Question

A small square loop of wire of side y is placed inside a large square loop of side `x(xgtgty)`. The loops are coplanar and their centres coincide. Find the mutual inductance of the system.

Answer 1

Let I be the current flowing through a square loop of side L.
Magnetic field at the centre of the loop, `B=4B_(1)`
[where `B_(1)` is magnetic field at the centre of the loop due to one of its sides] Now, for the large square loop, L = x
`therefore B_(1)=(mu_(0))/(4pi).(I)/((x)/(2))(sin45^(@)+sin45^(@))` [`because` distance of the centre from each side of the large loop = `(x)/(2)`]
`=(mu_(0))/(2pi).(I)/(x)((1)/(sqrt(2))+(1)/(sqrt(2)))=(mu_(0)I)/(sqrt(2)pix)`
`therefore B=(4mu_(0)I)/(sqrt(2)pix)=(2sqrt(2)mu_(0)I)/(pix)`
Now, magnetic flux linked with the small square loop,
`phi=Bxx` area of the small square loop
`=Bxxy^(2)=(2sqrt(2)mu_(0)Iy^(2))/(pix)" ... (1)"`
If M be the mutual inductance between the two loops,then
`phi=MI" ... (2)"`
From equations (1) and (2), `M=(2sqrt(2)mu_(0)y^(2))/(pix)`