A coil of n turns and radius R carries a curr | vedclass

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(C) The magnetic moment of a coil is given by $\mu = n I A = n I \pi R^2$.Let the initial number of turns be $n_1 = n$ and radius be $R_1 = R$.The ini

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$\frac{1}{3}$

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(C) The magnetic moment of a coil is given by $\mu = n I A = n I \pi R^2$.Let the initial number of turns be $n_1 = n$ and radius be $R_1 = R$.The initial magnetic moment is $\mu_1 = n I \pi R^2$.When the wire is unwound and rewound,the total length of the wire $L = n_1 (2 \pi R_1) = n_2 (2 \pi R_2)$ remains constant.Given $R_2 = \frac{R}{3}$,we have $n (2 \pi R) = n_2 (2 \pi \frac{R}{3})$,which gives $n_2 = 3n$.The new magnetic moment is $\mu_2 = n_2 I \pi R_2^2 = (3n) I \pi (\frac{R}{3})^2 = 3n I \pi \frac{R^2}{9} = \frac{n I \pi R^2}{3}$.Therefore,the ratio $\frac{\mu_2}{\mu_1} = \frac{\frac{n I \pi R^2}{3}}{n I \pi R^2} = \frac{1}{3}$.

$A$ coil of $n$ turns and radius $R$ carries a current $I$. It is unwound and rewound to make a new coil of radius $\frac{R}{3}$ and the same current is passed through it. The ratio of the magnetic moment of the new coil to that of the original coil is

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