A coil of copper wire (radius-r, self inductance-L) is bent in two concentric turns having radius r/2. The self inductance now is
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a
2L
b
4L
c
L/2
d
L
answer is A.
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Detailed Solution
magnetic field due to a coil is B=μ0ni2r , flux linked with the coil is Ï=nAB = nÏr2μ0ni2r=μ0n2Ïr2i=Li , so Lân2r , here n increases by two times and r decreases by 2 times , so L lincreases by two times