MHTCET 10 May 2024 Shift 2 Paper

Concept: The magnetic moment of a coil is $\mu = nIA$, and rewinding the same wire conserves its total length.Explanation:For the original coil, the magnetic moment is: $$\mu_1 = n I \pi R^2$$ Since the coil is unwound and rewound using the same wire, the total length of the wire remains unchanged. The original wire length is $n \times 2\pi R$, and the new wire length is $n' \times 2\pi\left(\frac{R}{3}\right)$. Equating the lengths: $$n \times 2\pi R = n' \times 2\pi\left(\frac{R}{3}\right)$$ Cancelling $2\pi$ from both sides gives $nR = n'\frac{R}{3}$, hence $n' = 3n$. The new magnetic moment is: $$\mu_2 = n' I \pi \left(\frac{R}{3}\right)^2 = 3n I \pi \frac{R^2}{9} = \frac{n I \pi R^2}{3}$$ Therefore, the required ratio is: $$\frac{\mu_2}{\mu_1} = \frac{\frac{n I \pi R^2}{3}}{n I \pi R^2} = \frac{1}{3}$$Answer: The ratio of the magnetic moment of the new coil to that of the original coil is $\frac{1}{3}$, which matches option C.