Question

A long solenoid has `500` turns. When a current of `2A` is passed through it, the resulting magnetic flux linked with each turn of the solenoid is `4xx10^(-3)Wb` . The self-inductance of the solenoid is
A. 1.0 henry
B. 4.0 henry
C. 2.5 henry
D. 2.0 henry

Answer 1

Correct Answer - A
`"Flux linkage "="Flux through each turn"xx"number of turn"`
`phi=500xx4xx10^(-3)=2Wb`
since `phi=Li`
`rArr L=(phi)/(i)=(2)/(2)=1H`