Download Physics 213 — Problem Set 8 —Solutions Spring 1998

Physics 213 — Problem Set 8 —Solutions
Spring 1998
1. Reading Assignment
Serway Sections 30.1–30.5. Next week Chapter 31.
2. Serway 30.12
Consider the current-carrying loop shown in Figure P30.12 of your text, formed of radial lines and
segments of circles whose centers are at point P . Find the magnitude and direction of B at P.
SOLUTION: We use the Biot-Savart law
µ0 I
B=
4π
Z
d` × r̂
r2
to compute the magnetic field at the origin. For the radial segments, d` is parallel to r̂ making the integrand
zero, so we only need to worry about the contribution from the circular arcs. Along the arc at radius a we
have that r = a and d` = a dφ φ̂, where φ is the angular (azimuthal) variable and φ̂ is a unit vector in the
azimuthal direction. Thus d` × r̂ = a dφ ẑ, where we have used φ̂ × r̂ = ẑ where ẑ is the unit vector pointing
up out of the page (in Figure P30.12). Similarly, along the arc at radius b we have r = b, d` = −b dφ φ̂
(since the current is flowing in the opposite direction), so d` × r̂ = −b ẑ. Putting this all together in the
Biot-Savart law gives
Z π/3
µ0 Iẑ 1
µ0 I 1
1
1
B=
ẑ.
dφ =
−
−
4π
a
b
12 a
b
0
3. Serway 30.16A
Two long, parallel wires, each having a mass per unit length µ, are supported in a horizontal plane by
strings of length L, as show in Figure P30.16 of your text. Each wire carries the same current I, causing
the wires to repel each other so that the angle between the supporting strings is θ. (a) Are the currents
in the same or opposite directions? (b) Find the magnitude of each current.
SOLUTION:
a)Suppose the current in wire 1 is directed out of the page. The B field from wire 1 is then directed
counterclockwise, i.e., up at wire 2. To have a repulsive force on wire 2, I2 ` × B tells us that the current
must be into the page.
b)At the equilibrium point the separation between the wires is d = 2L sin θ2 . With equal currents, the
magnetic force per unit length is therefore a repulsive force in the horizontal plane of the two wires of
magnitude FB = µ0 I 2 /(4πL sin θ2 ). In equilibrium, the horizontal component of the tension force per unit
length of the strings must cancel this magnetic force, so Tx = T sin θ2 = FB . Similarly, in the vertical
direction the component of the tension per unit length must balance that of gravity per unit length, giving
Ty = T cos θ2 = µg. Combining, we see that
µg sin(θ/2)
µ0 I 2
=
,
4πL sin(θ/2)
cos(θ/2)
r
so
I = sin(θ/2) ·
4πLµg
.
µ0 cos(θ/2)
4. Serway 30.22
A closely wound, long solenoid of overall length 30 cm has a magnetic field of 5 × 10−4 T at its center
produced by a current of 1 A through its windings. How many turns of wire are on the solenoid?
SOLUTION: The magnetic field inside a long solenoid is given by B = µo NI/L where N is the number of
turns, I is the current, and L is the overall length. Thus
N=
(5 × 10−4 T)(.3m)
BL
=
= 120 turns.
µ0 I
(4π × 10−7 Tm/A)(1A)
1
5. Serway 30.26
Imagine a long cylindrical wire of radius R that has a current density J(r) = J0 (1 − r2 /R2 ) for r ≤ R
and J(r) = 0 for r > R, where r is the distance from a point of interest to the central axis running along
the length of the wire. (a) Find the resulting magnetic field inside (r ≤ R) and outside (r > R) the wire.
(b) Plot the magnitude of the magnetic field as a function of r. (c) Find the location where the magnetic
field strength is a maximum, and the value of the maximum field.
SOLUTION: We solve this problem using Ampere’s law and choosing amperian loops as in Figure 30.9 of
the text.
a)From the symmetry of the problem, the magnetic field points in the azimuthal direction and its magnitude
depends only on r:
B = B(r)φ̂.
Ampere’s law states that
I
µ0 Iencl =
I
I
B · d` = B(r)
φ̂ · d` = B(r)
rdφ = 2πrB(r),
where we have used d` = r dφ φ̂. Thus
µ0 Iencl
φ̂.
2πr
So all that remains is to calculate the current “enclosed” by the amperian loop, Iencl , as a function of the
radius r of the loop:
B=
Z
Iencl (r)
=
Z
=
r
J dA =
πJ0 r2
J(r0 )2πr0 dr0 = 2πJ0
0
1−
r2
2R2
r
r0
1−
0
(r0 )2
R2
dr0
.
Thus, for r < R we have
B=
Z
µ0 J0
r
2
1−
r2
2R2
φ̂.
For r > R, the enclosed current is just the total current in the wire Iencl (R) = πJ0 R2 /2, and thus
B=
µ0 J0 R2
φ̂.
4r
b)See Figure 1.
Figure 1: The vertical axis increments are 2B/10 (µ0 J0 R) and the horizontal axis measures r/R in increments
of 0.5.
c)To find where the magnetic field is a maximum, look for where its derivative vanishes:
µ0 J0
dB(r)
=
0=
dr
2
3 r2
1−
2 R2
r
⇒ r=
At this value of r, the magnitude of the magnetic field is
√
p
2µ0 J0 R
√
.
B( 2/3R) =
3 3
2
2
R.
3
6. Serway 30.66
A large nonconducting belt with a uniform surface charge density σ moves with a speed v on a set of
rollers as shown in Figure P30.66 of your text. Consider a point just above the surface of the moving
belt. (a) Find an expression for the magnitude of the magnetic field B at this point. (b) If the belt is
positively charged what is the direction of B? (Note the belt may be considered as an infinite sheet.)
SOLUTION:
a)The moving charge density gives rise to a uniform current density J in a plane, which was solved using
Ampere’s law in example 30.6 in the text. There it was found that the magnitude of the magnetic field
is constant above the sheet and has the value B = µ0 J/2. Therefore we just have to express the current
density J in terms of the charge density σ and the speed of the belt v:
J=
current
charge
charge length
=
=
·
= σ · v.
length
time · length
area
time
Thus
B = µ0 σv/2.
b)The direction of the magnetic field can be read off from Figure 30.12 in the text. Translating to the
geometry shown in Figure P30.66, since the current is to the right, B is parallel to the sheet and pointing
out of the page (perpendicular to the velocity of the sheet).
7. Serway 30.70
Consider a thin disk of radius R mounted to rotate about the x axis in the yz plane. The disk has a
positive uniform surface charge density σ and angular velocity ω. Show that the magnetic field at the
center of the disk is B = µ0 σωR/2.
SOLUTION: The basic expression for the magnetic field due to moving charges is B = (µ0 /4π)qv × r̂/r2 .
Applying this to little area elements of the rotating disk and summing up gives
µ0
B=
4π
Z
disk
σv × r̂
µ0 σ
=
r2
4π
Z
R
0
v × r̂
· 2πrdr,
r2
where, since we are evaluating the magnetic field at the center of the disk, r̂ is just the radial unit vector in
the plane of the disk. Also we have used the fact that σ is constant and so can be taken out of the integral.
To evaluate this integral we must find v as a function of r. Recall that the angular velocity is given by
ω = v/r, so v = ωr; furthermore, the velocity is in the azimuthal direction, φ̂, so
v = ωrφ̂.
Plugging this into our expression for B and using φ̂ × r̂ = −x̂ (see note below) where x̂ is the unit vector
normal to the plane of the disk, gives
B=−
µ0 σω
2
Z
R
dr x̂ = −
0
µ0 σωR
x̂,
2
and thus the magnitude of B is indeed
µ0 σωR
.
2
Note on directions: We have taken the disk to be rotating in the counter-clockwise direction when viewed
from above, so that φ̂ points in the counter-clockwise direction. If the disk were taken to be rotating in
the opposite direction, all that would change would be the sign of B, which, of course, does not change its
magnitude.
B=
3