Physics 21
Fall, 2011
Solution to HW17
297
The current in the long, straight wire
AB
shown in
the figure is upward and is increasing steadily at a rate
di/dt
.
(a,b) At an instant when the current is
i
, what are the mag
nitude and direction of the field
B
at a distance
r
to the
right of the wire? (c) What is the flux
d
Φ
B
through the
narrow shaded strip? (d) What is the total flux through the
loop? (e) What is the induced emf in the loop? (f) Evalu
ate the numerical value of the induced emf if
a
= 12
.
0 cm,
b
= 36
.
0 cm,
L
= 24
.
0 cm, and
di/dt
= 9
.
60 A/s.
(a,b)
B
=
μ
0
i
(
t
)
2
πr
into page (by right hand rule)
(c)
d
Φ =
B dA
=
μ
0
i
(
t
)
2
πr
Ldr
(d) Integrate
Φ =
integraldisplay
b
a
d
Φ =
μ
0
i
(
t
)
L
2
π
integraldisplay
b
a
dr
r
=
μ
0
i
(
t
)
L
2
π
ln
b
a
(e)
E
=
d
Φ
dt
=
μ
0
L
2
π
di
dt
ln
b
a
(f)
(2
×
10

7
)(0
.
24m)(9
.
6)
parenleftbigg
ln
36
12
parenrightbigg
= 5
.
06
×
10

7
V
What will be the direction of the current? Counterclockwise.
298
A flat, circular, steel loop of radius 75 cm is at rest
in a uniform magnetic field, as shown in an edgeon view
in the figure. The field is changing with time, according to
B
(
t
) = (1
.
4T)
e

(0
.
057
s

1
)
t
. (a) Find the emf induced in the
loop as a function of time (assume
t
is in seconds). (b) When
is the induced emf equal to 1/20 of its initial value? (c) Find
the direction of the current induced in the loop, as viewed
from above the loop.
(a) We can find the emf induced in the loop using
contintegraldisplay
E
·
d
l
=

d
dt
integraldisplay
B
·
d
A
=

d
Φ
dt
,
where the integral of
E
·
d
l
is the induced emf. Because the
magnetic field makes a 60 degree angle with the loop, we can
write
integraldisplay
B
·
d
A
= Φ =
B
(
t
)
A
sin(60)
where
A
=
πr
2
is the area of the loop. Because the area
of the loop doesn’t change with time,
d
Φ
/dt
=
πr
2
dB/dt
,
where
d
dt
B
(
t
) = (1
.
4)(

0
.
057)
e

0
.
057
t
Putting these pieces together results in:
E
=

πr
2
sin(60)(1
.
4)(

0
.
057)
e

0
.
057
t
= 0
.
122
e

0
.
057
t
We usually take
E
to be the magnitude of the emf, and don’t
worry about the sign. The negative sign is related to the
polarity of the voltage.
(b) Now that we have an expression for the induced emf as
a function of time, we can solve for
t
at a particular emf.
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 Spring '08
 Hickman
 Current, Work, Flux, Magnetic Field, Lenz

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