Download CHAPTER 32 SOLUTION FOR PROBLEM 10 If the electric field is

CHAPTER 32
SOLUTION FOR PROBLEM 10
If the electric field is perpendicular to a region of a plane and has uniform magnitude over the
region then the displacement current through the region is related to the rate of change of the
electric field in the region by
dE
,
id = 0 A
dt
where A is the area of the region. The rate of change of the electric field is the slope of the
graph.
For segment a
dE 6.0 × 105 N/C − 4.0 × 105 N/C
=
= 5.0 × 1010 N/C · s
dt
4.0 × 10−6 s
and id = (8.85 × 10−12 F/m)(1.6 m2 )(5.0 × 1010 N/C · s = 0.71 A.
For segment b dE/dt = 0 and id = 0.
For segment c
dE 4.0 × 105 N/C − 0
=
= 2.0 × 1011 N/C · s
dt
2.0 × 10−6 s
and id = (8.85 × 10−12 F/m)(1.6 m2 )(2.0 × 1011 N/C · s = 2.8 A.
CHAPTER 32
SOLUTION FOR PROBLEM 27
(a) If the magnetization of the sphere is saturated, the total dipole moment is µtotal = N µ, where
N is the number of iron atoms in the sphere and µ is the dipole moment of an iron atom. We
wish to find the radius of an iron sphere with N iron atoms. The mass of such a sphere is N m,
where m is the mass of an iron atom. It is also given by 4πρR3/3, where ρ is the density of
iron and R is the radius of the sphere. Thus N m = 4πρR3/3 and
N=
4πρR3
.
3m
Substitute this into µtotal = N µ to obtain
µtotal =
Solve for R and obtain
4πρR3µ
.
3m
}
3mµtotal
R=
4πρµ
The mass of an iron atom is
]1/3
.
m = 56 u = (56 u)(1.66 × 10−27 kg/u) = 9.30 × 10−26 kg .
So
R=
^
3(9.30 × 10−26 kg)(8.0 × 1022 J/T)
4π(14 ×
3
103 kg/m )(2.1
(b) The volume of the sphere is
Vs =
×
10−23 J/T)
„1/3
= 1.8 × 105 m .
4π 3 4π
R =
(1.82 × 105 m)3 = 2.53 × 1016 m3
3
3
and the volume of Earth is
Ve =
4π
(6.37 × 106 m)3 = 1.08 × 1021 m3 ,
3
so the fraction of Earth’s volume that is occupied by the sphere is
2.53 × 1016 m3
= 2.3 × 10−5 .
1.08 × 1021 m3
The radius of Earth was obtained from Appendix C.
CHAPTER 32
HINT FOR PROBLEM 15
For each of the original levels there is a new level associated with each possible value of mf .
Thus one value of mf is associated with level E1 and three values are associated with level E2 .
Use µorb z = mf µB and U = −µorb z Bext to compute the difference in energy of the levels for
which mf = 0 and mf = 1, say.
J
ans: (a) 0; (b) −1, 0, 1; (c) 4.64 × 10−24 J
o
CHAPTER 32
HINT FOR PROBLEM 22
The magnitude of the dipole moment is iA where i is the current and A (= πr2 , where r is the
orbit radius) is the area bounded by the electron’s path. The current is e/T = ev/2πr, where T is
the period of the motion and v is the speed of the electron. The radius of the orbit is r = mv/eB
(see Chapter 28). Make substitutions to write the expression for the dipole moment in terms of
v, then use K = 12 mv 2 to write it in terms of the kinetic energy K. Since the magnetic force
must be inward toward the center of the path you can find the direction of travel for a given field
direction and hence can find the direction of the dipole moment.To find the magnetization of the
gas, vectorially add the dipole moments per unit volume of the electrons and ions.
J
ans: (b) Ki /B; (b) −z; (c) 0.31 kA/m
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CHAPTER 32
HINT FOR PROBLEM 26
The magnitude of the magnetic field inside a toroid, a distance r from the center, is given by
B0 = µ0 ip Np /2πr, where Np is the number of turns in the primary and i is the current (see
Eq. 29–24). Use the average of the inside and outside radii for r and solve for ip . The total
field is B = B0 + BM and the magnetic flux through one turn of the secondary coil is ΦB = BA,
where A is the cross-sectional area of the toroid. According to Faraday’s law the emf generated
in the secondary is E = dΦB /dt, so the current is is = E/R, where R is the resistance of the
secondary coil. The charge is the integral of the current with respect to time.
J
ans: (a) 0.14 A; (b) 79 µC
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