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(B) The magnetic field $B$ on the axis of a circular loop of radius $R$ at a distance $X$ from the center is given by:$B = \frac{\mu_0 i R^2}{2(R^2 +

Vedclass · Accepted Answer

$7$

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(B) The magnetic field $B$ on the axis of a circular loop of radius $R$ at a distance $X$ from the center is given by:$B = \frac{\mu_0 i R^2}{2(R^2 + X^2)^{3/2}}$Given $X = \sqrt{3} R$,the magnetic field at the center of the square loop is:$B = \frac{\mu_0 i R^2}{2(R^2 + 3R^2)^{3/2}} = \frac{\mu_0 i R^2}{2(4R^2)^{3/2}} = \frac{\mu_0 i R^2}{2 \cdot 8 R^3} = \frac{\mu_0 i}{16 R}$Since the square loop has $N = 2$ turns,area $A = a^2$,and the angle between the area vector and the magnetic field (which is along the $z$-axis) is $	heta = 45^{\circ}$,the magnetic flux $\phi$ is:$\phi = N B A \cos(45^{\circ}) = 2 \cdot \left(\frac{\mu_0 i}{16 R}ight) \cdot a^2 \cdot \frac{1}{\sqrt{2}} = \frac{\mu_0 i a^2}{8 \sqrt{2} R}$Since $\sqrt{2} = 2^{1/2}$ and $8 = 2^3$,the denominator is $2^3 \cdot 2^{1/2} = 2^{7/2}$.Thus,$\phi = \frac{\mu_0 i a^2}{2^{7/2} R}$.The mutual inductance $M$ is defined as $M = \frac{\phi}{i} = \frac{\mu_0 a^2}{2^{7/2} R}$.Comparing this with the given expression $\frac{\mu_0 a^2}{2^{p/2} R}$,we get $p = 7$.

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