Question

A small square loop of wire of side l is placed inside a large square loop of wire L(L>>l). As shown in figure, both loops are coplanar and their centers coincide at point O. The mutual inductance of the system is:

• A. $\frac{2\sqrt2 \mu_0L^2}{\pi l}$
• B. $\frac{\mu_0L^2}{2\sqrt2 \pi L}$
• C. $\frac{2\sqrt 2 \mu_0 l^2}{\pi L}$
• D. $\frac{\mu_0L^2}{2\sqrt2\pi l}$

Answer 1

Answer: (C)

$\frac{2\sqrt 2 \mu_0 l^2}{\pi L}$

Answer 2

B1=4B=$\frac{4μ_0i}{4π(\frac{L}{2})(2sin45°)}$
B1=$\frac{2\sqrt 2 \mu_0 l^2}{\pi L}$
M=$\frac{Flux inner loop}{i}$=$\frac{2\sqrt 2 \mu_0 il^2}{i\pi ^2}$
=$\frac{2\sqrt2 \mu_0l^2}{\pi L}$