Question
Charge q is uniformly distributed over the surface of a thin non-conducting annular disc of inner radius R1 and outer radius R2. The disc is made to rotate with constant frequency f, about an axis passing through the center of the annular disc and perpendicular to its plane. The magnetic moment of the disc is
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Solution
[Key Concept: A rotating charged ring of charge dq is a current f dq whose magnetic moment is current times area.]
$$\sigma = \frac{q}{\pi\left( R_{2}^{2} - R_{1}^{2} \right)}$$
For a small ring of radius x and width 'dx'
$$\begin{matrix} dM & \ = IA \\ & \ = fdq\text{ }A \\ & \ = f\sigma dA \times \pi x^{2} \\ & \ = \frac{q}{\pi\left( R_{2}^{2} - R_{1}^{2} \right)} \times 2\pi xdx \cdot \pi x^{2} \\ & \ = \frac{2\pi fq}{R_{2}^{2} - R_{1}^{2}}\int_{R_{1}}^{R_{2}}\mspace{2mu}\mspace{2mu} x^{3}dx\left( \frac{R_{2}^{4}}{4} - \frac{R_{1}^{4}}{4} \right) \\ & \ = \frac{2\pi fq\left( R_{2}^{2} + R_{1}^{2} \right)}{4\left( R_{2}^{2} - R_{1}^{2} \right)}\left( R_{2}^{2} - R_{1}^{2} \right) \\ & \ = \frac{\pi fq\left( R_{2}^{2} + R_{1}^{2} \right)}{2} \end{matrix}$$
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