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;
$M=\frac{N_2{\left(\varphi \right)}_2}{I_1}=\frac{N_2{B}_1{A}_{2}cos\theta }{I_1}..........(i)$
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
$B_P=\frac{{\left(\mu \right)}_0{I}_P{R}^2}{2{\left(R^2+x^2\right)}^{3/2}}=\frac{{\left(\mu \right)}_0{I}_P{R}^2}{2{\left(R^2+3R^2\right)}^{3/2}}\left(x=\sqrt{3}R\right)$
$B_P=\frac{{\mu }_0{I}_P}{16R}$
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:-
$\therefore M=\frac{2}{I_1}\frac{{\mu }_0{I}_1}{16R}{a}^{2}cos45^{\circ }=\frac{{\mu }_0{a}^2}{8\sqrt{2}R}$
$M=\frac{{\mu }_0{a}^2}{2^{7/2}R}$
$\Rightarrow p=7$