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The magnetic moments associated with two closely wound circular coils A and B of radius $r_A=10 \mathrm{~cm}$ and $r_B=20 \mathrm{~cm}$ respectively are equal if: (Where $\mathrm{N}_A, \mathrm{I}_A$ and $\mathrm{N}_{\mathrm{B}}, \mathrm{I}_{\mathrm{B}}$ are number of turn and current of $A$ and $B$ respectively)
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Verified Answer
The correct answer is:
$\mathrm{N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}=4 \mathrm{~N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}$
The magnetic moment associated with circular coil,
$\begin{aligned}
& M=N I A \\
& M_A=M_B \\
& \therefore N_A I_A A_A=N_B I_B A_B \\
& \therefore N_A I_A \pi(0.1)^2=N_B I_B \pi(0.2)^2\left(\because A=\pi r^2\right) \\
& \therefore N_A I_A=4 N_B I_B
\end{aligned}$
$\begin{aligned}
& M=N I A \\
& M_A=M_B \\
& \therefore N_A I_A A_A=N_B I_B A_B \\
& \therefore N_A I_A \pi(0.1)^2=N_B I_B \pi(0.2)^2\left(\because A=\pi r^2\right) \\
& \therefore N_A I_A=4 N_B I_B
\end{aligned}$
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