A straight solenoid has 50 turns per cm in primary and 200 turns in the secondary. The area of cross-section of the solenoid is 4 `cm^(2)`. Calculate the mutual inductance.

Explanation

The magnetic field at any point inside the straight solenoid of primary with `n_1` turns per unit length carrying a current `i_1` is given by the relation,
`B=mu_0n_1i_1`
The magnetic flux through the secondary of `N_2` turns each of area `S` is given as
`N_2phi_2=N_2(BS)=mu_0n_1N_2i_1S`
`:. M=(N_2phi_2)/i_1=mu_0n_1N_2S`
Substituting the values we get
`M=(4pixx10^-7)(50/10^-2)(200)(4xx10^-4)`
`=5.0xx10^-4H`