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IIT JEE Advanced 2011 Paper 1
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© examsnet.com
Question : 23
Total: 69
A long circular tube of length 10 m and radius 0.3 m carries a current I along its curved surface as shown. A wire-loop of resistance 0.005 Ω and of radius 0.1 m is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as I =
I
0
c
o
s
(
300
t
)
, where
I
0
is constant. If the magnetic moment of the loop is
N
μ
0
I
0
s
i
n
(
300
t
)
, then N is ______.
Your Answer:
Validate
Solution:
The flux through the ring is
ϕ =
B
π
r
2
Assuming the cylinder as a solenoid with close winding, we have
B =
µ
0
I
L
Therefore,
ϕ =
(
µ
0
I
0
L
)
π
r
2
c
o
s
300
t
The induced emf is
ε =
−
d
ϕ
d
t
=
300
(
µ
0
I
0
L
)
π
r
2
s
i
n
300
t
Therefore, the current induced is
i =
ε
R
=
(
π
r
2300
R
L
)
µ
0
I
0
s
i
n
300
t
The magnetic moment is
M = Current × Area of loop
Therefore,
m =
(
(
3.14
)
2
×
(
0.1
)
2
×
300
0.005
×
10
)
µ
0
I
0
s
i
n
300
t
=
6
µ
0
I
0
s
i
n
300
t
Hence, N = 6.
© examsnet.com
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