Example

Question:

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

Solution:

Let a current \( I_2 \) flow through the outer circular coil. The field at the centre of the coil is: \[ B_2 = \frac{\mu_0 I_2}{2 r_2} \] Since the small co-axially placed coil has radius \( r_1 \ll r_2 \), \( B_2 \) may be considered constant over its cross-sectional area:
Flux through small coil: \[ \Phi_1 = \pi r_1^2 B_2 = \pi r_1^2 \left( \frac{\mu_0 I_2}{2 r_2} \right) = \frac{\mu_0 \pi r_1^2}{2 r_2} I_2 \] Thus, \[ M_{12} = \frac{\mu_0 \pi r_1^2}{2 r_2} \] From symmetry, \( M_{21} = M_{12} \).
Note: This is valid since \( r_1 \ll r_2 \) and \( B_2 \) is nearly uniform over small coil area.