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JEE Main Physics Class 12 Alternating Currents Part 1 Questions
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© examsnet.com
Question : 58
Total: 100
A small circular loop of wire of radius
a
is located at the centre of a much larger circular wire loop of radius
b
. The two loops are in the same plane. The outer loop of radius
b
carries an alternating current
I
=
I
0
cos
(
ω
t
)
.
The emf induced in the smaller inner loop is nearly:
[Online April 8, 2017]
π
µ
0
I
0
2
.
a
2
b
ω
sin
(
ω
t
)
π
µ
0
I
0
2
.
a
2
b
ω
cos
(
ω
t
)
π
µ
0
I
0
a
2
b
ω
sin
(
ω
t
)
π
µ
0
I
0
b
2
a
ω
cos
(
ω
t
)
Validate
Solution:
For two concentric circular coil,
Mutual Inductance
M
=
µ
0
π
N
1
N
2
a
2
2
b
here,
N
1
=
N
2
=
1
Hence,
M
=
µ
0
π
a
2
2
b
..... (i)
and given
I
=
I
0
cos
ω
t
..... (ii)
Now according to Faraday's second law induced emf
e
=
−
M
d
I
d
t
From eq. (ii),
e
=
−
µ
0
π
a
2
2
b
d
d
t
(
I
0
cos
ω
t
)
e
=
µ
0
π
a
2
2
b
I
0
sin
ω
t
(
ω
)
e
=
π
µ
0
I
0
2
.
a
2
b
ω
sin
ω
t
© examsnet.com
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