A thin copper wire of length 100 m is wound as a solenoid of length l and radius r. Its self-inductance is found to be L. Now if the same length of wire is wound as a solenoid of length l but of radius r/2, then its self-inductance will be :
4L
2L
L/2
L
A
L
B
L/2
C
4L
D
2L
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Solution
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L=μN2arealength
L′=μN′2area′length′
N=1002πr, N′=1002πr/2=100πr=2N
length=length′=l
area=πr2 , area′=π(r/2)2
therefore,
L′=μ(2N)2πr2/4l=μN2πr2l=L
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