Two concentric circular coil,s 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

Let a current `I_(2)` flow through the outer circular coil. The field at the centre of the coil is `B_(2)=mu_(0) I_(2) // 2r_(2)`. Since the other co-axially placed coil has a very small radius, `B_(2)` may be considered constant over its cross- sectional area, Hence.
`Phi_(1)=pi r_(1)^(2)B_(2)=(mu_(0)pir_(1)^(2))/(2r_(2))I_(2)=M_(12)I_(2)` thus,
`:.` mutual inductance of solenoid `S_(1)` with respect to `S_(2)`
`M_(12)=(mu_(0)pir_(1)^(2))/(2r_(2))` -------(i) but `M_(12) = M_(21) = (mu_(0)pi r_(1)^(2))/(2r_(2))`,
Note that we calculated `M_(12)` from an approximate value of `Phi_(1)` assuming the magnetic field `B_(2)` to be uniform over the area. `pi r^(2)`. However we can accept this value because `r_(1) lt lt r_(2)`
`***` It would have been difficult to calculate the flux through the bigger coil of the nonuniform field due to the current in the smaller coil and hence the mutual inductance `M_(12)`. The equality `M_(12) = M_(21)` is helpful. Note also that mutual inductance depends solely on the geometry.