Question

A small square loop of wire of side `l` is placed inside a large square loop of wire of side `L (L gt gt l)`. The loops are coplanar and their centre coincide. What is the mutual inductance of the system ?
A. `2sqrt2(mu_(0))/(pi)(l^(2))/(L)`
B. `8sqrt2(mu_(0))/(pi)(l^(2))/(L)`
C. `2sqrt2(mu_(0))/(2pi)(l^(2))/(L)`
D. `2sqrt2(mu_(0)L^(2))/(pil)`

Answer 1

Correct Answer - A
Let the current I be flowing in the larger loop.

The larger loop is made up of four wires each of lengthL,n the field at the centre, i.e., at a distance `(L)/(2)` fom each wire, will be `B=4xx(mu_(0)I)/(4pi(L//2))(sin45^(@)+sin45^(@))`
`4xx(mu_(0))/(4pi)(2I)/(L)(2)/(sqrt(2))=2sqrt(2)(mu_(0))/(pi)(I)/(L)`
Flux linked with smaller loop
`phi_(0)=BA_(2)=2sqrt(2)(mu_(0))/(pi)(l^(2))/(L)`
Hence, `M=(phi_(0))/(I) rArrM=2sqrt(2)(mu_(0))/(pi)(l^(2))/(L)`