Problem
4.
A circular wire loop
40 cm
in diameter has 100-
!
resistance and lies in a horizontal plane. A
uniform magnetic field points vertically downward, and in
25 ms
it increases linearly from
5.0 mT to 55 mT.
Find the magnetic flux through the loop at (a) the beginning and (b) the end of the
25 ms
period. (c) What is the loop current during this time?
(d) Which way does this current flow?
Solution
(a) As in the previous solution,
!
B
=
B
"
A
=
1
4
#
d
2
B
=
1
4
#
(40 cm)
2
(5 mT)
=
6.28
$
10
%
4
Wb at
t
1
=
0, and
(b)
6.91
!
10
"
3
Wb at
t
2
=
25 ms.
(c) Since the field increases linearly,
d
!
B
=
dt
=
"
!
B
=
"
t
=
(
6
.
91
#
0
.
628
)
$
10
!
3
Wb
=
25 ms
=
0.251 V.
From Faraday’s law, this is equal to
the magnitude of the induced emf, which causes a current
I
=
E
=
R
=
0
.
251 V
=
100
!
=
2
.
51 mA
in the
loop. (d) The direction must oppose the increase of the external field downward, hence the induced field is
upward and
I
is CCW when viewed from above the loop.
Problem
6.
A conducting loop of area
A
and resistance
R
lies at right angles to a spatially uniform magnetic field.
At time
t
=
0
the magnetic field and loop current are both zero. Subsequently, the current increases
according to
I
=
bt
2
,
where
b
is a constant with the units
A/s
2
.
Find an expression for the magnetic
field strength as a function of time.
Solution
The induced current and the derivative of the magnetic field strength are related as in the previous problem,
dB
=
dt
=
IR
=
A
=
(
bR
=
A
)
t
2
.
Integration yields
B
(
t
)
=
(
bR
=
A
)
t
3
=
3
,
where
B
(0)
=
0
was specified.
Problem
9.
A square wire loop of side
l
and resistance
R
is pulled with constant speed
v
from a region of no
magnetic field until it is fully inside a region of constant, uniform magnetic field
B
perpendicular to the
loop plane. The boundary of the field region is parallel to one side of the loop. Find an expression for
the total work done by the agent pulling the loop.
Solution
The loop can be treated analogously to the situation analyzed in Section 31-3, under the heading “Motional
EMF and Lenz’s Law”; instead of exiting, the loop is entering the field region at constant velocity. All
quantities have the same magnitudes, except the current in the loop is CCW instead of CW, as in Fig. 31-
13. Since the applied force acts over a displacement equal to the side-length of the loop, the work done can
be calculated directly:
W
app
=
F
app
!
l
=
(
I
l
B
)
l
=
I
l
2
B
.
But,
I
=
E
=
R
=
d
!
B
=
dt
=
R
=
d
=
dt
(
B
l
x
)
=
R
=
B
l
v=
R
,
as before, so
W
app
=
B
2
l
3
v=
R
.
[Alternatively, the work
can be calculated from the conservation of energy:
I
=
B
l
v=
R
,
P
diss
=
I
2
R
=
(
B
l
v
)
2
=
R
,
and
W
app
=
P
diss
t
=
[(
B
l
v
)
2
=
R
](
l
=v
).]
Problem
14. A square wire loop
3.0 m
on a side is perpendicular to a uniform magnetic field of
2.0 T.
A 6-V light
bulb is in series with the loop, as shown in Fig. 31-45. The magnetic field is reduced steadily to zero
over a time
!
t
.
(a) Find
!
t
such that the light will shine at full brightness during this time. (b) Which
way will the loop current flow?

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