Download Home Work Solutions 12

Home Work Solutions 12
12-1 In Fig.1, an electric field is directed out of the page within a circular region of
radius R = 3.00 cm. The field magnitude is E = (0.500 V/m · s)(1 - r/R)t, where t is in
seconds and r is the radial distance (r ≤ R). What is the magnitude of the induced
magnetic field at radial distances (a) 2.00 cm and (b) 5.00 cm?
Sol
(a) Here, the enclosed electric flux is found by integrating
r
r
1
r
r3 
 E   E 2 rdr  t (0.500 V/m  s)(2 )  1   rdr  t  r 2 

0
0
3R 
 R
2
with SI units understood.
Then (after taking the derivative with respect to time) Eq.
32-3 leads to
1
r3 
B(2 r )   0 0  r 2 
.
3R 
2
For r = 0.0100 m and R = 0.0300 m, this gives B = 1.36  1021 T.
(b) The integral shown above will no longer (since now r > R) have r as the upper
limit; the upper limit is now R. Thus,

1
R3  1
2
 E  t  R 2 
  t R .
3R  6
2
Consequently, Eq. 32-3 becomes
1
B(2 r )   0 0 R 2
6
which for r = 0.0500 m, yields
B
 0 0 R 2
12r

(8.85 1012 )(4 107 )(0.030)2
 1.67 1020 T .
12(0.0500)
12-2 In Fig. 2, a parallel-plate capacitor has square plates of edge length L = 1.0 m. A
current of 2.0 A charges the capacitor, producing a uniform electric field
the plates, with
between
perpendicular to the plates. (a) What is the displacement current id
through the region between the plates? (b) What is dE/dt in this region? (c) What is the
displacement current encircled by the square dashed path of edge length d = 0.50 m?
(d) What is
around this square dashed path?
Sol
The electric field between the plates in a parallel-plate capacitor is changing, so there
is a nonzero displacement current id   0 (d  E / dt ) between the plates.
Let A be the area of a plate and E be the magnitude of the electric field between the
plates. The field between the plates is uniform, so E = V/d, where V is the potential
difference across the plates and d is the plate separation. The current into the positive
plate of the capacitor is
i
d E
dq d
dV  0 A d ( Ed )
dE
  CV   C

 0 A
 0
,
dt dt
dt
d
dt
dt
dt
which is the same as the displacement current.
(a) At any instant the displacement current id in the gap between the plates equals the
conduction current i in the wires. Thus id = i = 2.0 A.
(b) The rate of change of the electric field is
id
dE
1  d E 
2.0 A




dt  0 A  0 dt   0 A 8.85  1012 F m 1.0 m



2
 2.3  1011
V
.
m s
(c) The displacement current through the indicated path is
2
 d2 
 0.50m 
id  id  2    2.0A  
  0.50A.
 1.0m 
L 
(d) The integral of the field around the indicated path is


r r
16
7
B
—
  ds  0 id  1.26  10 H m 0.50A  6.3  10 T  m.

12-3 Assume that an electron of mass m and charge magnitude e moves in a circular
orbit of radius r about a nucleus. A uniform magnetic field
is then established
perpendicular to the plane of the orbit. Assuming also that the radius of the orbit does
not change and that the change in the speed of the electron due to field
is small,
find an expression for the change in the orbital magnetic dipole moment of the
electron due to the field.
Sol
An electric field with circular field lines is induced as the magnetic field is turned on.
Suppose the magnetic field increases linearly from zero to B in time t. According to
Eq. 31-27, the magnitude of the electric field at the orbit is given by
 r  dB  r  B
E 

,
 2  dt  2  t
where r is the radius of the orbit. The induced electric field is tangent to the orbit and
changes the speed of the electron, the change in speed being given by
v  at 
 e   r   B
eE
erB
t    t 
.
me
2me
 me   2   t 
The average current associated with the circulating electron is i = ev/2r and the
dipole moment is
 ev  1
  evr .
 2 r  2
  Ai   r 2  
The change in the dipole moment is
1
1  erB  e2 r 2 B
  erv  er 

.
2
2  2me 
4me
12-4 The magnetic field of Earth can be approximated as the magnetic field of a
dipole. The horizontal and vertical components of this field at any distance r from
Earth’s center are given by
where λm is the magnetic latitude (this type of latitude is measured from the
geomagnetic equator toward the north or south geomagnetic pole). Assume that
Earth’s magnetic dipole moment has magnitude μ = 8.00 × 1022 A‧m2. (a) Show that
the magnitude of Earth’s field at latitudeλm is given by
(b) Show that the inclination φi of the magnetic field is related to the magnetic
latitudeλm by .
(HRW 32-55)
Sol: (a) The Pythagorean theorem leads to


 

B  B  B   0 3 cos  m    0 3 sin  m   0 3 cos 2  m  4sin 2  m
4r
 4r
  2r

2
2
h

0 
4r 3
2
v
1  3sin 2  m ,
where cos2 m + sin2 m = 1 was used.
(b) We use Eq. 3-6:
3
Bv  0  2r  sin  m
tan i 

 2 tan  m .
Bh  0  4r 3  cos  m
2
12-5 A charge q is distributed uniformly around a thin ring of radius r. The ring is
rotating about an axis through its center and perpendicular to its plane, at an angular
speed ω. (a) Show that the magnetic moment due to the rotating charge has magnitude
(b) What is the direction of this magnetic moment if the charge is positive?
(HRW 32-60)
Sol: (a) The period of rotation is T = 2/ and in this time all the charge passes any
fixed point near the ring. The average current is i = q/T = q/2 and the
magnitude of the magnetic dipole moment is
  iA 
q 2 1
r  qr 2 .
2
2
(b) We curl the fingers of our right hand in the direction of rotation. Since the
charge is positive, the thumb points in the direction of the dipole moment. It is
the same as the direction of the angular momentum vector of the ring.
12-6 Consider a solid containing N atoms per unit volume, each atom having a
magnetic dipole momentμ. Suppose the direction ofμcan be only parallel or
antiparallel to an externally applied magnetic field B (this will be the case ifμis due to
the spin of a single electron). According to statistical mechanics, the probability of an
atom being in a state with energy U is proportional to e-U/kT, where T is the temperature
and k is Boltzmann’s constant. Thus, because energy U is -μ‧B, the fraction of atoms
whose dipole moment is parallel to B is proportional to eμB/kT and the fraction of atoms
whose dipole moment is antiparallel to is proportional to e-μB/kT. (a) Show that the
magnitude of the magnetization of this solid is M = Nμtanh(μB/kT). Here tanh is the
hyperbolic tangent function: tanh(x) = (ex – e-x)/( ex + e-x). (b) Show that the result
given in (a) reduces to M = Nμ2B/kT forμB<<kT. (c) Show that the result of (a)
reduces to M = Nμ forμB>>kT. (d) Show that both (b) and (c) agree qualitatively
with Fig. 3. (HRW32-45)
Fig. 3

Sol: (a) We use the notation P() for the probability of a dipole being parallel to B ,
and P(–) for the probability of a dipole being antiparallel to the field. The
magnetization may be thought of as a “weighted average” in terms of these
probabilities:
M
N  P     N  P   
P     P   


N  e  B KT  e  B KT
e
 B KT
e
  B KT
  N  tanh   B  .


 kT 
(b) For B  kT (that is, B / kT  1 ) we have e± B/kT  1 ± B/kT, so
 1  B kT g N B
FB I N b1  B kT gb
M  N tanhG J

.
HkT K b1  B kT gb
 1  B kT g
kT
2
B I
F
G
HkT J
K N .
(c) For B  kT we have tanh (B/kT)  1, so M  N tanh
(d) One can easily plot the tanh function using, for instance, a graphical calculator.
One can then note the resemblance between such a plot and Fig. 32-14. By
adjusting the parameters used in one’s plot, the curve in Fig. 32-14 can reliably
be fit with a tanh function.