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A circular coil of diameter $7 \mathrm{~cm}$ has 24 turns of wire carrying current of $0.75 \mathrm{~A}$. The magnetic moment of the coil is
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Verified Answer
The correct answer is:
$6.9 \times 10^{-2} a m p-m^2$
Given: \(d=7 \mathrm{~cm}, \mathrm{~N}=24, \mathrm{i}=0.75 \mathrm{~A}\)
The magnetic moment is given by, \(\mathrm{M}=\mathrm{NiA}\)
\(\begin{aligned}
& M=24 \times 0.75 \times 3.14 \times\left(3.5 \times 10^{-2}\right)^2 \\
& M=6.9 \times 10^{-2} \mathrm{Am}^2
\end{aligned}\)
Hence, option (A) is correct.
The magnetic moment is given by, \(\mathrm{M}=\mathrm{NiA}\)
\(\begin{aligned}
& M=24 \times 0.75 \times 3.14 \times\left(3.5 \times 10^{-2}\right)^2 \\
& M=6.9 \times 10^{-2} \mathrm{Am}^2
\end{aligned}\)
Hence, option (A) is correct.
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