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Physics 9 Fall 2009
Homework 8 - Solutions
1. Chapter 34 - Exercise 9.
The current in the solenoid in the figure is increasing. The solenoid is surrounded by a conducting
loop. Is there a current in the loop? If so, is the
loop current cw or ccw?
————————————————————————————————————
Solution
The magnetic field in the solenoid points to the right, by the right-hand rule for currents. The flux is increasing, making the magnetic field bigger. There is a current
induced in the loop in such a way that it resists the change in flux. This means that
the the magnetic field created by the loop has to point to the left. This requires a
current moving in the counter-clockwise direction.
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2. Chapter 34 - Exercise 13.
A 1000-turn coil of wire 2.0 cm in diameter is in a magnetic field that drops from 0.10
T to 0 T in 10 ms. The axis of the coil is parallel to the field. What is the emf of the
coil?
————————————————————————————————————
Solution
~ ~
We know that the induced emf in a coil of N turns is E = N dΦdtm , where
Φm = B · A =
BA cos θ is magnetic flux. Since the area is constant, then E = N A Ḃ , where a dot
over a quantity denotes the time derivative. For the coil, A = πR2 , and Ḃ =
Thus,
∆B 0 − 10 2
2
= π (0.01) (1000) E = πR N 10−3 = 3.14 V.
∆t 2
∆B
.
∆t
3. Chapter 34 - Exercise 21.
How much energy is stored in a 3.0 cm-diameter, 12 cm-long solenoid that has 200
turns of wire and carries a current of 0.80 A?
————————————————————————————————————
Solution
The energy stored in an inductor is E = 21 LI 2 , where L is the inductor. Now, for a
2
solenoid, L = µ0 N` A , where A = πR2 is the area of the solenoid, and ` is the length.
So,
1
µ0 N 2 A 2
E = LI 2 =
I .
2
2`
So, this solenoid has an energy
µ0 N 2 A 2 (4π × 10−7 ) × 2002 × π (0.015)2
I =
(0.8)2 = 9.5 × 10−5 J.
E=
2`
2 × 0.12
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4. Chapter 34 - Problem 34.
A circular loop made from a flexible, conducting wire is shrinking. Its radius as a
function of time is r = r0 e−βt . The loop is perpendicular to a steady, uniform magnetic
field B. Find an expression for the induced emf in the loop at time t.
————————————————————————————————————
Solution
d
~
~
For a given flux, Φm = B · A = BA (t), in this case, then the emf is E = dt Φm = B Ȧ,
where the dot denotes the time derivative, as we have seen before, Ȧ ≡ dA
. The area
dt
2 −2βt
2
2 −2βt
. Thus,
, and so Ȧ = −2βπr0 e
of the loop is A (t) = πr (t) = πr0 e
E = −2βπr02 Be−2βt = 2βπr02 Be−2βt .
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5. Chapter 34 - Problem 43.
A loop antenna, such as is used on a television to pick up UHF broadcasts, is 25
cm in diameter. The plane of the loop is perpendicular to the oscillating magnetic
field of a 150 MHz electromagnetic wave. The magnetic field through the loop is
B = (20nT) sin ωt.
(a) What is the maximum emf induced in the antenna?
(b) What is the maximum emf if the loop is turned 90◦ to be perpendicular to the
oscillating electric field?
————————————————————————————————————
Solution
(a) The area of the loop is fixed, and perpendicular
to the magnetic field, so Φm =
dΦ m
~
~
B · A = BA, and so the emf is E = dt = A Ḃ , where B = B0 sin (ωt). Thus,
E = πr2 ωB0 cos (ωt).
The cosine varies between ±1, and has a maximum at 1. So, Emax = πr2 ωB0 =
2
2π 2 r2 f B0 , since ω = 2πf . Thus, Emax = 2π 2 (12.5 × 10−2 ) (150 × 106 ) (20 × 10−9 ) =
0.93 V.
(b) Since the electric and magnetic fields are perpendicular to each other, and mutually perpendicular to the direction of propagation of the wave, when the plane of
the loop is perpendicular to the oscillating electric field, then it is parallel to the
magnetic field. So, the net flux through the antenna is zero, which gives a net
induced emf of zero.
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6. Chapter 34 - Problem 52.
You’ve decided to make a magnetic projectile
launcher for your science project. An aluminum
bar of length ` slides along metal rails through
a magnetic field B. The switch closes at t = 0
s, while the bar is at rest, and a battery of emf
Ebat starts a current flowing around the loop. The
battery has internal resistance r. The resistance
of the rails and the bar are effectively zero.
(a) Show that the bar reaches a terminal speed vterm , and find an expression for vterm .
(b) Evaluate vterm for Ebat = 1.0 V, r = 0.10Ω, ` = 6.0 cm, and B = 0.50 T.
————————————————————————————————————
Solution
(a) When the switch closes, the battery causes a current to flow. This causes the
bar to move since the current is moving in a magnetic field. The moving bar
changes the flux in the loop which induces a new emf that fights the battery’s
emf. This builds up as the velocity gets faster until the induced emf is equal to
that of the battery. Then, the net emf is zero and the bar coasts along at the
terminal velocity, vterm . So, E = `Bvterm . This gives
vterm =
E
.
`B
(b) For the given numbers,
vterm =
1
E
=
= 33 m/s.
`B
0.06 × 0.5
This is about 70 miles per hour!
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7. Chapter 34 - Problem 53.
A slide wire of length `, mass m, and resistance R slides down a U-shaped metal track
that is tilted upward at angle θ. The track has zero resistance and no friction. A
vertical magnetic field B fills the loop formed by the track and the slide wire.
(a) Find an expression for the induced current I when the slide wire moves at speed
v.
(b) Show that the slide wire reaches a terminal speed vterm , and find an expression
for vterm .
————————————————————————————————————
Solution
The setup is as seen in the diagram below.
(a) Here, all we are doing is tilting our usual current loop. This changes the flux
~ ·A
~ = BA cos θ. The induced current is I = E/R, where R is the
to Φm = B
resistance of the wire. So,
I=
E
1
B
B
B`v
= Φ̇m = cos θȦ = cos θ`ẋ =
cos θ,
R
R
R
R
R
where we set v = ẋ. Thus, I =
B`v
R
cos θ.
(b) The induced force on the bar is back to the left, pushing the bar up along the
loop (along −x as we’ve drawn things), while gravity pulls the bar down (along
the +x direction). The bar coasts along at a constant speed when the two forces
are equal to each other. Looking at the forces along the x−direction gives
X
Fx = −Fm cos θ + Fg sin θ = 0.
2 2
Now, Fm = I`B = B ` Rvterm cos θ, while Fg = mg. Setting the two equal and
solving for the terminal velocity gives
vterm =
7
mgR tan θ
.
`2 B 2 cos θ
8. Chapter 34 - Problem 65.
MRI (magnetic resonance imaging) is a medical technique that produces detailed “pictures” of the interior of the body. The patient is placed into a solenoid that is 40 cm
in diameter and 1.0 m long. A 100 A current creates a 5.0 T magnetic field inside
the solenoid. To carry such a large current, the solenoid wires are cooled with liquid
helium until they become superconducting (no electric resistance).
(a) How much magnetic energy is stored in the solenoid? Assume that the magnetic
field is uniform within the solenoid and quickly drops to zero outside the solenoid.
(b) How many turns of wire does the solenoid have?
————————————————————————————————————
Solution
(a) The energy density stored in the magnetic field is given by uB = 2µ1 0 B 2 . The total
energy is the energy density times the volume of the solenoid, UB = uB (Vol) =
2
. So,
uB A` = B2µA`
0
UB =
A` 2
π (0.22 )
B =
(5.0)2 = 1.25 × 106 J.
2µ0
2 × 4π × 10−7
(b) The magnetic field of a solenoid is B = µ0`N I , and so N =
100 A, we have
5×1
B`
=
≈ 4 × 105 ,
N=
−7
µ0 I
4π × 10 100
or about 40,000 turns!
8
B`
.
µ0 I
For a current of
9. Chapter 34 - Problem 66.
One possible concern with MRI (see problem 65) is turning the magnetic field on or off
too quickly. Bodily fluids are conductors, and a changing magnetic field could cause
electric currents to flow through the patient. Suppose a typical patient has a maximum
cross-section area of 0.060 m2 . What is the smallest time interval in which a 5.0 T
magnetic field can be turned on or off if the induced emf around the patient’s body
must be kept to less than 0.10 V?
————————————————————————————————————
Solution
The emf is E = Φ̇m . For a constant area, oriented along the direction of the magnetic
. So, the amount of time, ∆t ≈
field, then Φm = AB. So, Φ̇m = AḂ ≈ A ∆B
∆t
Plugging in the numbers give
∆t =
A∆B
.
E
0.06 × 5
A∆B
=
= 3 sec.
E
0.1
So, we should turn the magnetic field on or off over a time interval longer than 3
seconds.
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10. Chapter 34 - Problem 80.
The switch in the figure has been open for a long
time. it is closed at t = 0 s.
(a) After the switch has been closed for a long
time, what is the current in the circuit. Call
this current I0 .
(b) Find an expression for the current I as a
function of time. Write your expression in
terms of I0 , R, and L.
(c) Sketch a current-versus-time graph from t =
0 s until the current is no longer changing.
————————————————————————————————————
Solution
P
(a) From Kirchhoff’s loop law,
∆V = ∆Vbatt − IR − L dI
= 0. When the current
dt
has been flowing for a long time, it no longer changes, and so I˙ = 0. Thus,
I0 = ∆VRbatt .
(b) Now, going back to the loop law, we have, upon solving for
dI
,
dt
dI
R
= (I0 − I) ,
dt
L
after plugging in ∆Vbatt = I0 R. Separating this differential equation becomes
dI
dt. Now, integrating gives
= −R
I−I0
L
Z
I
I0
dI
R
=−
I − I0
L
Z
t
dt ⇒ ln
0
I − I0
−I0
Solving for I (t) gives
I(t) = I0 1 − e−t/τ ,
where τ ≡ L/R.
The graph is seen to the right. The current starts
(c) out at zero at time t = 0, and asymptotically
approaches the final current I0 as t → ∞.
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R
= − t.
L