A long solenoid of length `L`, cross section `A` having `N_(1)` turns has about its center a small coil of `N_(2)` turns as shows in Fig The mutual inductance of two circuits is

Options

A. `(mu_(0)A(N_(1)//N_(2)))/(L)`
B. `(mu_(0)A(N_(1)N_(2)))/(L)`
C. `mu_(0) A N_(1)N_(2) L`
D. `(mu_(0)AN_(1)^(2)N_(2))/(L)`

Answer

B

Explanation

`phi_(2) = N_(2)B_(1)A` or `phi_(2) = N_(2) (mu_(0)N_(1)I_(1))/(L)A`
or `phi_(2) = (mu_(0)N_(1)N_(2)A)/(L) I_(1)`
Comparing with `phi_(2) = MI_(1)`, we get `M_(2) = (mu_(0)N_(1)N_(2)A)/(L)`