B
5
m
0
I
p
r
.
a
B
i
D
s
5
B
1
2
p
r
2
.
2
I
.
r
.
R
2
:
B
5
m
0
I
2
p
r
.
a
B
i
D
s
5
B
1
2
p
r
2
.
R
1
,
r
,
R
2
:
R
3
I
I
R
2
R
1
B
i
5
B
.
B
5
0.
B
1
2
p
r
2
5
0
I
encl
5
0,
r
.
R
3
,
B
5
m
0
I
2
p
r
.
B
1
2
p
r
2
5 m
0
I
.
I
encl
5
I
,
R
1
,
r
,
R
2
,
a
B
i
D
s
5
B
1
2
p
r
2
.
B
S
B
5
m
0
I
2
p
r
.
B
1
2
p
r
2
5 m
0
I
I
encl
5
I
,
a
B
i
D
s
5
B
a
D
s
5
B
1
2
p
r
2
.
B
S
a
B
i
D
s
5 m
0
I
encl
.
I
5
2.80
3
10
2
5
T
#
m
4
p 3
10
2
7
T
#
m
/
A
5
22.3 A.
m
0
I
5
2.80
3
10
2
5
T
#
m
I
encl
m
0
I
encl
,
B
i
D
s
Magnetic Field and Magnetic Forces
20-21

20.78.
Set Up:
The toroidal solenoid is sketched in Figure 20.78. A few representative turns of wire are shown in
the sketch. These turns carry current into the page at the inner edge of the solenoid and carry current out of the page
at the outer edge of the solenoid. Apply Ampere’s law to a circular path of radius
r.
By symmetry the magnetic field
is constant on the path and tangent to it, so
Figure 20.78
Solve: (a)
No current passes through the path so the enclosed current is zero. Ampere’s
law gives
and
Each wire passes through the path twice, once carrying current into the page and once
carrying current out of the page. Therefore, the net current through the path is zero. Ampere’s law gives
and
(b)
Each of the
N
turns carries current through the area enclosed by the path and the
enclosed current is
Ampere’s law gives
and
20.79.
Set Up:
with
The magnetic field in the material is a factor of
greater than it is in vacuum.
Solve: (a)
(b)
Reflect:
The magnetic field inside the paramagnetic material is larger than the external field, but only slightly larger.
20.80.
Set Up:
Solve: (a)
(b)
m 5
K
m
m
0
5
1
1.0015
21
4
p 3
10
2
7
T
#
m
/
A
2
5
1.259
3
10
2
6
T
#
m
/
A
K
5
1.5023 T
1.5000 T
5
1.0015.
m 5
K
m
m
0
K
m
5
B
inside
/
B
outside
.
B
inside
5
K
m
B
external
5
1
1.00026
21
1.3500 T
2
5
1.3504 T
m 5
1
1.00026
21
4
p 3
10
2
7
T
#
m
/
A
2
5
1.257
3
10
2
6
T
#
m
/
A
K
m
m
0
5
4
p 3
10
2
7
T
#
m
/
A.
m 5
K
m
m
0
,
B
5
m
0
IN
2
p
r
.
B
1
2
p
r
2
5 m
0
NI
NI
.
a
B
i
D
s
5
B
1
2
p
r
2
.
a
,
r
,
b
:
B
5
0.
B
1
2
p
r
2
5
0
a
B
i
D
s
5
B
1
2
p
r
2
.
r
.
b
:
B
5
0.
B
1
2
p
r
2
5
0
a
B
i
D
s
5
B
1
2
p
r
2
.
r
,
a
:
I
I
I
I
I
I
r
B
i
5
B
.
20-22
Chapter 20

20.81.
Set Up:
Use
to calculate
With
downward,
and
The direction of the force is given by the right-hand rule and the magnitude is given by
The charge of the ball is
Solve:
The directions of
and
are shown in Figure 20.81, where the downward direction is into the page. The
right-hand rule (rhr) direction is north, but since the charge is negative the force is opposite the rhr direction and is to
the south. At the bottom of the shaft the speed of the ball is
so
Figure 20.81
Reflect:
The magnetic force is much less than the gravity force on the ball.
20.82.
Set Up:
The magnetic force
must be upward and equal to
mg.
The direction of
is determined by the
direction of
I
in the circuit.
with
where
V
is the battery voltage.
Solve: (a)
The forces are shown in Figure 20.82. The current
I
in the bar must be to the right to produce
upward.