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Physics Question on Inductance

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 4×10−34 \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

1.0 henry

Explanation

Solution

The correct answer is A:1.0 henry
For a long solenoid, B=μ0ni=μ0Nl.iB=\mu_0ni=\mu_0\frac{N}{l}.i
Flux =μ0Nl.i.A=\mu_0\frac{N}{l}.i.A
given flux per turn = 4 x 10−310^{-3}, i = 2 A
∴ Total flux=4×10−3\therefore \, Total \, flux = 4\times1 0^{-3}
L=(μ0Nl.NA)=4×10−3×5002=1L=\big(\mu_0\frac{N}{l}.NA\big)=\frac{4\times10^{-3}\times500}{2}=1 henry