Electro Magnetic Induction And Alternating Currents Question 552

Question: 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 [MP PET 1996; Pb PET 2000]

Options:

A) 25 millihenry

B) 25×10−3 millihenry

C) 50×10−3 millihenry

D) 50×10−3 henry

Show Answer

Answer:

Correct Answer: A

Solution:

φ=Li⇒NBA=Li

Since magnetic field at the centre of circular coil carrying current is given by

B=μ04π.2πNir

∴ N.μ04π.2πNir.πr2=Li⇒L=μ0N2πr2 Hence self inductance of a coil

=4π×10−7×500×500×π×0.052=25 mH

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