Download the app

Questions  

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

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

detailed solution

Correct option is A

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

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?


Similar Questions

Two different coils have self-inductance  L1=8mH, L2=2mH . The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same rate. At a certain instant of time, the power given to the two coils is the same. At that time the current, the induced voltage and the energy stored in the first coil are i1,V1 and W1 respectively. Corresponding values for the second coil at the same instant are i2,V2  and W2 respectively. Then


phone icon
whats app icon