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Q.

A circular coil of radius 5 cm has 500 turns of a wire. The approximate value of the coefficient of self-induction of the coil will be

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a

25 mH

b

25×10-3mH

c

50×10-3mH

d

50×10-3H

answer is A.

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Detailed Solution

ϕ=Li⇒NBA=LiSince magnetic field at the centre of circular coil carrying current is given by B=μ04π·2πNir∴  N·μ04π·2πNir·πr2=Li⇒L=μ0N2πr2Hence, self-inductance of a coil=4π×10-7×500×500×π×0.052=25 mH
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A circular coil of radius 5 cm has 500 turns of a wire. The approximate value of the coefficient of self-induction of the coil will be