Question

A circular wire loop of radius $R$ is placed in the $x-y$ plane centred at the origin $O$. A square loop of side $a(a\ll R)$ having two turns is placed with its centre at $z=\sqrt{3}R$ along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of ${45}^o$ with respect to the $z-$axis. If the mutual inductance between the loops in given by $\frac{{\mu }_0{a}^2}{2^{p/2}R}$, then the value of $p$ is:

Answer 1

The mutual inductance

$M=\frac{N_s\left(B_c{A}_{s}cos{45}^0\right)}{i_c}$

here $N_s$=no of turns in square coil and $B_c$ be the magnetic field of circular coil at the position of square coil.

$M=\frac{2\times \frac{{\mu }_0{i}_c{R}^2}{2(R^2+3R^2{)}^{\frac{3}{2}}}\times a^2\times \frac{1}{\sqrt{2}}}{i_c}\simeq \frac{{\mu }_0{a}^2}{8\sqrt{2}R}\simeq \frac{{\mu }_0{a}^2}{2^{\frac{7}{2}}R}$

$\therefore p=7$