Physics 9 Fall 2009
Homework 7 - Solutions
1. Chapter 33 - Exercise 10.
At what distance on the axis of a current loop is the magnetic field half the strength
of the field at the center of the loop? Give your answer as a multiple of R.
————————————————————————————————————
Solution
The magnetic field on the axis of a current loop is given by
B (z) =
IR2
µ0
.
2 (z 2 + R2 )3/2
We want a distance z, such that B (z) = 12 B (z = 0). The field at the center of the
0I
. So, we want
loop is B (z = 0) = µ2R
µ0
IR2
1 µ0 I
.
=
3/2
2
2
2 (z + R )
2 2R
3/2
Solving, we find 2R3 = (z 2 + R2 )
, or z =
1
√
41/3 − 1R. So, z ≈ 0.77R.
2. Chapter 33 - Exercise 25.
Magnetic resonance imaging needs a magnetic field strength of 1.5 T. The solenoid
is 1.8 m long and 75 cm in diameter. It is tightly wound with a single layer of 2.0
mm-diameter superconducting wire. What size current is needed?
————————————————————————————————————
Solution
. What’s
The magnetic field of a solenoid is B = µ0LN I , and so the current is I = µBL
0N
N ? It’s just the length of the wire, divided by the diameter of the wire - this is
how many times you can wind the coils together for a given length. So, there are
N = 1.8/0.002 = 900 turns of the wire.
Thus,
I=
BL
1.5 × 1.8
=
= 2400 A.
µ0 N
4π × 10−7 × 900
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3. Chapter 33 - Exercise 29.
Radio astronomers detect electromagnetic radiation at 45 MHz from an interstellar gas
cloud. They suspect this radiation is emitted by electrons spiraling in a magnetic field.
What is the magnetic field strength inside the gas cloud?
————————————————————————————————————
Solution
The frequency of the emitted radiation is equal to the frequency of the acceleration of
the electron (i.e., the frequency of the revolutions). The revolution of the frequency
qB
. Solving for the magnetic field gives
is just the cyclotron frequency, which is f = 2πm
2πf m
B = q . Plugging in the values gives
B=
2π45 × 106 × 9.11 × 10−31
= 1.61 × 10−3 T.
1.602 × 10−19
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4. Chapter 33 - Problem 45.
The element niobium, which is a metal, is a superconductor (i.e., no electrical resistance) at temperatures below 9 K. However, the superconductivity is destroyed if
the magnetic field at the surface of the metal reaches or exceeds 0.10 T. What is the
maximum current in a straight, 3.0 mm-diameter superconducting niobium wire?
————————————————————————————————————
Solution
0I
, where r is the distance away
The magnetic field of a long straight wire is B = µ2πr
from the center. So, at the surface of the wire, when r = d, the current in the wire is
I=
2πBd
.
µ0
Plugging in the values gives the maximum current as
I=
2πBd
2π × 0.10 × 0.0015
= 750 A.
=
µ0
4π × 10−7
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5. Chapter 33 - Problem 46.
(a) Find an expression for the magnetic field at
the center (point P) of the circular arc in
the figure.
(b) Does your result agree with the magnetic
field of a current loop when θ = 2π?
————————————————————————————————————
Solution
~ = µ0 I∆~s2×r̂ , where ∆~s points along the current, and
(a) Recall the Biot-Savart law, B
4π r
r̂ points long the direction from the current to the point P. Now, because the
current and r̂ point along the same line in both straight segments of the wire,
their magnetic field contributions to the point P are zero. So, we only need to
find the magnetic field from the curved part of the wire.
Along the arc, the current points along the wire, which is always at right angles
to r̂. So, along the arc, ∆~s × r̂ = |∆~s||r̂| = ∆s, since r̂ is a unit vector. So, along
the arc the little bit of the magnetic field from the little section ∆s is
∆B =
µ0 I ∆s
.
4π r2
Now, the segment is always a distance r = R from the point P. So, we just have
to add up the distance, ∆s, along the arc, which is ∆s = Rθ. Thus,
B=
µ0 Iθ
.
4πR
(b) Recall that the magnetic field of a loop of wire is
B=
µ0
IR2
,
2 (z 2 + R2 )3/2
0I
which gives the field at the center, when z = 0, as Bcenter = µ2R
, which is precisely
the correct result in part a when θ = 2π. So, our results agree.
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6. Chapter 33 - Problem 56.
An electron orbits in a 5.0 mT field with angular momentum 8.0 × 10−26 kg m2 /s.
What is the diameter of the orbit?
————————————————————————————————————
Solution
Recall that the electron makes an orbit in a magnetic field of radius r = mv
. The
eB
angular momentum of a particle in an orbit is L = mvr. So, multiplying the expression
for the radius of the orbit by r gives
r2 =
L
mvr
=
,
eB
eB
and so the radius can be expressed as
r
r=
L
.
eB
With numbers, we find that
r
r=
8.0 × 10−26
= 0.01 m.
1.602 × 10−19 × 5.0 × 10−3
So, the electron orbits at a distance of about 1 cm.
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7. Chapter 33 - Problem 61.
An antiproton (same properties as a proton except
that q = −e) is moving in the combined electric
and magnetic fields of the figure.
(a) What are the magnitude and direction of the
antiproton’s acceleration at this instant?
(b) What would be the magnitude and direction
of the acceleration if ~v were reversed?
————————————————————————————————————
Solution
~ and
(a) The antiproton is negatively charged, so the electric force is F~E = −eE,
~ and points down. So the
points up, while the magnetic force is F~M = −e~v × B,
two forces work
against each other. The net force is just the sum of the forces,
~
~
~ . The magnitude of the force is Fnet = e (E − vB) =
Fnet = −e E − ~v × B
1.602 × 10−19 (1000 − 500 × 2.5) ≈ 4 × 10−16 N. Because the magnetic force is
stronger, the force points down. Finally, since a = F/m, where m is the mass of
the proton, the acceleration is a = 2.4 × 1010 m/s2 , and still points down.
(b) If ~v was reversed,
then both forces would point up. The forces would add, such
~
~ + ~v × B
~ . In this case, the magnitudes would add, instead of
that Fnet = −e E
subtract, giving Fnet = 3.6 × 10−16 N, and points up. Again, the acceleration is
just a = F/m, giving a = 2.2 × 1011 m/s2 upwards.
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8. Chapter 33 - Problem 73.
In the semiclassical Bohr model of the hydrogen atom, the electron moves in a circular
orbit of radius 5.3 × 10−11 m with speed 2.2 × 106 m/s. According to this model, what
is the magnetic field at the center of a hydrogen atom? Hint: Determine the average
current of the orbiting electron.
————————————————————————————————————
Solution
The moving charge orbiting the nucleus constitutes a current, I. The magnitude of
the current is just the charge on the electron, divided by the time it takes to orbit
the nucleus, i.e., it’s orbital period, T . The period is the total distance around 2πR,
divided by the speed of the electron, v. So, T = 2πR/v, which gives
I=
ev
e
=
.
T
2πR
The magnetic field at the center of the loop is, as we’ve seen, B =
upon plugging in for the current,
B=
With numbers, we find B =
field!
µ0 ev
4πR2
=
which gives
µ0 ev
.
4πR2
10−7 ×1.602×10−19 ×2.2×106
(5.3×10−11 )2
8
µ0 I
,
2R
= 12.5 T, which is a huge
9. Chapter 33 - Problem 79.
A flat, circular disk of radius R is uniformly charged with total charge Q. The disk
spins at angular velocity ω about an axis through its center. What is the magnetic
field strength at the center of the disk?
————————————————————————————————————
Solution
Q
The plate is uniformly charged, so it’s surface charge density, η = Q
= πR
2 , is constant.
A
Consider a small bit of charge, dq. From the Biot-Savart law, the tiny bit of magnetic
field from the this charge is
~ =
dB
µ0 ~v × r̂
µ0 v
dq 2 =
dq k̂,
4π
r
4π r2
since ~v points along the direction of spin, which is tangent to the disk, while r̂ points
towards the z−axis. This means that ~v and r̂ are perpendicular. Now, for an object
spinning in a circle, v = rω, where ω is the angular velocity. Therefore,
dB =
µ0 ω
dq .
4π r
Next, we integrate, writing dq = ηdA = 2πrηdr, since we are on a disk, as we’ve seen
before. Then, dB = µ02ωη dr. Then, integrating both sides gives
µ0 ωη
B=
2
Z
R
dr =
0
µ0 ωηR
.
2
We can express this in terms of the total charge, Q, on the disk by recalling that
Q
η = πR
2 , which gives
µ0 ωQ
,
B=
2πR
and points up.
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10. Chapter 33 - Problem 81.
A long, straight conducting wire of radius R has a nonuniform current density J =
J0 r/R, where J0 is a constant. The wire carries total current I.
(a) Find an expression for J0 in terms of I and R.
(b) Find an expression for the magnetic field strength inside the wire at radius r.
(c) At the boundary, r = R, does your solution match the known field outside a long,
straight current-carrying wire?
————————————————————————————————————
Solution
R
(a) For a varying current density, the total current in the wire is I = JdA. For a
disk, dA = 2πrdr, as we’ve seen before. Integrating over the total surface area of
the disk gives the total current, I, on the wire. So,
Z
Z
2πJ0 R3
2πR2
2πJ0 R 2
r dr =
=
J0 ,
I = JdA =
R 0
R 3
3
and so J0 =
3
I.
2πR2
H
~ · d~s = µ0 Iencl . Because of the
(b) Here we will use Ampere’s law, which says that B
symmetry of the wire, the magnetic field circles around inside the wire. So, we
choose a circle of radius r ≤ R for our Amperian loop. Since the magnetic field
is constant
H along that loop, and points along d~s, the left-hand side of Ampere’s
~ · d~s = B (2πr). Now we just need the enclosed current.
law says B
We can use the same method as in part (a) to determine the enclosed current only now we integrate out to radius r, since we only want the enclosed current.
Thus,
2
Z
Z
2πr2
r
2πJ0 r 2
Iencl = JdA =
r dr =
J0 = I
,
R 0
3
R2
where in the last step we inserted our value for J0 . Thus, Ampere’s law gives
B (2πr) = µ0 Ir2 /R2 , or
µ0 I
r.
B=
2πR2
0I
(c) The field outside a long straight wire with current I is given by B = µ2πr
. At the
µ0 I
boundary, we have r = R, and so B = 2πR , which is precisely the correct answer
that we get from part (b) at the boundary.
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