis.
Induced current flows through the ring as shown. The magnet
A.
B.
C.
D.
E.
must be moving away from the ring
must be moving toward the ring
must be moving either away from or toward the ring
is not necessarily moving
must remain stationary to keep the current flowing
6
Name: ________________________
ID: A
15. A 3.0 cm by 5.0 cm rectangular coil has 100 turns and its axis (normal) makes an angle of 55º with a
uniform magnetic field of 0.35 T. If it takes 0.33 s to turn the coil until its plane is parallel to the
magnetic field (no flux). What is the average magnitude of the induced emf?
A. 160 mV
B. 29 mV
C. 91 mV
D. 68 mV
16. Two coaxial solenoids have diameters R and 2R respectively and each have identical winding
densities (turns per unit length). The solenoids each carry the same current I in the same direction
as shown in two views below. If the magnetic field strength is B in the region between the two coils
(R < r < 2R), what is the magnetic field inside the inner solenoid?
A. B / 2
B. 2 B
C. B / 3
D. 3 B
E.
zero
17. A straight wire carrying current I = 3.75 A is bent into a semi-circle of radius R = 3.30 mm. What is
the magnitude of the magnetic field at the center of the semicircle?
A. 71.6 mT
B. 35.8 mT
C. 358 mT
7
D. 716 mT
Name: ________________________
ID: A
18. Two concentric semicircles of wire R1 = 13 mm and R2 = 20 mm are connected as shown in the
diagram below. A current I = 23 A passes through the circuit as shown. What is the magnetic field
at the center of the two semicircles?
A.
B.
C.
D.
E.
200 mT
0.20 mT
0.92 mT
92 mT
zero
19. During a demonstration of "60 Cycle Resonance", an alternating current is passed through a coil
of wire surrounding a cylindrical iron core. An aluminum ring placed at the bottom of the core is
launched vertically by an induced magnetic force. What principle best explains this phenomenon?
A.
B.
C.
D.
E.
Harley's Law
Faraday's Law
Gauss's Law
Maxwell's Law
Lenz's Law
8
Name: ________________________
ID: A
20. According to Faraday's Law, a necessary and sufficient condition for an electromotive force to be
induced in a closed circuit loop is the presence in the loop of
A.
B.
C.
D.
E.
a magnetic field
magnetic materials
an electric current
a time-varying magnetic flux
a time-varying magnetic field
21. A square coil of side s = 3.6 cm has 450 turns and its axis makes an angle of 35º with a uniform
magnetic field B. It takes 13 ms to rotate the coil until its plane is parallel to the magnetic field (no
flux). If the average induced emf is 110 mV, what is the strength of the magnetic field?
A. 4.3 mT
B. 68 mT
C. 0.11 mT
D. 3.0 mT
22. An proton of mass m p = 1.67 × 10−27 kg is fired into a chamber where the perpendicular magnetic field
strength B = 3.22 × 10−3 T . If the proton’s velocity v = 3.7 × 103
A. 0.12 mm
B. 0.81 mm
m
, what is the radius of its cyclotron orbit?
s
C. 81 mm
D. 12 mm
23. What is the magnetic field at the center of a circular loop with a diameter 15.0 cm that carries a
current of 1.50 A?
A. 1.68 x 10-4 T
B. 6.28 x 10-6 T
C. 1.26 x 10-5 T
9
D. 2.51 x 10-8 T
Name: ________________________
ID: A
24. The LR circuit shown below has a resistor R = 25 Ω , an inductor L = 5.4 mH , and a battery V = 9.0
V. How much energy is stored in the inductor of this circuit after the switch S is closed for a long
time and a steady current is achieved?
A. zero
B. 0.97 mJ
C. 0.35 mJ
D. 0.70 mJ
25. What is the time constant of an RL circuit with a resistance R = 25.0 Ω and an inductance L = 3.40
mH?
A. 7350 s
B. 3.43 s
C. 0.136 ms
D. 77.5 ms
26. What is the self inductance of a solenoid of length l = 15.0 cm, cross sectional area A = 30.0 cm2 and number
of windings N = 3200?
A. 0.258 H
B. 0.806 H
C. 2580 H
10
D. 8060 H
ID: A
UCSD Physics 2B
Answer Section
Practice Exam 4
Magnetism & Induction
MULTIPLE CHOICE
1. ANS: C
The vector resulting from any cross product is perpendicular to both input vectors
TOP: FORCE
2. ANS: B
F = qv × B
The right-hand-rule gives west × north = down (into the page), so q must be
negative. Now solve for the magnitude (sin 90° = 1)
|F | = | q v × B | = qvB sin θ
ÊÁ
ˆ
Á 7.8 × 10 −16 N ˜˜
|F |
Ë
¯
|q | = | | =
= 4.0 × 10 −18 C
Ê
ˆ
Ê
6
| vB | ÁÁ 3.5 × 10 m / s ˜˜ ÁÁ 5.6 × 10−5 T ˆ˜˜
Ë
¯Ë
¯
TOP: FORCE
3. ANS: D
F = Il×B
j × i = −k so force is in negative z-direction (or just apply the right hand rule). Solve
for the magnitude:
F
N
= I B = (6.3A) (1.5T ) = 9.5 N/m Hence F = −9.5 k8
l
m
TOP: FORCE | WIRE
4. ANS: C
Force between two parallel wires of length l separated by distance d carrying different currents :
μ0 I 1 I 2 l
F=
. So force per unit length for two wires carrying the same current:
2π d
Ê
ˆ
ÁÁ 1.26 × 10 −6 H / m˜˜ (1.0A)
2
F μ0 I
N
Ë
¯
=
=
= 2.0 × 10−7
l
m
2π d
(6.28) (1.0m)
2
TOP: PARALLEL WIRES
5. ANS: D
ˆ2
ˆ
Ê
Ê
μ = N I A = N I ÁÁ π R 2 ˜˜ = (20) (2.5A) (3.14) ÁÁ 5.0 × 10 −2 m˜˜ = 0.39A ⋅ m2
¯
¯
Ë
Ë
TOP: DIPOLE MOMENT
1
ID: A
6. ANS: C
Ê
ˆ
The magnetic moment of the coil μ = N I A = N I ÁÁ π R 2 ˜˜
Ë
¯
Ê
ˆ
2˜
Á
The torque on the coil τ = μ × B = μ B sinϑ = Á N Iπ R ˜ B sin ϑ
Ë
¯
2
ÊÁ
−2 ˆ
τ = (20) (2.5A) (3.14) Á 5.0 × 10 m˜˜ (0.15 T) sin 30 = 2.9 × 10 −2 N ⋅ m
Ë
¯
TOP: DIPOLE TORQUE
7. ANS: B
B=
ˆ
ˆ2
Ê
Ê
(30) ÁÁ 1.26 × 10 −6 H / m˜˜ (5.0A) ÁÁ 6.0 × 10 −2 m˜˜
¯
¯
Ë
Ë
=
= 5.6 × 10 −4 T
3/2
ÈÍ
˘
ÍÊ
ˆ2 Ê
ˆ 2 ˙˙
2 ÍÍÍÍ ÁÁ 6.0 × 10 −2 m˜˜ + ÁÁ 6.0 × 10 −2 m˜˜ ˙˙˙˙
¯
¯ ˙˚
Ë
ÍÎ Ë
μ0 Ia 2
ˆ3/2
Ê
2 ÁÁ x 2 + a 2 ˜˜
¯
Ë
TOP: LOOP FIELD
8. ANS: B
Point your right thumb in the direction of the bottom wire current and your fingers curl pointing into
the page. Do the same with the top wire and your fingers also curl into the page.
TOP: FIELD CONCEPT
9. ANS: D
Since a changing magnetic field induces an emf (voltage), calculate the power using
ÊÁ Δ (B x h ) / Δt ˆ˜ 2
ÊÁ B h (Δx / Δt) ˆ˜ 2
2
2
(B h v)
V 2 (ΔΦ / Δt)
Ë
¯
Ë
¯
P=
=
=
=
=
.
R
R
R
R
R
ÊÁ (3.2T ) (2.2m) (3.3m / s) ˆ˜ 2
Ë
¯
P=
= 3.6W
150Ω
TOP: FARADAY LOOP
10. ANS: A
μ0 I
. To calculate the total flux, we integrate:
2π r
ˆ
μ0 IL ÊÁ ˆ˜
dr ˜˜˜˜ μ0 IL
=
(ln B − lnA) =
lnÁÁ b ˜˜
˜
˜
2π
2π
r ˜
Ë a¯
The field from a long, straight wire B =
ÊÁ
Á
Φ = B • dA = ÁÁÁÁ
ÁË
ÿ
∫
L
0
ˆ˜ ÁÊÁ μ I
˜ 0
dl ˜˜˜˜ ÁÁÁÁ
˜¯ Á 2π
Ë
∫
b
a
¯
TOP: FLUX
11. ANS: C
Since a point at distance r = 5.0 cm is outside the wire, an Amperean Loop surrounds all the current.
Hence the diameter of the wire itself is of no consequence.
Ê
ˆ
ÁÁ 1.26 × 10 −6 H / m˜˜ (75A)
μ0 I
Ë
¯
B=
=
= 3.0 × 10 −4 T = 0.30mT
ÊÁ
−2 ˆ
2π r
˜
(6.28) Á 5.0 × 10 m˜
Ë
¯
TOP: WIRE FIELD OUTSIDE
2
ID: A
12. ANS: D
ÊÁ
ˆ
Á 1.26 × 10 −6 H / m˜˜ (3.0A) (350)
Ë
¯
B = μ0 In = μ0 IN / L =
= 8.8 × 10 −3 T
−2
15 × 10 m
TOP: SOLENOID
13. ANS: B
The magnetic flux is the product of the coil area A, the number of turns N, the magnetic field
strength B and the cosine of the angle at which the field lines penetrate the coil:
2
Φ = NB • A = NBA cos ϑ = (300) (0.5T ) (10cm) (cos 60) = 0.75 Wb
TOP: FLUX
14. ANS: B
Two ways to view this problem:
Ê
Ë
ˆ
¯
1. By Faraday's Law, the induced emf = − ÁÁÁ dΦ dt ˜˜˜ so curl the finger of the right hand in the direction
of the current (emf) and the thumb points to the right. Hence the flux is increasing to the left and so
the magnet (North pole to the left) must be moving to the left.
2. Lenz's Law says the current tries to oppose the change in flux. Since the current loop makes
magnetic North pole pointing to the right, it is trying to repel the bar magnet so it must be moving to
the left.
TOP: LENZ
15. ANS: C
ˆ
−4 2 ˜
ÁÊ
| dΦ | Δ (B • A) (0.35T ) (100) ÁË 15 × 10 m ˜¯ cos (55)
|EMF | = | −
=
= 9.1 × 10 −2 V
|=
0.33s
Δt
| dt |
TOP: FARADAY
16. ANS: B
Since the fields point in the same direction, they reinforce at the center of the inner solenoid, that is,
B + B = 2B.
TOP: SOLENOID CONCEPT
17. ANS: C
Field at the center of a full circle is B =
μ0 I
2R
so the field for a semicircle must be half that.
ˆ
ÊÁ
Á 1.26 × 10 −6 N / m˜˜ (3.75A)
¯
Ë
B=
=
= 0.358T
Ê
ˆ
−
3
4R
4 ÁÁ 3.30 × 10 m˜˜
Ë
¯
μ0 I
TOP: FIELD SUPERPOSITION
3
ID: A
18. ANS: C
Field at the center of a full circle is B =
μ0 I
2R
so the field for a semicircle must be half that. The two
semicircular currents flow counter-clockwise so the field contributions add:
ˆ
ÊÁ
Á 1.26 × 10 −6 N / m˜˜ (23A)
ˆ
˜
¯
1 ˜˜ Ë
B=
+
Á
˜=
4
4 ÁÁË R 1 R 2 ˜˜¯
μ0 I ÊÁÁ 1
Á
ÁÊÁ 1
1 ˜ˆ˜˜
−4
ÁÁ
Á 0.013m + 0.020m ˜˜ = 9.2 × 10 T
Ë
¯
TOP: FIELD SUPERPOSITION
19. ANS: E
Lenz's Law states that the induced current produces a magnetic which tends to oppose the change in
the magnetic field which created it. Since the current alternates 60 times per second, this is faster
than the time for an induced current in the ring to die out. Therefore, when the core B points up, the
ring B points down and vice-versa. Hence N-N or S-S oppose each other i.e., "like poles repel" and
the ring flies up by the repulsive force.
TOP: LENZ DEMO
20. ANS: D
By Definition.
TOP: FARADAY CONCEPT
21. ANS: D
The change in Flux is the difference between the initial value and zero:
| dΦ | Δ ( NB • A) NBA cos θ
|EMF | = | −
=
|=
t
Δt
| dt |
ˆÊ
ˆ
ÊÁ
Á 110 × 10 −3 V ˜˜ ÁÁ 13 × 10 −3 s ˜˜
¯Ë
¯
Ë
⇒B=
=
= 3.0mT
2
2
ˆ
Ê
Ns cos θ
(450) ÁÁ 3.6 × 10−2 m˜˜ cos (35)
¯
Ë
|EMF | × t
TOP: FARADAY GENERATOR
22. ANS: D
ÊÁ
ˆÊ
ˆ
Á 1.67 × 10−27 kg ˜˜ ÁÁ 3.7 × 103 m / s ˜˜
¯Ë
¯
mv Ë
r=
= Ê
= 12 mm
ˆ
ˆ
Ê
−19
−3
qB
ÁÁ 1.60 × 10 C ˜˜ ÁÁ 3.22 × 10 T ˜˜
¯Ë
¯
Ë
TOP: CYCLOTRON RADIUS
4
ID: A
23. ANS: C
The field at the center of a current loop is
B=
μ0 NI
2r
ÊÁ
ˆ
Á 1.26 × 10 −6 H / m˜˜ (1.50A)
Ë
¯
=
=
= 1.26 × 10 −5 T
ÊÁ
−2 ˆ
D
˜
Á 15.0 × 10 m˜
Ë
¯
μ0 I
TOP: LOOP FIELD
24. ANS: C
When steady current is achieved, the magnetic field in the inductor is no longer changing, so the
inductor becomes "invisible" to the current flow. Hence, the current I = V/R and the stored energy is
thus
U=
2
1 2 1 ÊÁÁÁ V ˆ˜˜˜
1Ê
ˆ ÊÁ 9.0V
L I = L ÁÁ ˜˜ = ÁÁ 5.4 × 10 −3 H ˜˜ ÁÁÁÁ
2
2 ËR¯
2Ë
¯ Ë 25Ω
ˆ˜ 2
˜˜ = 3.5 × 10 −4 J
˜˜
¯
TOP: INDUCTOR ENERGY + CONCEPT
25. ANS: C
The time constant (see text p.840) for an RL circuit is
τ=
L 3.40 × 10−3 H
=
= 1.36 × 10 −4 s = 0.136ms
R
25.0 Ω
TOP: LR CIRCUIT
26. ANS: A
ˆ2
Ê
ÁÊÁ 1.26 × 10 −6 H / m˜ˆ˜ (3200) 2 ÁÊÁ 30.0cm2 ˜ˆ˜ ÁÁÁÁ m ˜˜˜˜
¯
¯ ÁË 100cm ˜¯
Ë
μ0 N 2 A Ë
2
L = μ0 n Al =
=
= 0.258H
ÊÁ
ˆ
l
Á 15.0 × 10 −2 m˜˜
Ë
¯
TOP: SELF-INDUCTANCE
5