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

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(B) The length of the wire remains constant. Let $N_1 = n$ and $N_2$ be the number of turns in the new coil.Since the total length of the wire $L = N_

Vedclass · Accepted Answer

$1: 3$

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(B) The length of the wire remains constant. Let $N_1 = n$ and $N_2$ be the number of turns in the new coil.Since the total length of the wire $L = N_1 (2 \pi R) = N_2 (2 \pi R_2)$,where $R_2 = \frac{R}{3}$:$N_1 (2 \pi R) = N_2 (2 \pi \frac{R}{3})$$N_1 = \frac{N_2}{3} \implies N_2 = 3 N_1$.The magnetic moment $\mu$ of a coil is given by $\mu = N I A$,where $A = \pi R^2$.For the original coil: $\mu_1 = N_1 I (\pi R^2)$.For the new coil: $\mu_2 = N_2 I (\pi R_2^2) = (3 N_1) I \pi (\frac{R}{3})^2 = 3 N_1 I \pi \frac{R^2}{9} = \frac{1}{3} N_1 I \pi R^2$.Therefore,the ratio $\frac{\mu_2}{\mu_1} = \frac{\frac{1}{3} N_1 I \pi R^2}{N_1 I \pi R^2} = \frac{1}{3}$.

$A$ coil of $n$ turns and radius $R$ carries a current $I$. It is unwound and rewound again to make another coil of radius $\frac{R}{3}$,with the current remaining the same. The ratio of the magnetic moment of the new coil to that of the original coil is:

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