Question

A circular wire loop of radius $R$ is placed in an $X-Y$ plane centered at the origin $O$ . A square loop of side $a\left(a < < R\right)$ having two turns is placed with its center at $z=\sqrt{3}R$ along the axis of 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. $7$
• C. $4$
• D. $5$

Answer 1

Answer: (B)

Mutual inductance of second coil due to the magnetic field of first coil is given by;
$M=\frac{N_{2} \left(\phi\right)_{2}}{I_{1}}=\frac{N_{2} B_{1} A_{2} cos \theta }{I_{1}}..........\left(\right.i\left.\right)$
Where $B_{1}$ is the magnetic field of the first coil.
If circular coil is primary coil then the magnetic field on the axis of a circular loop is given by
$B_{P}=\frac{\left(\mu \right)_{0} I_{P} R^{2}}{2 \left(R^{2} + x^{2}\right)^{3/2}}=\frac{\left(\mu \right)_{0} I_{P} R^{2}}{2 \left(R^{2} + 3 R^{2}\right)^{3/2}}\left(x = \sqrt{3} R\right)$
$B_{P}=\frac{\mu _{0} I_{P}}{16 R}$
Where $B_{P}$ is the magnetic field due to the primary coil and $I_{P}$ is the current in the primary coil.
It is given in the question that, $N_{2}=2$ , $A_{2}=a^{2}$ and $\theta =45^\circ $ .
Plugging these values in (i), we obtain:-
$\therefore M=\frac{2}{I_{1}}\frac{\mu _{0} I_{1}}{16 R}a^{2}cos45^\circ =\frac{\mu _{0} a^{2}}{8 \sqrt{2} R}$
$M=\frac{\mu _{0} a^{2}}{2^{7/2} R}$
$\Rightarrow p=7$