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Magnetic field at the centre of a circular loop of area \(A\) is \(B\). Then the magnetic moment of the loop is
(\(\mu_0\)-permeability of the free space)
Options:
(\(\mu_0\)-permeability of the free space)
Solution:
2298 Upvotes
Verified Answer
The correct answer is:
\(\frac{2 B A \sqrt{A}}{\mu_0 \sqrt{\pi}}\)
As we know, the magnetic field due to a current carrying circular loop at the centre,
\(B=\frac{\mu_0 i}{2 R}\) ...(i)
So, the magnetic moment of the Loop,
\(M=i A\)
From Eq. (i), we get
\(\begin{aligned}
& & M & =\frac{2 B R}{\mu_0} \cdot A \\
& & A & =\pi R^2 \\
& & R & =\sqrt{\frac{A}{\pi}} \\
& \text {Hence, } & M & =\frac{2 B A \sqrt{A}}{\mu_0 \sqrt{\pi}}
\end{aligned}\)
Hence, the correct option is (d).
\(B=\frac{\mu_0 i}{2 R}\) ...(i)
So, the magnetic moment of the Loop,
\(M=i A\)
From Eq. (i), we get
\(\begin{aligned}
& & M & =\frac{2 B R}{\mu_0} \cdot A \\
& & A & =\pi R^2 \\
& & R & =\sqrt{\frac{A}{\pi}} \\
& \text {Hence, } & M & =\frac{2 B A \sqrt{A}}{\mu_0 \sqrt{\pi}}
\end{aligned}\)
Hence, the correct option is (d).
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