Question

A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3}$ 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

Answer: (A)

For a long solenoid, $B=\mu_0ni=\mu_0\frac{N}{l}.i$
Flux $=\mu_0\frac{N}{l}.i.A$
given flux per turn = 4 x $10^{-3}$, i = 2 A
$\therefore \, Total \, flux = 4\times1 0^{-3}$
$L=\big(\mu_0\frac{N}{l}.NA\big)=\frac{4\times10^{-3}\times500}{2}=1$ henry