Two concentric circular coils, one of small radius `r_1` and the other of large radius `r_2`, such that `r_1 lt lt r_2`, are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement.

Explanation

Consider two extremely long co-axial circular coils (or solenoids) of equal length l, with smaller one inside the bigger one. Let small and big coils be coil no. 1 and coil no. 2 respectively. Their common area of cross section is suppose `a=pir_1^2`.
When `I_1` amount of current is passed through `N_1` turns of small coil, magnetic flux linked with big coil is,
`Phi_2=N_2B_1 a cos 0^@`
(Here, magnetic field is present only inside the small coil and so cross-sectional area of small coil is considered).
`therefore Phi_2=N_2((mu_0N_1I_1)/l)a`
`therefore Phi_2/I_1=(mu_0N_1N_2a)/l`
Here `Phi_2/I_1=M_21`=mutual inductance of second coil towards first coil.
`therefore M_21=(mu_0N_1N_2a)/l`....(1)
Similarly when `I_2` amount of current is passed through `N_2` no. of turns of big coil, magnetic flux linked with small coil is,
`Phi_1=N_1B_2a`
`=N_1((mu_0N_2I_2)/l)a`
`therefore Phi_1/I_2=(mu_0N_1N_2a)/l`
Here `Phi_1/I_2=M_12`=mutual inductance of first coil towards second coil.
`therefore M_12 =(mu_0N_1N_2a)/l` ....(2)
Equations (1) and (2) prove that
`M_12=M_21`...(3)
From equations (1),(2) and (3)
`M_12=M_21=M=(mu_0N_1N_2a)/l`
(where `a=pir_1^2`=cross-sectional area of small coil)