ID: A
UCSD Physics 2B
Answer Section
Unit Exam 4A
Magnetism & Induction
MULTIPLE CHOICE
1. ANS: B
rc =
mv
qB
ÊÁ
ˆ˜ ÊÁ
−19
−3 ˆ
−3 ˆ
˜ ÊÁ
˜
qBr c ÁË 1.60 × 10 C ˜¯ ÁË 3.22 × 10 T ˜¯ ÁË 28.1 × 10 m˜¯
m
⇒ v=
=
= 8.67 × 10 3
s
m
ÁÊÁ 1.67 × 10 −27 kg ˆ˜˜
Ë
¯
TOP: CYCLOTRON RADIUS
2. ANS: D
For the trajectory to be undisturbed, the combined net force must be zero:
F = q (E + vB)
F = 0 ⇒ E = −vB =
ÊÁ
ˆ
ÁÁ 7.33 × 10 6 m ˜˜˜ ÊÁ 8.31 × 10 −5 T ˆ˜˜ = 609 V
ÁÁ
m
s ˜˜¯ ÁË
¯
Ë
TOP: CYCLOTRON RADIUS
3. ANS: B
μ0 NIa 2
B=
Ê
ˆ3/2
2 ÁÁ z 2 + a 2 ˜˜
Ë
¯
ˆ
ˆÊ
ˆ2
ÊÁ
Ê
Á 1.26 × 10−6 H / m˜˜ (150) ÁÁ 125 × 10−3 A ˜˜ ÁÁ 3.0 × 10 −2 m˜˜
¯
¯Ë
¯
Ë
Ë
=
= 1.39 × 10 −4 T = 139μ T
3/2
ÈÍ
˘
ÍÊ
ˆ2 Ê
ˆ 2 ˙˙
2 ÍÍÍÍ ÁÁ 3.0 × 10 −2 m˜˜ + ÁÁ 3.0 × 10 −2 m˜˜ ˙˙˙˙
¯
Ë
¯ ˙˚
ÍÎ Ë
TOP: LOOP FIELD
4. ANS: D
The magnitude is |F| = I | l × B | = I l B sin θ = (1.5A) (3.0m) (1.0T )sin (58) = 3.8N
TOP: FORCE | WIRE
5. ANS: C
The magnetic moment is the number of turns * current * area
Ê
Ë
ˆ
¯
ˆ2
¯
Ê
Ë
μ = N I A = N I ÁÁ π R 2 ˜˜ = (100) (5.20A) (3.14) ÁÁ 12.0 × 10 −2 m˜˜ = 23.5 A ⋅ m2
TOP: DIPOLE MOMENT
1
ID: A
6. ANS: A
B = μ0 In = μ0 I
⇒L=
μ0 IN
B
N
L
ÊÁ
ˆ
Á 1.26 × 10 −6 H / m˜˜ (3.42A) (169)
¯
Ë
=
= 1.94m
375 × 10 −6 T
TOP: SOLENOID
7. 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
(decreasing to the right) and so the magnet (South pole to the left) must be moving to the right.
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, the loop is trying to attract the bar magnet so it must be
moving to the right.
TOP: LENZ
8. ANS: D
Ê
ˆ
The magnetic moment of the coil μ = N I A = N I ÁÁ π R 2 ˜˜
Ë
¯
ÊÁ
2ˆ
˜
The torque on the coil τ = μ × B = μ B sinθ = Á NIπ R ˜ B sin θ
Ë
¯
2
ˆ
Ê
τ = (200) (57.0A) (3.14) ÁÁ 25 × 10 −2 m˜˜ (0.265 T) sin13 = 133 N • m
¯
Ë
TOP: DIPOLE TORQUE
9. ANS: B
We can use the “wire” formula which comes from making a circular Amperean Loop at radius
r = 0.20 cm. But first, we must find the current enclosed. To do this, we integrate the current density
over the enclosed area:
r Ê
π J0
2π J 0 r 3
Á r ˆ˜
I Enclosed = ∫ J • d A = ∫ ÁÁÁÁ J 0 ˜˜˜˜ 2π r dr =
∫ r dr = 2 r 4
2
0
ÁË
R ˜¯
R
0
2R
Now apply the enclosed current to the “wire” formula:
μ0 I Enc ÊÁÁÁ μ0 ˆ˜˜˜ π J 0 4
˜˜
B=
= ÁÁÁ
r
2π r
Á 2π r ˜˜ 2R 2
Ë
¯
ÊÁ
ˆÊ
ˆ
ÁÁ 1.26 × 10 −6 T ⋅ m ˜˜˜ ÁÁÁ 7600 A ˜˜˜ ÊÁÁ 2.50 × 10 −2 mˆ˜˜ 3
Á
˜
Á
˜
A ˜¯ ÁË
¯
μ0 J 0 r 3 ÁË
m2 ˜¯ Ë
=
=
= 3.05 × 10 −5 T = 30.5 μ T
2
2
Ê
ˆ
−2
4R
4ÁÁ 3.50 × 10 m˜˜
Ë
¯
TOP: WIRE FIELD INSIDE
2
ID: A
10. ANS: D
Applying Ampere's Law in a loop of radius 10 mm around the wire,
Ê
ˆ
−6
Á
˜
μ0 I ÁË 1.26 × 10 H / m˜¯ (52.0A)
B=
=
= 2.69 × 10 −4 T = 269 μ T
ÊÁ
−2 ˆ
2π r
˜
(6.28) Á 3.88 × 10 m˜
Ë
¯
TOP: WIRE FIELD
11. ANS: A
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:
Φ = NB • A = NBA cos θ
⇒ N=
Φ
=
BA cos θ
0.75 Wb
= 300
2
ÊÁ
−2 ˆ
˜
(0.5T ) Á 10 × 10 ˜ (cos 60)
Ë
¯
TOP: FLUX
12. 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 in opposite directions, so the fields subtract:
μ0 I ÊÁÁ 1
Á
1
B=
−
Á
4 ÁÁË R 1 R 2
ˆ
Ê
−6
ˆ˜ ÁÁË 1.26 × 10 N / m˜˜¯ (595A)
˜˜ =
˜˜
4
˜¯
ÊÁ 1
1 ˆ˜˜
−3
ÁÁ
ÁÁ 0.023m − 0.032m ˜˜˜ = 2.29 × 10 T = 2.29mT
Ë
¯
TOP: FIELD SUPERPOSITION
13. ANS: A
Lenz's Law says that the induced current flows in such a way as to oppose the change in magnetic
field. As the magnet falls through the copper tube, current begins to swirl beneath it making its own
magnetic field which points in the opposite direction, slowing the magnet's fall.
TOP: LENZ DEMO
3
ID: A
14. ANS: A
F = qv × B whose magnitude F = qvB sin θ .
Lay a coordinate system with east = i, north = j and up = k. Curl right-hand fingers from vector v to
B. If the thumb points out of the page (k), then the angle θ = 180° + 35° = 215°. If the thumb points
into the page (- k), then the angle θ = 180° − 35° = 145°. Now solve for magnitude:
Ê
ˆÊ
ˆ
F = qvB = ÁÁ −1.6 × 10 −19 C ˜˜ ÁÁ 2.5 × 10 5 m / s ˜˜ (0.45T ) = −1.80 × 10−14 N
Ë
¯Ë
¯
For the projection & direction, sin (215)k = −5.74 × 10 −1 k while the other choice of orientation gives
the same result: sin (145) (−k ) = 5.74 × 10 −1 (−k ) = −5.74 × 10 −1 (k )
Ê
Ë
ˆÊ
¯Ë
ˆ
¯
The final answer is thus F = ÁÁ −5.74 × 10 −1 ˜˜ ÁÁ −1.80 × 10 −14 N ˜˜ k = 1.03 × 10−14 N (k )
so the direction is "up" or out of the page.
TOP: FORCE
15. ANS: E
Since the fields point in opposite directions, they cancel at the center of the inner solenoid.
TOP: SOLENOID CONCEPT
16. ANS: C
Force between two parallel wires of length L separated by distance r carrying different currents :
μ0 I 1 I 2 L
F=
. So force per unit length for two wires carrying the same current:
2π r
2
ÊÁ
ˆ
−6
2
ÁË 1.26 × 10 H / m˜˜¯ (10A)
F μ0 I
N
=
=
= 2.01 × 10 −3
L 2π r
m
2 (3.14) (0.01m)
TOP: PARALLEL WIRES
17. ANS: C
Point your right thumb in the direction of the bottom wire current and your fingers curl pointing out
of the page at point P. Do the same with the top wire and your fingers point out of the page. Since
the distance from P to each wire is the same, the two fields cancel.
TOP: FIELD CONCEPT
4
ID: A
18. ANS: A
First we use Faraday’s Law to get the EMF. The flux through the loop is the field B times the area
A = h × x. The change in the flux is only due to the penetration depth x since the field B and height h
are fixed. The power dissipated by the resistor is
P=
V2
=
R
ÊÁ d Φ
ÁÁ
ÁÁ d t
Ë
ˆ˜ 2
˜˜
˜˜
¯
R
=
ÊÁ d
ˆ2
ÁÁ (B x h ) ˜˜˜
ÁÁ d t
˜˜
Ë
¯
R
2
ÊÁ
ÁÊÁ d x ˜ˆ˜ ˆ˜˜˜
ÁÁ
˜˜ ˜˜
ÁÁ B h ÁÁÁ
˜
Á
2
Ë d t ¯ ˜¯
(B h v)
Ë
=
=
R
R
Since the system is conservative, the mechanical power to pull the loop is the same:
Power =
force × dis tan ce
work
dis tance
=
= force ×
= force × velocity
time
time
time
2
v (B h )
P (B h v )
=
Hence, the force F = =
v
vR
R
=
2
2
ÁÊÁ
˜ˆ
ÁÁ 2.80 m ˜˜˜ ÊÁ (2.30T ) (3.60m) ˆ˜
Á
˜
¯
s ¯Ë
Ë
15.0 Ω
= 12.8N
Note that the actual penetration depth x is not needed, just the fact that part of the loop is still inside
the field region.
TOP: FARADAY LOOP
19. ANS: B
Let the loop orientation angle be θ (t) = ω t
emf = −
d
dΦ
d Ê
ˆ
= − ÁÁ Bs 2 cos θ (t) ˜˜ = −Bs 2
(cos ωt) = Bs 2 ω sin ωt
dt
dt
dt Ë
¯
Since the maximum of the sine function is 1,
ˆ2
Ê
maxÊÁË emf ˆ˜¯ = Bs 2 ω = (2.26T ) ÁÁ 13.7 × 10 −2 m˜˜ (120 / s) = 5.09V
¯
Ë
TOP: FARADAY GENERATOR
5