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A coil of ' $n$ ' turns and radius ' $R$ ' carries a current ' $I$ '. It is unwound and rewound again to make another coil of radius $\left(\frac{\mathrm{R}}{3}\right)$, current remaining the same. The ratio of magnetic moment of the new coil to that of original coil is
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Verified Answer
The correct answer is:
$1: 3$
As the length of the wire remains the same, we can write
$$
\begin{aligned}
& \mathrm{N}_1 2 \pi \mathrm{R}=\mathrm{N}_2 \times 2 \pi \frac{\mathrm{R}}{3} \\
& \mathrm{~N}_1=\frac{\mathrm{N}_2}{3} \\
& \mathrm{~N}_2=3 \mathrm{~N}_1
\end{aligned}
$$
$\begin{array}{ll} & \text { Magnetic moment of coil } \mu=\text { NIA } \\ \therefore & \mu_1=N_1 I A_1=N_1 I \pi R^2 \\ \therefore & \mu_2=N_2 I A_2=3 \frac{N_1 I \pi R^2}{9} \\ \therefore & \frac{\mu_2}{\mu_1}=3 \frac{N_1 I \pi R^2}{9} \times \frac{1}{N_1 I \pi R^2}=\frac{1}{3}\end{array}$
$$
\begin{aligned}
& \mathrm{N}_1 2 \pi \mathrm{R}=\mathrm{N}_2 \times 2 \pi \frac{\mathrm{R}}{3} \\
& \mathrm{~N}_1=\frac{\mathrm{N}_2}{3} \\
& \mathrm{~N}_2=3 \mathrm{~N}_1
\end{aligned}
$$
$\begin{array}{ll} & \text { Magnetic moment of coil } \mu=\text { NIA } \\ \therefore & \mu_1=N_1 I A_1=N_1 I \pi R^2 \\ \therefore & \mu_2=N_2 I A_2=3 \frac{N_1 I \pi R^2}{9} \\ \therefore & \frac{\mu_2}{\mu_1}=3 \frac{N_1 I \pi R^2}{9} \times \frac{1}{N_1 I \pi R^2}=\frac{1}{3}\end{array}$
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