Two circular coils, one of smaller radius `r_(1)` and the other of very large radius `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) = µ_(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)=pir_(1)^(2)B_(2)`
`=(mu_(0)pir_(1)^(2))/(2r_(2))I_(2)`
`M_(12)I_(2)`
Thus,
`M_(12)=(mu_(0)pir_(1)^(2))/(2r_(2))`
From Eq. (6.14)
`M_(12)=M_(21)=(mu_(0)pir_(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 `pir_(1)^(2)`. However, we can accept this value because `r_(1) gtgt r_(2)`.