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

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