________________________________________ Name: Code # _________
Summer Session 2008
ID: X
UCSD Physics 2B
Practice Exam 4
READ ME !!!
Magnetism & Induction
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 = 50 to the direction of a
magnetic field B. The component of the resulting force in the direction of B is
A. evB sin(50) cos(50) B. evB sin(50)
C. zero D. evB cos(50)
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
D. -1.2 10-14 C
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Name: ________________________
ID: A
8 3. A long straight wire parallel to the y-axis carries a current of I = 6.3A j in a region of uniform 8 magnetic field B = 1.5T i . 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. 2 10 -6
N m
B. 2.0 10 -5
N m
C. 2.0 10 -7
N m
D. 2 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 Am2
B. 0.15 Am2
C. 3.9 kAm2
D. 0.39 Am2
6. A circular 20 turn coil of radius 5.0 cm is so oriented that its axis makes a 30 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
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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 T
B. 0.56 m T
C. 0.11 T
D. 56 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
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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
D. 3.6 W
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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. 0 B. 0 C. 0 D. 0 E. 0
~ / 4 ^ 2 I L ~ / 4 ^ 4 I L ~ / 4 ^ 2 I L ^ / 4 ~ 2 I L ~ / 4 ^ 4 I L
ln(b / a ) ln(b / a ) ln (b + a ) / (b - a ) (b - a ) / (b + a ) ln (b - a ) / (b + a ) ln
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
D. 0.56 mT
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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 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 = 60 with the x-axis. What is the approximate magnetic flux 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
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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 55 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
D. 716 mT
7
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
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Name: ________________________
ID: A
20. According to Faraday's Law, a necessary and sufficient condition an for 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 35 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
m , what is the radius of its cyclotron orbit? s
A. 0.12 mm
B. 0.81 mm
C. 81 mm
D. 12 mm
23. What is the magnetic field at the center of a circular loop with a diameter of 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
D. 2.51 x 10-8 T
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ID: A
UCSD Physics 2B Answer Section
MULTIPLE CHOICE 1. ANS: C
Practice Exam 4
Magnetism & Induction
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
7.8 10 -16 N ^ ~ ~ |F | |q | = | | = = 4.0 10 -18 C ^ 6 ~ | vB | 3.5 10 m / s ~ 5.6 10-5 T ^ ~ ~
TOP: FORCE 3. ANS: D F = IlB j i = -k so force is in negative z-direction (or just apply the right hand rule). Solve
for the magnitude:
F N 8 = I B = (6.3A) (1.5T ) = 9.5 N/m Hence F = -9.5 k l m
TOP: FORCE | WIRE 4. ANS: C
Force between two parallel wires of length l separated by distance d carrying different currents : 0 I 1 I 2 l F= . So force per unit length for two wires carrying the same current: 2 d
1.26 10 H / m~ (1.0A) 2 ~ F 0 I N = = = 2.0 10-7 l m 2 d (6.28) (1.0m)
2
-6
^
TOP: PARALLEL WIRES 5. ANS: D ^2 ^ ~ ~ = N I A = N I R 2 ~ = (20) (2.5A) (3.14) 5.0 10 -2 m~ = 0.39A m2 TOP: DIPOLE MOMENT
1
ID: A 6. ANS: C
^ ~ The magnetic moment of the coil = N I A = N I R 2 ~
N I R 2 ^ B sin ~ ~ The torque on the coil = B = B sin = 2 -2 ^ = (20) (2.5A) (3.14) 5.0 10 m~ (0.15 T) sin 30 = 2.9 10 -2 N m ~
TOP: DIPOLE TORQUE 7. ANS: B
B=
0 Ia 2
^3/2 ~ 2 x 2 + a 2 ~
^ ^2 ~ ~ (30) 1.26 10 -6 H / m~ (5.0A) 6.0 10 -2 m~ = = 5.6 10 -4 T 3/2 ^2 ^2 ~ ~ 2 6.0 10 -2 m~ + 6.0 10 -2 m~
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
(B x h ) / t ^ 2 B h (x / t) ^ 2 2 2 ~ ~ (B h v) V 2 ( / t) P= = = = = . R R R R R (3.2T ) (2.2m) (3.3m / s) ^ 2 ~ P= = 3.6W 150
TOP: FARADAY LOOP 10. ANS: A
The field from a long, straight wire B =
= B dA =
L
0
^ I ~ 0 ~ ~ dl ~ ~ 2 ~ ~
b
a
0 I . To calculate the total flux, we integrate: 2 r ^ 0 IL ^ ~ dr ~ 0 IL ~= ~ (ln B - lnA) = ln b ~ ~ ~ a~ ~ 2 2 r ~
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.
1.26 10 H / m~ (75A) ~ 0 I B= = = 3.0 10 -4 T = 0.30mT 2 r 5.0 10 -2 m^ ~ (6.28) ~
TOP: WIRE FIELD OUTSIDE 2
-6
^
ID: A 12. ANS: D
1.26 10 -6 H / m^ (3.0A) (350) ~ ~ B = 0 In = 0 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 = NB A = NBA cos = (300) (0.5T ) (10cm) (cos 60) = 0.75 Wb
TOP: FLUX 14. ANS: B
Two ways to view this problem:
~ 1. By Faraday's Law, the induced emf = - d 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
-4 2 ^ ~ | d | (B A) (0.35T ) (100) 15 10 m ~ cos (55) |EMF | = | - = = 9.1 10 -2 V |= 0.33s t | 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 =
0 I
0 I
2R
so the field for a semicircle must be half that.
~ 1.26 10 -6 N / m^ (3.75A) ~ B= = = 0.358T -3 ^ 4R ~ 4 3.30 10 m~
TOP: FIELD SUPERPOSITION
3
ID: A 18. ANS: C
Field at the center of a full circle is B =
0 I
2R
so the field for a semicircle must be half that. The two
semicircular currents flow counter-clockwise so the field contributions add:
^ -6 ~ ^ 1.26 10 N / m~ (23A) ~ 1 ~ ~= B= + ~ 4 4 R1 R2 ~ ~
0 I 1
1 ~ 1 ^ -4 ~ 0.013m + 0.020m ~ = 9.2 10 T ~ ~
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:
| d | ( NB A) NBA cos |EMF | = | - = |= t t | dt | ~ ~ 110 10 -3 V ^ 13 10 -3 s ^ ~ ~ B= = = 3.0mT 2 2 ^ Ns cos (450) 3.6 10-2 m~ cos (35) ~ |EMF | t
TOP: FARADAY GENERATOR 22. ANS: D ^ ^ 1.67 10-27 kg ~ 3.7 103 m / s ~ ~ ~ mv r= = = 12 mm -19 ^ -3 ^ qB ~ ~ 1.60 10 C ~ 3.22 10 T ~ TOP: CYCLOTRON RADIUS
4
ID: A 23. ANS: C
The field at the center of a current loop is
B=
0 NI
2r
1.26 10 -6 H / m^ (1.50A) ~ ~ = = = 1.26 10 -5 T D 15.0 10 -2 m^ ~ ~
0 I
TOP: LOOP FIELD
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