291
E
LECTROMAGNETIC
I
NDUCTION
29.1.
I
DENTIFY
:
Altering the orientation of a coil relative to a magnetic field changes the magnetic flux through the
coil. This change then induces an emf in the coil.
S
ET
U
P
:
The flux through a coil of
N
turns is
Φ
=
NBA
cos
φ
, and by Faraday°s law the magnitude of the induced
emf is
E
=
d
Φ
/dt
.
E
XECUTE
:
(a)
ΔΦ
=
NBA
= (50)(1.20 T)(0.250 m)(0.300 m) = 4.50 Wb
(b)
E
=
d
Φ
/dt =
(4.50 Wb)/(0.222 s) = 20.3 V
E
VALUATE
:
This induced potential is certainly large enough to be easily detectable.
29.2.
I
DENTIFY
:
B
t
ΔΦ
=
Δ
E
.
cos
B
BA
φ
Φ
=
.
B
Φ
is the flux through each turn of the coil.
S
ET
U
P
:
i
0
φ
=
°
.
f
90
φ
=
°
.
E
XECUTE
:
(a)
5
4
2
8
,i
cos0
(6.0
10
T)(12
10
m
)(1)
7.2
10
Wb.
B
BA
−
−
−
Φ
=
=
×
×
=
×
°
The total flux through the coil is
8
5
,i
(200)(7.2
10
Wb)
1.44
10
Wb
B
N
−
−
Φ
=
×
=
×
.
,f
cos90
0
B
BA
Φ
=
=
°
.
(b)
5
4
i
f
1.44
10
Wb
3.6
10
V
0.36 mV
0.040 s
N
N
t
−
−
Φ −
Φ
×
=
=
×
=
Δ
=
E
.
E
VALUATE
:
The average induced emf depends on how rapidly the flux changes.
29.3.
I
DENTIFY
and
S
ET
U
P
:
Use Faraday°s law to calculate the average induced emf and apply Ohm°s law to the coil
to calculate the average induced current and charge that flows.
(a) E
XECUTE
:
The magnitude of the average emf induced in the coil is
av
.
B
I
N
t
ΔΦ
=
Δ
E
Initially,
i
cos
.
B
BA
BA
φ
Φ
=
=
The final flux is zero, so
f
i
av
.
B
B
NBA
N
t
t
Φ
− Φ
=
=
Δ
Δ
E
The average induced current is
av
.
NBA
I
R
R t
=
=
Δ
E
The total charge that flows through the coil is
.
NBA
NBA
Q
I
t
t
R t
R
⎛
⎞
=
Δ =
Δ =
⎜
⎟
Δ
⎝
⎠
E
VALUATE
:
The charge that flows is proportional to the magnetic field but does not depend on the time
.
t
Δ
(b)
The magnetic stripe consists of a pattern of magnetic fields. The pattern of charges that flow in the reader coil
tell the card reader the magnetic field pattern and hence the digital information coded onto the card.
(c)
According to the result in part (a) the charge that flows depends only on the change in the magnetic flux and it
does not depend on the rate at which this flux changes.
29.4.
I
DENTIFY
and
S
ET
U
P
:
Apply the result derived in Exercise 29.3:
/
.
Q
NBA R
=
In the present exercise the flux
changes from its maximum value of
B
BA
Φ
=
to zero, so this equation applies.
R
is the total resistance so here
60.0
45.0
105.0
.
R
=
Ω +
Ω =
Ω
E
XECUTE
:
5
4
2
(3.56
10
C)(105.0
)
says
0.0973 T.
120(3.20
10
m )
NBA
QR
Q
B
R
NA
−
−
×
Ω
=
=
=
=
×
E
VALUATE
:
A field of this magnitude is easily produced.
29.5.
I
DENTIFY
:
Apply Faraday°s law.
S
ET
U
P
:
Let +
z
be the positive direction for
A
!
. Therefore, the initial flux is positive and the final flux is zero.
E
XECUTE
:
(a)
and
(b)
2
3
0
(1.5 T)
(0.120 m)
34 V.
2.0
10
s
B
t
π
−
ΔΦ
−
= −
= −
= +
Δ
×
E
Since
E
is positive and
A
!
is toward us,
the induced current is counterclockwise.
E
VALUATE
:
The shorter the removal time, the larger the average induced emf.
29
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292
Chapter 29
29.6.
I
DENTIFY
:
Apply Eq.(29.4).
/ .
I
R
=
E
S
ET
U
P
:
/
/
.
B
d
dt
AdB dt
Φ
=
E
XECUTE
:
(a)
(
)
5
4
4
(
)
(0.012 T/s)
(3.00
10
T/s )
B
Nd
d
d
NA
B
NA
t
t
dt
dt
dt
−
Φ
=
=
=
+
×
E
(
)
4
4
3
4
3
3
(0.012 T/s)
(1.2
10
T/s
)
0.0302 V
(3.02
10
V/s )
.
NA
t
t
−
−
=
+
×
=
+
×
E
(b)
At
5.00 s,
t
=
4
3
3
0.0302 V
(3.02
10
V/s
)(5.00 s)
0.0680 V.
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 Spring '08
 Jabbour
 Materials Science And Engineering, Magnetic Field, Faraday

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