Question

A long solenoid has 500 turns. When a current of 2 A is passed through it, the

resulting magnetic flux linked with each turn of the solenoid is 4×10

−3 Wb. find

self-inductance.

Answer 1

$$\phi=NLI$$

$$L=\frac{\phi}{IN}=\frac{4*10^{-3}}{2*500}=4*10^{-6}\:\rm H$$