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(D) The magnetic field on the axis of a current-carrying circular loop at a distance $d$ from its center is given by:$B = \frac{\mu_{0} I R^{2}}{2(R^{

Vedclass · Accepted Answer

$m=3, n=3$

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(D) The magnetic field on the axis of a current-carrying circular loop at a distance $d$ from its center is given by:$B = \frac{\mu_{0} I R^{2}}{2(R^{2} + d^{2})^{3/2}}$Since the rod is rotated with a time period $T$, the equivalent current $I$ is:$I = \frac{q}{T} = \frac{\lambda (2 \pi R)}{T}$Substituting $I$ into the magnetic field formula:$B = \frac{\mu_{0} (2 \pi R \lambda / T) R^{2}}{2(R^{2} + d^{2})^{3/2}}$For $d >> R$, we can approximate $(R^{2} + d^{2})^{3/2} \approx (d^{2})^{3/2} = d^{3}$.Thus, the magnetic field becomes:$B \approx \frac{\mu_{0} (2 \pi R \lambda) R^{2}}{2 T d^{3}} = \frac{\mu_{0} \pi \lambda}{T} \frac{R^{3}}{d^{3}}$Comparing this with the given form $B \propto \frac{R^{m}}{d^{n}}$, we get $m = 3$ and $n = 3$.

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