Question

A circular wire loop of radius $R$ is placed in the $xy$-plane centred at the origin $\text{O}$. A square loop of side $a(a<<R)$ 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 ${45}^{\circ }$ with respect to the $z$-axis. If the mutual inductance between the loops is given by $\frac{{\mu }_0{a}^2}{2^{\frac{\rho }{2}}R}$, then the value of $\rho$ is(integer only)

Answer 1

Let us assume that a current $i$ is flowing through the circular loop.

The magnetic field due to circular loop along its axis at the location of the square loop is,

$B=\frac{{\mu }_{0}i}{4\pi }\times \frac{2\pi R^2}{[z^2+R^2{]}^{\frac{3}{2}}}$

$B=\frac{{\mu }_{0}i{R}^2}{2(3R^2+R^2{)}^{\frac{3}{2}}}[\because z=\sqrt{3}R]$

$\Rightarrow B=\frac{{\mu }_{0}i}{16R}$


Now, Magnetic flux linked with the square loop due to field produced by the circular loop is,

$\varphi =NBA\cos \theta$

Here, $N=2;\theta ={45}^{\circ }$

$\Rightarrow \varphi =2\times \frac{{\mu }_{0}i}{16R}\times a^2\cos {45}^{\circ }$

$\varphi =\frac{{\mu }_{0}i{a}^2}{8\sqrt{2}R}$

$\therefore$ The mutual inductance between the loops is,

$M=\frac{\varphi }{i}=\frac{{\mu }_{0}i{a}^2}{8\sqrt{2}R\times i}=\frac{{\mu }_0{a}^2}{2^{\left(\frac{7}{2}\right)}R}$

Given, $M=\frac{{\mu }_0{a}^2}{2^{\frac{\rho }{2}}R}$

$\therefore \rho =7$