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1. (a) The flux through the top is +(0.30 T)π*r*2 where *r* = 0.020 m. The flux through the
bottom is +0.70 mWb as given in the problem statement. Since the *net* flux must be zero

then the flux through the sides must be negative and exactly cancel the total of the

previously mentioned fluxes. Thus (in magnitude) the flux though the sides is 1.1 mWb.

(b) The fact that it is negative means it is inward.

2. We use
*n
*

*Bn*=
=

1

6 0Φ to obtain

Φ Φ
*B
*

*n
*

*Bn*6

1

5

1 2 3 4 5 3= − = − − + − + − = + =

Wb Wb Wb Wb Wb Wb .b g

3. (a) We use Gauss’ law for magnetism: z ⋅ =*B dA *0 . Now, z ⋅ = + +*B dA C*Φ Φ Φ1 2 ,
where Φ1 is the magnetic flux through the first end mentioned, Φ2 is the magnetic flux
through the second end mentioned, and Φ*C* is the magnetic flux through the curved
surface. Over the first end the magnetic field is inward, so the flux is Φ1 = –25.0 µWb.
Over the second end the magnetic field is uniform, normal to the surface, and outward, so

the flux is Φ2 = *AB* = π*r*2 *B*, where *A* is the area of the end and *r* is the radius of the
cylinder. It value is

Φ2 2 3 50120 160 10 7 24 10 72 4= × = + × = +− −π . . . . .m T Wb Wbb g c h µ

Since the three fluxes must sum to zero,

Φ Φ Φ

*C
*= − − = − = −1 2 25 0 72 4 47 4. . . .µ µ µWb Wb Wb

Thus, the magnitude is | | 47.4 Wb.
*C
*

µΦ =

(b) The minus sign in
*c
*

Φ indicates that the flux is inward through the curved surface.

4. From Gauss’ law for magnetism, the flux through *S*1 is equal to that through *S*2, the

portion of the *xz* plane that lies within the cylinder. Here the normal direction of *S*2 is +*y*.

Therefore,

( ) ( ) ( ) ( ) 01 2 left 0

1 2 2

2 2

ln 3 .

*r r r
*

*B B
r r r
*

*i
S S B x L dx B x L dx L dx
*

*r x
*

*iL
*

µ

µ − − −

Φ = Φ = = = π −

= π

5. We use the result of part (b) in Sample Problem 32-1:

( ) 2

0 0 for 2

*R dE
B r R
*

*r dt
*

µ ε= ≥

to solve for *dE*/*dt*:

( )( ) ( )( )( )2 2

7 3

13

22 12 3C

0 0 N m

2 2.0 10 T 6.0 10 m2 V 2.4 10 .

m s4 T m A 8.85 10 3.0 10 m

*dE Br
*

*dt R*µ ε

− −

−7 − − ⋅

× × = = = ×

⋅π×10 ⋅ × ×

6. From Sample Problem 32-1 we know that *B *∝* r* for *r *≤* R* and *B *∝* r*–1 for *r *≥* R*. So the
maximum value of *B* occurs at *r = R*, and there are two possible values of *r* at which the

magnetic field is 75% of *B*max. We denote these two values as *r*1 and *r*2, where *r*1 < *R* and

*r*2 > *R*.

(a) Inside the capacitor, 0.75 *B*max/*B*max = *r*1/*R*, or *r*1 = 0.75 *R* = 0.75 (40 mm)=30 mm.

(b) Outside the capacitor, 0.75 *B*max/*B*max = (*r*2/*R*)
–1

, or *r*2 = *R*/0.75 = 4*R*/3* = *(4/3)(40 mm)

= 53 mm..

(c) From Eqs. 32-15 and 32-17,

*B
i
*

*R
*

*i
*

*R
*

*d
*

max

.

. .= = =

× ⋅ = ×

− −µ µ0 0

7

5

2 2

4 10 6 0

2 0 040 3 0 10

π π π

π T m A A

m T.

c hb g b g

7. (a) Noting that the magnitude of the electric field (assumed uniform) is given by *E =
*

*V*/*d* (where *d* = 5.0 mm), we use the result of part (a) in Sample Problem 32-1

*B
r dE
*

*dt
*

*r
*

*d
*

*dV
*

*dt
r R*= = ≤µ ε µ ε0 0 0 0

2 2 forb g.

We also use the fact that the time derivative of sin (ω*t*) (where ω = 2π*f* = 2π(60) ≈ 377/s
in this problem) is ω cos(ω*t*). Thus, we find the magnetic field as a function of *r* (for *r *≤*
R*; note that this neglects “fringing” and related effects at the edges):

*B
r
*

*d
V t B
*

*rV
*

*d
*= =µ ε ω ω µ ε ω0 0 0 0

2 2 max max

maxcosb g

where *V*max = 150 V. This grows with *r* until reaching its highest value at *r = R* = 30 mm:

( )( )( )( )( ) ( )

12 3

0 0 max max

3

12

4 H m 8.85 10 F m 30 10 m 150V 377 s

2 2 5.0 10 m

1.9 10 T.

*r R
*

*RV
B
*

*d
*

µ ε ω −7 − − = −

−

π×10 × × = =

×

= ×

(b) For *r* ≤ 0.03 m, we use the 0 0 maxmax
2

*rV
B
*

*d
*

µ ε ω = expression found in part (a) (note the

*B *∝* r* dependence), and for *r* ≥ 0.03 m we perform a similar calculation starting with the
result of part (b) in Sample Problem 32-1:

( )

( )

2 2 2

0 0 0 0 0 0 max max

max max max

2

0 0 max

cos 2 2 2

for 2

*R R RdE dV
B V t
*

*r dt rd dt rd
*

*R V
r R
*

*rd
*

µ ε µ ε µ ε ω ω

µ ε ω

= = =

= ≥

(note the *B *∝* r*–1 dependence — See also Eqs. 32-16 and 32-17). The plot (with SI units
understood) is shown below.

8. (a) Inside we have (by Eq. 32-16) *B* = µo*id r*1 /2π*R
*2
, where *r*1 = 0.0200, *R* = 0.0300,

and the displacement current is given by Eq. 32-38: *id *= εo* d*Φ*E /dt* = εo(0.00300), in SI
units. Thus we find *B* = 1.18 × 10−19 T.

(b) Outside we have (by Eq. 32-17) *B* = µo*id* /2π*r*2 where *r*2 = 0.0500 in SI units. Here we
obtain *B* = 1.06 × 10−19 T.

9. (a) Application of Eq. 32-3 along the circle referred to in the second sentence of the

problem statement (and taking the derivative of the flux expression given in that sentence)

leads to

*B* (2π*r* ) = εo µo (0.60 V.m/s)
*r
*

*R
* .

Using *r* = 0.0200 (which, in any case, cancels out) and *R* = 0.0500 (SI units understood)

leads to *B* = 3.54 × 10−17 T.

(b) For a value of *r* larger than *R*, we must note that the flux enclosed has already reached

its full amount (when *r = R* in the given flux expression). Referring to the equation we

wrote in our solution of part (a), this means that the final fraction (
*r
*

*R
* ) should be replaced

with unity. On the left hand side of that equation, we set *r* = 0.0500 m and solve. We

now find *B* = 2.13 × 10−17 T.

10. (a) Application of Eq. 32-7 with *A* = π*r*2 (and taking the derivative of the field
expression given in the problem) leads to

*B* (2π*r*) = εo µo π*r*2 (0.00450 V/m.s) .

With *r* = 0.0200 m, this gives *B* = 5.01 × 10−22 T.

(b) With *r* > *R*, the expression above must replaced by

*B* (2π*r*) = εo µo π*R
*2
(0.00450 V/m

. s) .

Substituting *r* = 0.050 m and *R* = 0.030 m, we obtain *B* = 4.51 × 10−22 T.

11. (a) Here, the enclosed electric flux is found by integrating

Φ*E* =
* r
*

o *E* 2π*r *dr = *t*(0.500 V/m
.
s)(2π)

* r
*

o ( )1 – *rR **r**dr* = *t *π
1

2
*r
*

2 –

*r
*3

3*R
*

with SI units understood. Then (after taking the derivative with respect to time) Eq. 32-3

leads to

*B* (2π*r*) = εo µo π 1 2 *r
*2
–

*r
*3

3*R
* .

With *r* = 0.0200 m and *R* = 0.0300 m, this gives *B* = 3.09 × 10−20 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, Φ*E* = *t *π 1 2 *R
*2
–

*R
*3

3*R
* =

1

6
*t *π *R*2. Consequently, Eq. 32-3

becomes

*B* (2π*r*) = 1
6
εo µo π*R
*

2

which yields (for *r* = 0.0500 m) *B* = 1.67 × 10−20 T.

12. Let the area plate be *A* and the plate separation be *d*. We use Eq. 32-10:

*i
d
*

*dt
*

*d
*

*dt
AE A
*

*d
*

*dt
*

*V
*

*d
*

*A
*

*d
*

*dV
*

*dt
d
*

*E*= = = FHG
I
KJ =

F HG I KJε ε ε

ε 0 0 0

0Φ b g ,

or

*dV
*

*dt
*

*i d
*

*A
*

*i
*

*C
*

*d d*= = =
×

= ×−ε 0 6

515 7 5 10 .

. . A

2.0 10 F V s

Therefore, we need to change the voltage difference across the capacitor at the rate of 57.5 10 V/s× .

13. The displacement current is given by 0 ( / ),*di A dE dt*ε= where *A* is the area of a plate
and *E* is 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. Thus

*i
A
*

*d
*

*dV
*

*dt
d
*= ε 0 .

Now, ε0*A*/*d* is the capacitance *C* of a parallel-plate capacitor (not filled with a dielectric),
so

*i C
dV
*

*dt
d
*= .

14. We use Eq. 32-14: 0 ( / ).*di A dE dt*ε= Note that, in this situation, *A* is the area over
which a changing electric field is present. In this case *r > R*, so *A = *π*R*2. Thus,

*dE
*

*dt
*

*i
*

*A
*

*i
*

*R
*

*d d*= = =
×

= × ⋅

⋅ ε ε0 0

2 2

122 0

010 7 2 10

π π 8.85 10−12 .

. . .

A

m

V

m sC N m

2

2d ib g

15. Consider an area *A*, normal to a uniform electric field *E *. The displacement current

density is uniform and normal to the area. Its magnitude is given by *Jd = id*/*A*. For this

situation , 0 ( / )*di A dE dt*ε= , so

*J
A
*

*A
dE
*

*dt
*

*dE
*

*dt
d
*= =1 0 0ε ε .

16. (a) From Eq. 32-10,

( ) ( ) ( )

( )( )

5 4 4

0 0 0 0

2 12 2 2 4

2

8

4.0 10 6.0 10 6.0 10 V m s

C 8.85 10 4.0 10 m 6.0 10 V m s

N m

2.1 10 A.

*E
*

*d
*

*d dE d
i A A t A
*

*dt dt dt
*ε ε ε ε

− −

−

Φ= = × − × = − × ⋅

= − × × × ⋅ ⋅

= − ×

Thus, the magnitude of the displacement current is 8| | 2.1 10 A.
*d
i
*

−= ×

(b) The negative sign in
*d
i *implies that the direction is downward.

(c) If one draws a counterclockwise circular loop *s* around the plates, then according to

Eq. 32-18

*s
d
*

*B ds i*z ⋅ = <µ 0 0,

which means that *B ds*⋅ < 0 . Thus *B * must be clockwise.

17. (a) We use *B ds I*⋅ =z µ 0 enclosed to find

( ) ( )( )( ) 2

0 6 2 30 enclosed 0

7

1 1 1.26 10 H m 20A m 50 10 m

2 2 2 2

6.3 10 T.

*d
*

*d
*

*J rI
B J r
*

*r r
*

µµ µ − −

−

π = = = = × ×

π π = ×

(b) From *i J r
d
*

*dt
r
*

*dE
*

*dt
d d
*

*E*= = =π π2 0 0
2ε εΦ , we get

*dE
*

*dt
*

*J
d*= =

× = ×

⋅−ε 0 12

1220

885 10 2 3 10

A m

F m

V

m s

2

. . .

18. (a) Since *i = id* (Eq. 32-15) then the portion of displacement current enclosed is

*i i
R
*

*i
d
*

*R
*

, .enc A .= = = π π

3

2

2

1

9 133

b g

(b) We see from Sample Problems 32-1 and 32-2 that the maximum field is at *r = R* and

that (in the interior) the field is simply proportional to *r*. Therefore,

*B
*

*B
*

*r
*

*R*max

.= =3 00 mT 12.0 mT

which yields *r = R*/4 = (1.20 cm)/4 = 0.300 cm.

(c) We now look for a solution in the exterior region, where the field is inversely

proportional to *r* (by Eq. 32-17):

*B
*

*B
*

*R
*

*r*max

.= =3 00 mT 12.0 mT

which yields *r* = 4*R* = 4(1.20 cm) = 4.80 cm.

19. (a) In region *a* of the graph,

*i
d
*

*dt
A
*

*dE
*

*dt
d
*

*E*= =

= × × − × ×

=− −

ε ε0 0

12 5 5

6 885 10 16

4 5 10 6 0 10

4 0 10 0 71

Φ

. . . .

. .F m m

N C N C

s A.2c hc h

(b) *id *∝* dE*/*dt* = 0.

(c) In region *c* of the graph,

( )( ) 5

12 2

0 6

4.0 10 N C | | 8.85 10 F m 1.6m 2.8A.

2.0 10 s
*d
*

*dE
i A
*

*dt
*ε − −

− ×= = × = ×

20. From Eq. 28-11, we have *i* = (ε / *R* ) *e*− *t/*τ since we are ignoring the self-inductance of
the capacitor. Eq. 32-16 gives

*B* =
µo*id r
*2π*R*2 .

Furthermore, Eq. 25-9 yields the capacitance

*C* =
εoπ(0.05 m)2

0.003 m = 2.318 ×10

−11 F,

so that the capacitive time constant is τ = (20.0 × 106 Ω)(2.318 × 10−11 F) = 4.636 × 10−4 s.

At *t* = 250 × 10
−6

s, the current is

*i* =
12.0 V

20.0 x 10
6 Ω *e
*

− *t/*τ
= 3.50 × 10

−7 A .

Since *i = id* (see Eq. 32-15) and *r* = 0.0300 m, then (with plate radius *R* = 0.0500 m) we

find

*B* =
µo*id r
*2π*R*2 =

µo(3.50 x 10-7)(0.03) 2π(0.05)2 = 8.40 × 10

−13 T .

21. (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

*dE
*

*dt A
*

*d
*

*dt
*

*i
*

*A
*

*E d*= FHG
I
KJ = = × = × ⋅−

1 2 0

85 10 10 2 3 10

0

0

0 12 2

11

ε ε

ε Φ .

. . . .

A

8 F m m

V

m sc hb g

(c) The displacement current through the indicated path is

( ) 2

2

2

0.50m 2.0A 0.50A.

1.0m
*d d
*

*d
i i
*

*L
*

′ = × = =

(d) The integral of the field around the indicated path is

*B ds i
d
*

⋅ = ′ = × = × ⋅− −z µ 0 16 7126 10 050 6 3 10. . .H m A T m.c hb g

**2**2. (a) Fig. 32-34 indicates that *i* = 4.0 A when *t* = 20 ms. Thus, *Bi* = µo*i*/2π*r* = 0.89 mT.

(b) Fig. 32-34 indicates that *i* = 8.0 A when *t* = 40 ms. Thus, *Bi* ≈ 0.18 mT.

(c) Fig. 32-34 indicates that *i* = 10 A when *t* > 50 ms. Thus, *Bi * ≈ 0.220 mT.

(d) Eq. 32-4 gives the displacement current in terms of the time-derivative of the electric

field: *id* = εo*A*(*dE/dt*), but using Eq. 26-5 and Eq. 26-10 we have *E* = ρ*i/A* (in terms of the
real current); therefore, *id* = εoρ(*di/dt*). For 0 < *t* < 50 ms, Fig. 32-34 indicates that *di/dt* =
200 A/s. Thus, *Bid* = µo*id */2π*r* = 6.4 × 10−22 T.

(e) As in (d), *Bid* = µo*id */2π*r* = 6.4 × 10−22 T.

(f) Here *di/dt* = 0, so (by the reasoning in the previous step) *B* = 0.

(g) By the right-hand rule, the direction of
*i
*

*B *at *t* = 20 s is out of page.

(h) By the right-hand rule, the direction of
*id
*

*B *at *t* = 20 s is out of page.

23. (a) Eq. 32-16 (with Eq. 26-5) gives

*B* =
µo*id r
*2π*R*2 =

µo* Jd A r
*2π*R*2 = 75.4 nT

where we set *A* = π*R*2 (which led to several cancellations).

(b) Similarly, Eq. 32-17 gives *B* =
µo*id
*2π*r* =

µo *Jd *π*R
*2

2π*r * = 67.9 nT.

24. (a) Eq. 32-16 gives *B* =
µo*id r
*2π*R*2 = 2.22 µT.

(b) Eq. 32-17 gives *B* =
µo*id
*2π*r* = 2.00 µT.

25. (a) Eq. 32-11 applies (though the last term is zero) but we must be careful with *id*,enc .

It is the enclosed portion of the displacement current, and if we related this to the

displacement current density *Jd *, then

*id*,enc =
* r
*

o *Jd* 2π*r *dr = (4.00 A/m
.
s)(2π)

* r
*

o ( )1 – *rR **r**dr* = 8π
1

2
*r
*

2 –

*r
*3

3*R
*

with SI units understood. Now, we apply Eq. 32-17 (with *id* replaced with *id*,enc) or start

from scratch with Eq. 32-11, to get *B*=
µo *id* enc

2π*r * = 27.9 nT.

(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,

*id*,enc = *id* = 8π 1 2 *R
*2
–

*R
*3

3*R
* =

4

3
π*R*2.

Now Eq. 32-17 gives *B* =
µo*id
*2π*r* = 15.1 nT.

26. (a) Eq. 32-11 applies (though the last term is zero) but we must be careful with *id*,enc .

It is the enclosed portion of the displacement current. Thus Eq. 32-17 (which derives

from Eq. 32-11) becomes, with *id* replaced with *id*,enc,

*B* =
µo *id* enc

2π*r * =
µo (3.00 A) *r
*

2π*r R *

which yields (after canceling *r*, and setting *R* = 0.0300 m) *B* = 20.0 µ T.

(b) Here *id* = 3.00 A, and we get *B* =
µo*id
*2π*r* =12.0 µ

T.

27. The horizontal component of the Earth’s magnetic field is given by *B
h i
*= *B *cosφ ,

where *B* is the magnitude of the field andφ
*i
* is the inclination angle. Thus

*B
B
h
*

*i
*

= = ° =

cos cosφ µ µ16 73

55 T

T .

28. (a) The flux through Arizona is

Φ = − = − × = − ×−*B A
r
*

43 10 295 000 10 13 106 3 2

7T km m km Wb2c hc hc h, . ,

inward. By Gauss’ law this is equal to the negative value of the flux Φ*'* through the rest of
the surface of the Earth. So Φ*'* = 1.3 × 107 Wb.

(b) The direction is outward.

29. We use Eq. 32-31: µ orb, *z* = –*m *µ*B*.

(a) For *m * = 1, µorb,*z* = –(1) (9.3 × 10–24 J/T) = –9.3 × 10–24 J/T.

(b) For *m * = –2, µorb,*z* = –(–2) (9.3 × 10–24 J/T) = 1.9 × 10–23 J/T.

30. We use Eq. 32-27 to obtain ∆*U* = –∆(µ*s,zB*) = –*B*∆µ*s,z*, where µ µ*s z e Beh m*, = ± = ±4π
(see Eqs. 32-24 and 32-25). Thus,

∆*U B B
B B B
*

= − − − = = × = ×− −µ µ µb g c hb g2 2 9 27 10 0 25 4 6 1024 24. . . .J T T J

31. (a) Since *m * = 0, *L*orb,*z* = *m h*/2π = 0.

(b) Since *m * = 0, µorb,*z* = –*m *µ*B* = 0.

(c) Since *m * = 0, then from Eq. 32-32, *U* = –µorb,*zB*ext = –*m *µ*BB*ext = 0.

(d) Regardless of the value of *m *, we find for the spin part

*U B B
s z B
*

= − = ± = ± × = ± ×− −µ µ, . . .9 27 10 35 3 2 10 24 25J T mT Jc hb g

(e) Now *m * = –3, so

( ) ( )27 34 34

orb,

3 6.63 10 J s 3.16 10 J s 3.2 10 J s

2 2
*z
*

*m h
L
*

− − −

− × ⋅ = = = − × ⋅ ≈ − × ⋅

π π

(f) and

( ) ( )24 23 23orb, 3 9.27 10 J T 2.78 10 J T 2.8 10 J T .*z Bm*µ µ − − −= − = − − × = × ≈ ×

(g) The potential energy associated with the electron’s orbital magnetic moment is now

( )( )23 3 25orb, ext 2.78 10 J T 35 10 T 9.7 10 J.*zU B*µ − − −= − = − × × = − ×

(h) On the other hand, the potential energy associated with the electron spin, being

independent of *m *, remains the same: ±3.2 × 10–25 J.

32. Combining Eq. 32-27 with Eqs. 32-22 and 32-23, we see that the energy difference is

∆*U* = 2 µB *B*

where µB is the Bohr magneton (given in Eq. 32-25). With ∆*U* = 6.00 × 10−25 J, we
obtain *B* = 32.3 mT.

33. (a) The potential energy of the atom in association with the presence of an external

magnetic field *B*ext is given by Eqs. 32-31 and 32-32:

orb ext orb, ext ext .*z BU B B m B*µ µ µ= − ⋅ = − = −

For level *E*1 there is no change in energy as a result of the introduction of *B*ext , so *U *∝*m *
= 0, meaning that *m *= 0 for this level.

(b) For level *E*2 the single level splits into a triplet (i.e., three separate ones) in the

presence of *B*ext , meaning that there are three different values of *m *. The middle one in

the triplet is unshifted from the original value of *E*2 so its *m * must be equal to 0. The

other two in the triplet then correspond to *m * = –1 and *m * = +1, respectively.

(c) For any pair of adjacent levels in the triplet |∆*m *| = 1. Thus, the spacing is given by

24 24| ( ) | | | (9.27 10 J/T)(0.50T) 4.64 10 J.
*B B B
*

*U m B m B B*µ µ µ − −∆ = ∆ − = ∆ = = × = ×

34. (a) A sketch of the field lines (due to the presence of the bar magnet) in the vicinity of

the loop is shown below:

(b) The primary conclusion of §32-9 is two-fold: *u * is opposite to *B *, and the effect of *F *

is to move the material towards regions of smaller *B * values. The direction of the

magnetic moment vector (of our loop) is toward the right in our sketch, or in the +*x*

direction.

(c) The direction of the current is clockwise (from the perspective of the bar magnet.)

(d) Since the size of *B * relates to the “crowdedness” of the field lines, we see that *F * is

towards the right in our sketch, or in the +*x* direction.

35. 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

*E
r dB
*

*dt
*

*r B
*

*t
*= FHG
I
KJ =

F HG I KJ2 2 ,

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 eE
m
*

*t
e
*

*m
*

*r B
*

*t
t
*

*erB
*

*m
e e e
*

= = = F HG I KJ F HG I KJ F HG I KJ =2 2 .

The average current associated with the circulating electron is *i = ev*/2π*r* and the dipole
moment is

µ = = FHG
I
KJ =*Ai r
*

*ev
*

*r
evr*π

π 2

2

1

2 c h .

The change in the dipole moment is

∆ ∆µ = = F HG

I KJ =

1

2

1

2 2 4

2 2

*er v er
erB
*

*m
*

*e r B
*

*m
e e
*

.

36. Reviewing Sample Problem 32-3 before doing this exercise is helpful. Let

*K kT B B B*= = ⋅ − − ⋅ =3
2

2µ µ µd i

which leads to

*T
B
*

*k
*= =

×

× =

−

−

4

3

4 10 10 050

3 138 10 0 48

23

23

µ . . .

. . J T T

J K K

c hb g c h

37. The magnetization is the dipole moment per unit volume, so the dipole moment is

given by µ = *M* , where *M* is the magnetization and is the volume of the cylinder
( = π*r L*2 , where *r* is the radius of the cylinder and *L* is its length). Thus,

µ = = × × × = ×− −*M r L*π π 0.500 10−22 3
2

2 2530 10 500 10 2 08 10. . . .A m m m J Tc h c h c h

38. (a) From Fig. 32-14 we estimate a slope of *B*/*T* = 0.50 T/K when *M*/*M*max = 50%. So

*B* = 0.50 T = (0.50 T/K)(300 K) = 1.5×10
2
T.

(b) Similarly, now *B*/*T* ≈ 2 so *B* = (2)(300) = 6.0×102 T.

(c) Except for very short times and in very small volumes, these values are not attainable

in the lab.

39. For the measurements carried out, the largest ratio of the magnetic field to the

temperature is (0.50 T)/(10 K) = 0.050 T/K. Look at Fig. 32-14 to see if this is in the

region where the magnetization is a linear function of the ratio. It is quite close to the

origin, so we conclude that the magnetization obeys Curie’s law.

40. Section 32-10 explains the terms used in this problem and the connection between *M*

and µ. The graph in Fig. 32-37 gives a slope of

*M/M*max

*B*ext */T
* =

0.15

0.20 = 3/4

in Kelvins per Tesla. Thus we can write

µ µmax

= ¾ (0.800 T)/(2.00 K) = 0.30 .

41. (a) A charge *e* traveling with uniform speed *v* around a circular path of radius *r* takes

time *T* = 2π*r*/*v* to complete one orbit, so the average current is

*i
e
*

*T
*

*ev
*

*r
*= =

2π .

The magnitude of the dipole moment is this multiplied by the area of the orbit:

µ = =*ev
r
*

*r
evr
*

2 2

2

π π .

Since the magnetic force with magnitude *evB* is centripetal, Newton’s law yields *evB =
*

*mev
*2
/*r*, so / .

*e
r m v eB*= Thus,

µ = FHG I KJ = F HG I KJ F HG

I KJ =

1

2

1 1

2

2
*ev
*

*m v
*

*eB B
m v
*

*K
*

*B
*

*e
*

*e
*

*e*b g .

The magnetic force − ×*ev B * must point toward the center of the circular path. If the
magnetic field is directed out of the page (defined to be +z direction), the electron will

travel counterclockwise around the circle. Since the electron is negative, the current is in

the opposite direction, clockwise and, by the right-hand rule for dipole moments, the

dipole moment is into the page, or in the –*z* direction. That is, the dipole moment is

directed opposite to the magnetic field vector.

(b) We note that the charge canceled in the derivation of µ = *Ke*/*B*. Thus, the relation µ =
*Ki*/*B* holds for a positive ion.

(c) The direction of the dipole moment is –*z*, opposite to the magnetic field.

(d) The magnetization is given by *M = *µ*ene + *µ*ini*, where µ*e* is the dipole moment of an
electron, *ne* is the electron concentration, µ*i* is the dipole moment of an ion, and *ni* is the
ion concentration. Since *ne = ni*, we may write *n* for both concentrations. We substitute µ*e*
= *Ke*/*B* and µ*i* = *Ki*/*B* to obtain

( ) ( ) 21 3

20 21 25.3 10 m 6.2 10 J+7.6 10 J 3.1 10 A m. 1.2T

*e i
*

*n
M K K
*

*B
*

− − −×= + = × × = ×

42. The Curie temperature for iron is 770°C. If *x* is the depth at which the temperature

has this value, then 10°C + (30°C/km)*x* = 770°C. Therefore,

*x *= ° − °
°

=770 10 25C C 30 C km

km.

43. (a) The field of a dipole along its axis is given by Eq. 30-29:

*B
z
*

= µ µ0 32π

,

where µ is the dipole moment and *z* is the distance from the dipole. Thus,

*B
*

*A
*

= × ⋅ ×

× = ×

− − −

4 10 15 10

2 30 10

7 23

6 π

π 10 10−9 T m J T

m T.

c hc h c h

. .

(b) The energy of a magnetic dipole µ in a magnetic field *B * is given by
*U B B*= − ⋅ = −µ µ φcos , where φ is the angle between the dipole moment and the field.
The energy required to turn it end-for-end (from φ = 0° to φ = 180°) is

∆*U B*= = × × = × ×− − − −2 2 15 10 3 0 10 9 0 1023 6 29 10µ . . .J T T J = 5.6 10 eV.c hc h

The mean kinetic energy of translation at room temperature is about 0.04 eV. Thus, if

dipole-dipole interactions were responsible for aligning dipoles, collisions would easily

randomize the directions of the moments and they would not remain aligned.

44. (a) The number of iron atoms in the iron bar is

*N *=
×

= × 7 9 50 10

55847 6 022 10 4 3 10

23

23 . . .

. . . .

g cm cm cm

g mol mol

3 2c hb gc h b g c h

Thus the dipole moment of the iron bar is

µ = × × = ⋅−2 1 10 4 3 10 8 923 23. . . .J T A m2c hc h

(b) τ = µ*B* sin 90° = (8.9 A · m2)(1.57 T) = 13 N · m.

45. The saturation magnetization corresponds to complete alignment of all atomic dipoles

and is given by *M*sat = µ*n*, where *n* is the number of atoms per unit volume and µ is the
magnetic dipole moment of an atom. The number of nickel atoms per unit volume is *n =
*

ρ/*m*, where ρ is the density of nickel. The mass of a single nickel atom is calculated using
*m = M*/*NA*, where *M* is the atomic mass of nickel and *NA* is Avogadro’s constant. Thus,

( )( )3 23 22 3

28 3

8.90g cm 6.02 10 atoms mol 9.126 10 atoms cm

58.71g mol

9.126 10 atoms m .

*A
N
*

*n
M
*

ρ ×= = = ×

= ×

The dipole moment of a single atom of nickel is

µ = = × ×

= × ⋅−*M
n
*

sat

3

2A m

m A m

4 70 10

9126 10 515 10

5

28

24.

. . .

46. (a) Eq. 29-36 gives

τ = µrod *B* sinθ= (2700 A/m)(0.06 m)π(0.003 m)2(0.035 T)sin(68°)= 1.49 × 10−4 N.m.

We have used the fact that the volume of a cylinder is its length times its (circular) cross

sectional area.

(b) Using Eq. 29-38, we have

∆*U* = – µrod *B*(cos θ*f * – cos θ*i*)
= –(2700 A/m)(0.06 m)π(0.003m)2(0.035T)[cos(34°) – cos(68°)]

= –72.9 µJ.

47. (a) The magnitude of the toroidal field is given by *B*0 = µ0*nip*, where *n* is the number
of turns per unit length of toroid and *ip* is the current required to produce the field (in the

absence of the ferromagnetic material). We use the average radius (*r*avg = 5.5 cm) to

calculate *n*:

3

2

avg

400 turns 1.16 10 turns/m .

2 2 m)

*N
n
*

*r
*−= = = ×π π(5.5×10

Thus,

*i
B
*

*n
p
*= = ×

× ⋅ × =

−

− 0

0

3

7

0 20 10

4 014

µ .

( / .

T

T m / A)(1.16 10 m) A .

3π 10

(b) If Φ is the magnetic flux through the secondary coil, then the magnitude of the emf
induced in that coil is ε = *N*(*d*Φ/*dt*) and the current in the secondary is *is = *ε/*R*, where *R* is
the resistance of the coil. Thus,

*i
N
*

*R
*

*d
*

*dt
s
*= FHG

I KJ

Φ .

The charge that passes through the secondary when the primary current is turned on is

0 .

*s
*

*N d N N
q i dt dt d
*

*R dt R R
*

ΦΦ Φ= = = Φ =

The magnetic field through the secondary coil has magnitude *B = B*0 + *BM* = 801*B*0,

where *BM* is the field of the magnetic dipoles in the magnetic material. The total field is

perpendicular to the plane of the secondary coil, so the magnetic flux is Φ = *AB*, where *A*
is the area of the Rowland ring (the field is inside the ring, not in the region between the

ring and coil). If *r* is the radius of the ring’s cross section, then *A = *π*r*2. Thus,

Φ = 801 2 0π*r B *.

The radius *r* is (6.0 cm – 5.0 cm)/2 = 0.50 cm and

Φ = × × ×− − −801 0 20 102 3π(0.50 10 m) T) = 1.26 10 Wb .2 5( .

Consequently,

5

550(1.26 10 Wb) 7.9 10 C . 8.0

*q
*

− −×= = ×

Ω

48. From Eq. 29-37 (see also Eq. 29-36) we write the torque as τ = −µ*Bh* sinθ where the
minus indicates that the torque opposes the angular displacement θ (which we will
assume is small and in radians). The small angle approximation leads to

*h
B*τ µ θ≈ − , which is an indicator for simple harmonic motion (see section 16-5,

especially Eq. 16-22). Comparing with Eq. 16-23, we then find the period of oscillation

is

*T* = 2π
*I
*

µ*Bh
*

where *I* is the rotational inertial that we asked to solve for. Since the frequency is given

as 0.312 Hz, then the period is *T* = 1/*f *= 1/0.312 in SI units. Similarly, *Bh* = 18.0 × 10−6
and µ = 6.80 × 10−4. The above relation then yields *I* = 3.19 × 10−9 kg.m2.

49. (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 *Nm*, where *m* is the mass of an iron atom. It is also given by 4πρ*R*3/3,
where ρ is the density of iron and *R* is the radius of the sphere. Thus *Nm* = 4πρ*R*3/3 and

*N
R
*

*m
*= 4

3

3πρ .

We substitute this into µtotal = *N*µ to obtain

µ ρ µtotal = 4

3

3π *R
m
*

.

We solve for *R* and obtain

*R
m*=

F HG

I KJ

3

4

1 3

µ ρµ

total

π .

The mass of an iron atom is *m *= = × = ×− −56 56 166 10 9 30 1027 26u u kg u kg.b gc h. .
Therefore,

*R *=
× ×

× ×

L N M M

O Q P P

= × −

−

3 9 30 10 8 0 10

4 2 1 10 18 10

26 22

23

1 3

5 . .

. .

kg J T

kg m J T m.

3

c hc h c hc hπ 14 103

(b) The volume of the sphere is *V R
s
*= = × = ×4 4 182 10 2 53 103 5

3 16π

3 π 3

. .m m3c h and the volume of the Earth is

*V
e
*= × = ×4 6 37 10 108 106

3 21π

3 . . ,m m3c h

so the fraction of the Earth’s volume that is occupied by the sphere is

2 53 10

108 10 2 3 10

16

21

5.

. . .

× ×

= × −m m

3

3

50. The integral of the field along the indicated path is, by Eq. 32-18 and Eq. 32-19, equal

to

µo*id*(enclosed area)/(total area) = µo(0.75 A)(4 cm ×2 cm)/(12 cm)2 = 52 nT.m.

51. (a) Inside the gap of the capacitor, *B*1 = µo*id r*1 /2π*R
*2
(Eq. 32-16); outside the gap the

magnetic field is *B*2 = µo*id* /2π*r*2 (Eq. 32-17). Consequently, *B*2 = *B*1*R
*2
/*r*1 *r*2 = 16.7 nT.

(b) The displacement current is *id * = 2π*B*1*R
*2
/µo*r*1 = 5.00 mA.

52. (a) The period of rotation is *T* = 2π/ω and in this time all the charge passes any fixed
point near the ring. The average current is *i = q*/*T = q*ω/2π and the magnitude of the
magnetic dipole moment is

µ ω ω= = =*iA q r q r
*2

1

2

2 2

π π .

(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.

53. (a) We use the result of part (a) in Sample Problem 32-1:

*B
r dE
*

*dt
r R*= ≤µ ε0 0

2 forb g ,

where *r* = 0.80*R* and

*dE
*

*dt
*

*d
*

*dt
*

*V
*

*d d
*

*d
*

*dt
V e
*

*V
*

*d
e
*

*t t*= FHG
I
KJ = = −

− −1 0

0τ τ

τc h .

Here *V*0 = 100 V. Thus

*B t
r V
*

*d
e
*

*V r
*

*d
e
*

*e
*

*e
*

*t t
*

*t
*

*t
*

b g

c hd ib gb gb g c hb g

c h

= FHG I KJ − F HG

I KJ = −

= − × ⋅ ×

×

= − ×

− −

− − ⋅

− −

− −

µ ε τ

µ ε τ

τ τ0 0 0 0 0 0

7 12

3

12

13 12

2 2

4 10 885 10 100 080 16

2 12 10 50

12 10

2π T m A V mm

s mm

T

C

N m ms

ms

2

. .

.

. .

The magnitude is ( ) ( )13 12 ms1.2 10 T .*tB t e*− −= ×

(b) At time *t* = 3τ, *B*(*t*) = –(1.2 × 10–13 T)*e*–3τ/τ = –5.9 × 10–15 T, with a magnitude |*B*(*t*)|=
5.9 × 10–15 T.

54. (a) Eq. 30-22 gives *B* =
µo *i r
*2π*R*2 =222 µT.

(b) Eq. 30-19 (or Eq. 30-6) gives *B* =
µo*i
*2π*r* =167 µT.

(c) As in part (b), we obtain a field of
µo*i
*2π*r* =22.7 µT.

(d) Eq. 32-16 (with Eq. 32-15) gives *B* =
µo*id r
*2π*R*2 =1.25 µT.

(e) As in part (d), we get
µo*id r
*2π*R*2 = 3.75 µT.

(f) Eq. 32-17 yields *B* = 22.7 µT.

(g) Because the displacement current in the gap is spread over a larger cross-sectional

area, values of *B* within that area are relatively small. Outside that cross-sectional area,

the two values of *B* are identical. See Fig. 32-23*b*.

55. (a) Again from Fig. 32-14, for *M*/*M*max = 50% we have *B*/*T* = 0.50. So *T = B*/0.50 =

2/0.50 = 4 K.

(b) Now *B*/*T* = 2.0, so *T* = 2/2.0 = 1 K.

56. (a) The complete set of values are {−4, −3, −2, −1,0,+1,+2,+3,+4} nine values in all.

(b) The maximum value is 4µB = 3.71 × 10−23 J/T.

(c) Multiplying our result for part (b) by 0.250 T gives *U* = +9.27 × 10−24 J.

(d) Similarly, for the lower limit, *U* = −9.27 × 10−24 J.

57. (a) Using Eq. 27-10, we find *E J
i
*

*A
*= = =

× ⋅ ×

= −

−ρ ρ 162 10 100

500 10 0 324

8

6

.

. . .

Ω m A m

V m 2

c hb g

(b) The displacement current is

( )( )( )12 80 0 0 0 16

8.85 10 F 1.62 10 2000A s

2.87 10 A.

*E
*

*d
*

*d dE d i di
i A A
*

*dt dt dt A dt
*

ρε ε ε ε ρ − −

−

Φ= = = = = × × Ω

= ×

(c) The ratio of fields is
*B i
*

*B i
*

*i r
*

*i r
*

*i
*

*i
*

*d d d
*due to

due to

A

100A

b g b g = = =

× = × −

−µ µ

0

0

16 182

2

2 87 10 2 87 10

π π

. . .

58. (a) Using Eq. 32-31, we find

µorb,*z* = –3µ*B* = –2.78 × 10–23 J/T.

(That these are acceptable units for magnetic moment is seen from Eq. 32-32 or Eq. 32-

27; they are equivalent to A·m 2 ).

(b) Similarly, for *m *= −4 we obtain µorb,*z* = 3.71 × 10–23 J/T.

59. Let the area of each circular plate be *A* and that of the central circular section be *a*,

then

*A
*

*a
*

*R
*

*R
*

= =π π

2

2 2

4 b g

.

Thus, from Eqs. 32-14 and 32-15 the total discharge current is given by *i = id* = 4(2.0 A)

= 8.0 A.

60. The interacting potential energy between the magnetic dipole of the compass and the

Earth’s magnetic field is *U B B
e e
*

= − ⋅ = −µ µ θcos , where θ is the angle between µ and
*B
e
*. For small angle θ

*U B B B
e e e
*

θ µ θ µ θ κθ µb g = − ≈ − −FHG I KJ = −cos 1 2

1

2

2 2

where κ = µ*Be*. Conservation of energy for the compass then gives

2

21 1 const. 2 2

*d
I
*

*dt
*

θ κθ+ =

This is to be compared with the following expression for the mechanical energy of a

spring-mass system:

1

2

1

2

2

2
*m
*

*dx
*

*dt
kx
*

F HG I KJ + = const. ,

which yields ω = *k m *. So by analogy, in our case

ω κ µ µ= = =
*I
*

*B
*

*I
*

*B
*

*ml
*

*e e
*

2 12 ,

which leads to

µ ω= = ×

× = ×

−

−

*ml
*

*B
e
*

2 2 2 2 2

6

2

12

0 050 4 0 10 45

12 16 10 8 4 10

. . . .

kg m rad s

T J T

b gc h b g c h

61. (a) At any instant the displacement current *id* in the gap between the plates equals the

conduction current *i* in the wires. Thus *i*max = *id* max = 7.60 µA.

(b) Since *id* = ε0 (*d*Φ*E*/*dt*),

*d
*

*dt
*

*i
E d
*

ΦF HG

I KJ = =

× ×

= × ⋅ −

− max

max . . .

ε 0

6

12

57 60 10 859 10 A

8.85 10 F m V m s

(c) According to problem 13,

*i C
dV
*

*dt
*

*A
*

*d
*

*dV
*

*dt
d
*= = ε 0 .

Now the potential difference across the capacitor is the same in magnitude as the emf of

the generator, so *V* = εm sin ω*t* and *dV*/*dt* = ωεm cos ω*t*. Thus, 0 m( / ) cos*di A d t*ε ωε ω=
and max 0 m / .*di A d*ε ωε= This means

( ) ( ) ( )( )212 30 m

6

max

8.85 10 F m 0.180 m 130rad s 220 V 3.39 10 m,

7.60 10 A
*d
*

*A
d
*

*i
*

ε ωε − − −

× π = = = ×

×

where *A = *π*R*2 was used.

(d) We use the Ampere-Maxwell law in the form *B ds I
d
*

⋅ =z µ 0 , where the path of
integration is a circle of radius *r* between the plates and parallel to them. *Id* is the

displacement current through the area bounded by the path of integration. Since the

displacement current density is uniform between the plates, *Id* = (*r
*2
/*R
*

2
)*id*, where *id* is the

total displacement current between the plates and *R* is the plate radius. The field lines are

circles centered on the axis of the plates, so *B * is parallel to *ds *. The field has constant

magnitude around the circular path, so *B ds rB*⋅ =z 2π . Thus,

2

0 0 2 2

2 . 2

*d
*

*d
*

*i rr
rB i B
*

*R R
*

µµπ = = π

The maximum magnetic field is given by

*B
i r
*

*R
*

*d
*

max max

. . .= =

× ⋅ × = ×

− −µ 0

2

6

2

12

2

4 7 6 10 0110

2 0 516 10

π π 10

π 0.18

−7 T m A A m

m T.

c hc hb g b g

62. The definition of displacement current is Eq. 32-10, and the formula of greatest

convenience here is Eq. 32-17:

( )( )6 7

0

2 0.0300 m 2.00 10 T2 0.300 A .

4 10 T m A
*d
*

*r B
i
*

µ

−

−

π ×π= = = π× ⋅

63. (a) For a given value of *, m * varies from – to + . Thus, in our case = 3, and the

number of different *m *’s is 2 + 1 = 2(3) + 1 = 7. Thus, since *L*orb,*z* ∝*m *, there are a total
of seven different values of *L*orb,*z*.

(b) Similarly, since µorb,*z* ∝*m *, there are also a total of seven different values of µorb,*z*.

(c) Since *L*orb,*z* = *m h*/2π, the greatest allowed value of *L*orb,*z* is given by | *m *|max*h*/2π =
3*h*/2π.

(d) Similar to part (c), since µorb,*z* = – *m *µ*B*, the greatest allowed value of µorb,*z* is given by
| *m *|maxµ*B* = 3*eh*/4π*me*.

(e) From Eqs. 32-23 and 32-29 the *z* component of the net angular momentum of the

electron is given by

net, orb, , . 2 2

*s
*

*z z s z
*

*m hm h
L L L*= + = +

π π

For the maximum value of *L*net,*z* let *m * = [ *m *]max = 3 and *ms *= 12 . Thus

*L
h h
*

*z*net ,
max

. .= +FHG

I KJ =3

1

2 2

35

2π π

(f) Since the maximum value of *L*net,*z* is given by [*mJ*]max*h*/2π with [*mJ*]max = 3.5 (see the
last part above), the number of allowed values for the *z* component of *L*net,*z* is given by

2[*mJ*]max + 1 = 2(3.5) + 1 = 8.

64. Ignoring points where the determination of the slope is problematic, we find the

interval of largest ∆ ∆*E t * is 6 µs < *t* < 7 µs. During that time, we have, from Eq. 32-14,

*i A
*

*E
*

*t
d
*= = ×ε ε0 0

2 62 0 2 0 10 ∆

∆ . .m V mc hc h

which yields *id* = 3.5 × 10–5 A.

65. (a) A sketch of the field lines (due to the presence of the bar magnet) in the vicinity of

the loop is shown below:

(b) For paramagnetic materials, the dipole momentµ is in the same direction as *B *. From
the above figure,µ points in the –*x* direction.

(c) Form the right-hand rule, since µ points in the –*x* direction, the current flows
counterclockwise, from the perspective of the bar magnet.

(d) The effect of *F * is to move the material towards regions of larger *B * values. Since

the size of *B * relates to the “crowdedness” of the field lines, we see that *F * is towards

the left, or –*x*.

66. (a) From Eq. 21-3,

*E
e
*

*r
*= =

× × ⋅

× = ×

−

−4

160 10 8 99 10

52 10 53 10

2

19 9 2

11 2

11

π 0ε . .

. . .

C N m C

m N C

2c hc h c h

(b) We use Eq. 29-28:

*B
r
*

*p*= =
× ⋅ ×

× = ×

− −

−

−µ µ0 3

7 26

11 3

2

2

4 10 14 10

2 52 10 2 0 10

π π

π

T m A J T

m T

c hc h c h

.

. . .

(c) From Eq. 32-30,

µ µ µ

µ µ

orb J T

J T
*p
*

*e
*

*p
*

*B
*

*p
*

*eh m*= = = ×
×

= × −

−

4 9 27 10

14 10 6 6 10

24

26

2π . .

. .

67. (a) From 2
*e
*

*iA i R*µ = = π we get

*i
R
e
*

= = × ×

= ×µ π π(6.37 102

22

6

88 0 10 6 3 10 .

. J / T

m) A .

2

(b) Yes, because far away from the Earth the fields of both the Earth itself and the current

loop are dipole fields. If these two dipoles cancel each other out, then the net field will be

zero.

(c) No, because the field of the current loop is not that of a magnetic dipole in the region

close to the loop.

68. (a) Using Eq. 32-13 but noting that the capacitor is being *discharged*, we have

*d E
*

*dt
*

*i
*

*A
*

| | .= − = − ×

ε 0 1588 10

where *A* = (0.0080)
2
and SI units are understood.

(b) Assuming a perfectly uniform field, even so near to an edge (which is consistent with

the fact that fringing is neglected in §32-4), we follow part (a) of Sample Problem 32-2

and relate the (absolute value of the) line integral to the portion of displacement current

enclosed.

7

0 ,enc 0 2 5.9 10 Wb/m.

*d
*

*WH
B ds i i
*

*L
*µ µ −⋅ = = = ×

69. (a) We use the notation *P*(µ) for the probability of a dipole being parallel to *B *, and
*P*(–µ) 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:

( ) ( ) ( ) ( )

( ) tanh .

*B KT B KT
*

*B KT B KT
*

*N e eN P N P B
M N
*

*P P e e kT
*

µ µ

µ µ

µµ µ µ µ µµ µ µ

−

−

−− − = = =

+ − +

(b) For µ*B kT*<< (that is, µ*B kT*/ << 1) we have *e±*µ*B*/*kT* ≈ 1 ± µ*B*/*kT*, so

*M N
B
*

*kT
*

*N B kT B kT
*

*B kT B kT
*

*N B
*

*kT
*= FHG

I KJ ≈

+ − − + + −

=µ µ µ µ µ

µ µ µ

tanh . 1 1

1 1

2b g b g b g b g

(c) For µ*B kT*>> we have tanh (µ*B*/*kT*) ≈ 1, so *M N B
kT
*

*N*= FHG
I
KJ ≈µ

µ µtanh .

(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.

70. (a) From Eq. 32-1, we have

( ) ( ) ( )( )2 3in out 0.0070Wb 0.40T 9.2 10 Wb.*B B r *−Φ = Φ = + π = ×

Thus, the magnetic of the magnetic flux is 9.2 mWb.

(b) The flux is inward.

71. (a) The Pythagorean theorem leads to

2 2

2 2 2 20 0 0

3 3 3

20

3

cos sin cos 4sin 4 2 4

1 3sin , 4

*h v m m m m
*

*m
*

*B B B
r r r
*

*r
*

µ µ µ µ µ µ

µ µ

= + = λ + λ = λ + λ π π π

= + λ π

where cos
2
λ*m* + sin2 λ*m* = 1 was used.

(b) We use Eq. 3-6:

tan sin

cos tan .φ

µ µ
µ µ*i
*

*m
*

*m
*

*m
*

*Bv
*

*Bh
*

*r
*

*r
*

= = =0 3

0

3

2

4 2

π λ

π λ λ

c h c h

72. (a) At the magnetic equator (λ*m* = 0), the field is

( ) ( ) ( )

7 22 2

50

33 6

4 10 T m A 8.00 10 A m 3.10 10 T.

4 4 6.37 10 m
*B
*

*r
*

µ µ − −π× ⋅ × ⋅= = = × π π ×

(b) φ*i* = tan–1 (2 tan λ*m*) = tan–1 (0) = 0° .

(c) At λ*m* = 60.0°, we find

( )2 5 2 50 3 1 3sin 3.10 10 1 3sin 60.0 5.59 10 T.4 *mB r
*µ µ − −= + λ = × + ° = ×
π

(d)φ*i* = tan–1 (2 tan 60.0°) = 73.9°.

(e) At the north magnetic pole (λ*m* = 90.0°), we obtain

( ) ( )22 5 50 3 1 3sin 3.10 10 1 3 1.00 6.20 10 T.4 *mB r
*µ µ − −= + λ = × + = ×
π

(f) φ*i* = tan–1 (2 tan 90.0°) = 90.0°.

73. (a) At a distance *r* from the center of the Earth, the magnitude of the magnetic field is

given by

*B
r
*

*m
*= +µ µ0

3

2

4 1 3

π λsin ,

where µ is the Earth’s dipole moment and λ*m* is the magnetic latitude. The ratio of the
field magnitudes for two different distances at the same latitude is

*B
*

*B
*

*r
*

*r
*

2

1

1

3

2

3 = .

With *B*1 being the value at the surface and *B*2 being half of *B*1, we set *r*1 equal to the

radius *Re* of the Earth and *r*2 equal to *Re + h*, where *h* is altitude at which *B* is half its

value at the surface. Thus,

1

2

3

3 =

+
*R
*

*R h
*

*e
*

*e*b g
.

Taking the cube root of both sides and solving for *h*, we get

( ) ( )( )1 3 1 3 32 1 2 1 6370km 1.66 10 km.*eh R*= − = − = ×

(b) We use the expression for *B* obtained in problem 6, part (a). For maximum *B*, we set

sin λ*m* = 1.00. Also, *r* = 6370 km – 2900 km = 3470 km. Thus,

( ) ( ) ( )

( ) 7 22 2

220 max 33 6

4

4 10 T m A 8.00 10 A m 1 3sin 1 3 1.00

4 4 m

3.83 10 T.

*m
B
*

*r
*

µ µ −

−

π× ⋅ × ⋅ = + λ = +

π π 3.47×10

= ×

(c) The angle between the magnetic axis and the rotational axis of the Earth is 11.5°, so

λ*m* = 90.0° – 11.5° = 78.5° at Earth’s geographic north pole. Also *r = Re* = 6370 km. Thus,

( ) ( ) ( )

7 22 2

20

33

5

4 10 T m A 8.0 10 J T 1 3sin 78.5 1 3sin

4 4 m

6.11 10 T.

*m
*

*E
*

*B
R
*

µ µ −

6

−

π× ⋅ × + ° = + λ =

π π 6.37×10

= ×

(d)φ
*i
*= ° = °−tan tan . . .1 2 78 5 84 2b g

(e) A plausible explanation to the discrepancy between the calculated and measured

values of the Earth’s magnetic field is that the formulas we obtained in problem 6 are

based on dipole approximation, which does not accurately represent the Earth’s actual

magnetic field distribution on or near its surface. (Incidentally, the dipole approximation

becomes more reliable when we calculate the Earth’s magnetic field far from its center.)

74. Let *R* be the radius of a capacitor plate and *r* be the distance from axis of the capacitor.

For points with *r *≤* R*, the magnitude of the magnetic field is given by

*B
r dE
*

*dt
*= µ ε0 0

2 ,

and for *r *≥* R*, it is

*B
R
*

*r
*

*dE
*

*dt
*= µ ε0 0

2

2 .

The maximum magnetic field occurs at points for which *r = R*, and its value is given by

either of the formulas above:

*B
R dE
*

*dt
*max .=

µ ε0 0 2

There are two values of *r* for which *B = B*max/2: one less than *R* and one greater.

(a) To find the one that is less than *R*, we solve

µ ε µ ε0 0 0 0 2 4

*r dE
*

*dt
*

*R dE
*

*dt
*=

for *r*. The result is *r = R*/2 = (55.0 mm)/2 = 27.5 mm.

(b) To find the one that is greater than *R*, we solve

µ ε µ ε0 0 2

0 0

2 4

*R
*

*r
*

*dE
*

*dt
*

*R dE
*

*dt
*=

for *r*. The result is *r* = 2*R* = 2(55.0 mm) = 110 mm.

75. (a) Since the field lines of a bar magnet point towards its South pole, then the *B *

arrows in one’s sketch should point generally towards the left and also towards the

central axis.

(b) The sign of *B dA*⋅ for every *dA * on the side of the paper cylinder is negative.

(c) No, because Gauss’ law for magnetism applies to an *enclosed* surface only. In fact, if

we include the top and bottom of the cylinder to form an enclosed surface *S* then

*s
*

*B dA*z ⋅ = 0 will be valid, as the flux through the open end of the cylinder near the
magnet is positive.