Question

A circular wire loop of radius $R$ is placed in the $XY-$plane, centered at the origin $O$. A square loop of side $a$ $(a<<R)$ having two turns is placed with its center 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 $\frac{{\mu }_0{a}^2}{2^{P/2}R}$, then the value of $P$ is -

• A. $3$
• B. $4$
• C. $5$
• D. $7$

Answer 1

The correct option is D $7$


Magnetic flux through the square loop is given by,

$\varphi =N\overrightarrow{B}\cdot \overrightarrow{A}$

$\Rightarrow \varphi =2BA\cos {45}^{\circ }$

$\Rightarrow \varphi =\sqrt{2}BA$

Magnetic field due to the current carrying circular loop at a point on its axis,

$B=\frac{{\mu }_{0}i}{2}\frac{R^2}{(R^2+Z^2{)}^{3/2}}$

$=\frac{{\mu }_{0}i}{2}\frac{R^2}{(R^2+3R^2{)}^{3/2}}$

$=\frac{{\mu }_{0}i}{2\times 2^{3}R}=\frac{{\mu }_{0}i}{2^{4}R}$


$\therefore \varphi =\sqrt{2}\times \frac{{\mu }_{0}i}{2^{4}R}\times a^2$

$\Rightarrow \varphi =\frac{{\mu }_{0}i{a}^2}{2^{7/2}R}$

Also, $\varphi =M\cdot i$

Mutual inductance,

$M=\frac{{\mu }_0{a}^2}{2^{7/2}R}=\frac{{\mu }_0{a}^2}{2^{P/2}R}$

$\Rightarrow P=7$

Hence, option $\left(D\right)$ is the correct answer.