Question

A small square loop of wire of side I is placed inside a large square loop of wire of side L (> > l). The loops are coplanar and their centres coincide. What is the mutual inductance of the system?

• A. $2\sqrt{2}\frac{\mu_{0}}{\pi}\frac{l^{2}}{L}$
• B. $8\sqrt{2}\frac{\mu_{0}}{\pi}\frac{l^{2}}{L}$
• C. $2\sqrt{2}\frac{\mu_{0}}{2\pi}\frac{l^{2}}{L}$
• D. $2\sqrt{2}\frac{\mu_{0}L^{2}}{\pi l}$

Answer 1

Answer: (A)

$2\sqrt{2}\frac{\mu_{0}}{\pi}\frac{l^{2}}{L}$

Answer 2

Let the current I be flowing in the larger loop.

The larger loop is made up of four wires each of length L, the field at the centre i.e, at a distance $\frac{L}{2}$ from each wire will be

$B = 4 \times \frac{\mu_{0}I}{4\pi(L/2)}(\sin 45º + \sin 45º)$

$= 4 \times \frac{\mu_{0}}{4\pi}\frac{2I}{L}\frac{2}{\sqrt{2}} = 2\sqrt{2}\frac{\mu_{0}}{\pi}\frac{l}{L}$

Flux linked with smaller loop

$\varphi_{2} = BA_{2} = 2\sqrt{2}\frac{\mu_{0}}{\pi}\frac{I}{L} \times l^{2}$

Hence $M = \frac{\varphi_{0}}{I} \Rightarrow M = 2\sqrt{2}\frac{\mu_{0}}{\pi}\frac{l^{2}}{L}$