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 millihenry
b
25×10−3 millihenry
c
50×10−3 millihenry
d
50×10−3henry
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