Electro Magnetic Induction And Alternating Currents Question 499

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

Options:

A) 25mH

B) 25×10−3,mH

C) 50×10−3,mH

D) 50×10−3,H

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=25mH

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