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

Suppose a current `I_(2)` flows through the outer circular coil. The field at the centre of the coil is

`B_(2) = (mu_(0)I_(2))/(2r_(2))`
The second co-axially placed coil has very small radius. So `B_(2)` may be considered constant over its cross-sectional area.
Now, `phi_(1) = pi r_(1)^(2)B_(2) = pir_(1)^(2)((mu_(0)I_(2))/(2r_(2)))` or `phi_(1) = (mu_(0)pir_(1)^(2))/(2r_(2))I_(2)`
Comparing with `phi_(1) = M_(12)I_(2)`, we get , `M_(12) = (mu_(0)pi r_(1)^(2))/(2r_(2))`
Aslo, `M_(21) = M_(12) = (mu_(0)pir_(1)^(2))/(2r_(2)) rArr M prop (r_(1)^(2))/(r_(2))`
It would have been difficult to calculate the flux through the bigger coil nouniform 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.