Mutual Inductance Between Concentric Tilted Square and Circular Loops

To find the mutual inductance $M$ between the circular wire loop and the small square loop, we can pass a current $I$ through the circular loop and calculate the magnetic flux linked with the square loop.

The magnetic field at a distance $z$ along the axis of a circular loop of radius $R$ carrying current $I$ is given by: $B = \frac{\mu_0 I R^2}{2(R^2 + z^2)^{3/2}}$

Given $z = \sqrt{3}R$, substituting this into the equation gives: $B = \frac{\mu_0 I R^2}{2(R^2 + 3R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(4R^2)^{3/2}} = \frac{\mu_0 I R^2}{2 \times 8R^3} = \frac{\mu_0 I}{16 R}$

Since $a \ll R$, the magnetic field $\vec{B}$ can be considered uniform over the area of the small square loop.

The area of the square loop is $A = a^2$. The normal to the plane of the square loop makes an angle $\theta = 45^\circ$ with the $z$-axis (the direction of the magnetic field $\vec{B}$).

The magnetic flux $\Phi_1$ through a single turn of the square loop is: $\Phi_1 = \vec{B} \cdot \vec{A} = B A \cos(45^\circ)$ $\Phi_1 = \left(\frac{\mu_0 I}{16 R}\right) a^2 \left(\frac{1}{\sqrt{2}}\right) = \frac{\mu_0 I a^2}{16 \sqrt{2} R}$

Since the square loop consists of $N = 2$ turns, the total magnetic flux $\Phi_{\text{total}}$ linked with the square loop is: $\Phi_{\text{total}} = N \Phi_1 = 2 \times \frac{\mu_0 I a^2}{16 \sqrt{2} R} = \frac{\mu_0 I a^2}{8 \sqrt{2} R}$

The mutual inductance $M$ is given by: $M = \frac{\Phi_{\text{total}}}{I} = \frac{\mu_0 a^2}{8 \sqrt{2} R}$

Rewriting the denominator in terms of powers of 2: $8 \sqrt{2} = 2^3 \times 2^{1/2} = 2^{7/2}$

Therefore: $M = \frac{\mu_0 a^2}{2^{7/2} R}$

Comparing this expression with the given formula $M = \frac{\mu_0 a^2}{2^{p/2} R}$, we get: $\frac{p}{2} = \frac{7}{2} \implies p = 7$