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Answer: \(M \approx 5.184\times10^{-2}\) A m\(^2\)

Given \(R = 12\) cm \(= 0.12\) m, \(B = 6\times10^{-6}\) T, \(\mu_0 = 4\pi\times10^{-7}\) Wb/A m.

Field at the centre: \(B = \dfrac{\mu_0 I}{2R}\), so \(I = \dfrac{2RB}{\mu_0}\).

Magnetic moment:

\[M = I\pi R^2 = \frac{2RB}{\mu_0}\pi R^2 = \frac{2\pi R^3 B}{4\pi\times10^{-7}} = \frac{R^3B}{2\times10^{-7}}\]\[M = \frac{(0.12)^3\times6\times10^{-6}}{2\times10^{-7}} = \frac{1.728\times10^{-3}\times6\times10^{-6}}{2\times10^{-7}} = 5.184\times10^{-2}\ \mathrm{A\,m^2}\]

(The current in the loop is \(I = \dfrac{2\times0.12\times6\times10^{-6}}{4\times3.142\times10^{-7}} \approx 1.146\) A.)

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report
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