A long solenoid has `1000` turns. When a current of `4A` flows through it, the magnetic flux linked with each turn of the solenoid is `4xx10^(-3)Wb`. The self-inductance of the solenoid is

Options

A. 2H
B. 1H
C. 4H
D. 3H

Answer

B

Explanation

Here , `N = 1000 , I = 4A , phi_0 = 4 xx 10^(-3) Wb`
Total flux linked with the solenoid , `phi = Nphi_0`
` = 1000 xx 4 xx 10^(-3)Wb = 4Wb`
Since `phi = LI`
`therefore `Self - inductance of solenoid
`L = phi/I = (4Wb)/(4A) = 1H`