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 two-loop 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
right-hand side and the left-hand 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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