song (shs546) – Practice HW 16 Solutions – weathers – (22202)
1
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001
0.0points
A coil is wrapped with 224 turns of wire on the
perimeter of a square frame of sides 21
.
6 cm.
Each turn has the same area, equal to that of
the frame, and the total resistance of the coil
is 2
.
42 Ω. A uniform magnetic field is turned
on perpendicular to the plane of the coil.
If
the
field
changes
linearly
from
0
to
−
0
.
237 Wb
/
m
2
in a time of 1
.
16 s, find the
magnitude of the induced emf in the coil while
the field is changing.
Correct answer: 2
.
13524 V.
Explanation:
BasicConcept:
Faraday’s Law is
E
=
−
d
Φ
B
dt
.
Solution:
The magnetic flux through the
loop at
t
= 0 is zero since
B
= 0.
At
t
=
1
.
16 s
,
the magnetic flux through the loop is
Φ
B
=
B A
=
−
0
.
0110575 Wb
.
Therefore the
magnitude of the induced emf is
E
=
N
·
ΔΦ
B
Δ
t
=
(224 turns) [(
−
0
.
0110575 Wb)
−
0]
(1
.
16 s)
=
−
2
.
13524 V
E
= 2
.
13524 V
.
002
0.0points
The twoloop wire circuit is 43
.
7559 cm wide
and 29
.
1706 cm high.
The wire circuit in
the figure is located in a magnetic field whose
magnitude varies with time according to the
expression
B
= (0
.
001 T
/
s)
t
and its direction
is into the page.
Assume The resistance per length of the
wire is 0
.
113 Ω
/
m.
B
B
P
Q
14
.
5853 cm
29
.
1706 cm
29
.
1706 cm
When the magnetic field is 0
.
5 T, find the
magnitude of the current through middle leg
PQ
of the circuit.
Correct answer: 215
.
122
μ
A.
Explanation:
Let :
ℓ
= 0
.
291706 m
,
A
l
=
1
2
ℓ
2
= 0
.
0850924 m
2
/
s
,
A
r
=
ℓ
2
= 0
.
170185 m
2
/
s
,
δ
= 0
.
113 Ω
/
m
,
and
d B
dt
=
d
dt
α t
=
α
= 0
.
001 T
/
s
.
BasicConcept:
Faraday’s Law is
E
=
−
d
Φ
B
dt
,
where
Φ
B
=
A
B .
Ohm’s Law is
V
=
I R .
Solution:
The instantaneous value of the
magnetic field (
B
= 0
.
5 T) is not germane to
this problem.
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song (shs546) – Practice HW 16 Solutions – weathers – (22202)
2
B
B
P
Q
ℓ
2
ℓ
ℓ
I
r
I
l
I
PQ
The resistance for the wire is proportional
to the length of the wire.
For a length of
29
.
1706 cm, the resistance is
R
=
δ ℓ
= (0
.
113 Ω
/
m) (0
.
291706 m) = 0
.
0329628 Ω
.
Using Ohm’s law for the right perimeter and
left perimeter of the left loops in the circuit,
we have
E
r
= 4
ℓ δ I
r
= 4
R
I
r
,
and
(1)
E
l
= 3
ℓ δ I
l
= 3
R
I
l
.
(2)
Note:
When the magnetic field changes
with time, there is an induced emf in both the
righthand side and the lefthand side. From
Faraday’s law, the magnitude of the induced
emf in the loops equals
E
r
=
A
r
d B
dt
=
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 Spring '00
 Littler
 Magnetic Field, Faraday, iL, Faraday's law of induction

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