A small square loop of wire of side 'I' is placed inside a large square loop of side L (L > I). If the loops are coplanar and their centres coincide, the mutual inducation of the system is directly proportional to

Explanation

Considering the large loop to be made up of four rods each of length L, the field at the centre, i.e., at a distance `(L//2)` from each rod, will be
`B=4 xx (mu_(0))/(4 pi)(I)/(d)[sin alpha+sin beta]`
i.e., `B=4 xx (mu_(0))/(4phi)(I)/((L//2)) xx 2sin45`
i.e., `B_(1)=(mu_(0))/(4pi)(8sqrt2)/(L)I`

So the flux linked with smaller loop
`Phi_(2)=B_(1)S_(2)=(mu_(0))/(4pi)(8sqrt2 I)/(LO) l^(2)`
and hence, `M=(phi_(2))/(I) =2sqrt2 (mu_(0))/(pi)(l^(2))/(L) implies M prop (l^(2))/(L)`