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Question

A non conducting disc of radius $$R$$, charge $$q$$ is rotating about an axis passing through its centre and perpendicular to its plane with an angular velocity $$\omega $$, charge $$q$$ is uniformly distributed over its surface. The magnetic moment of the disc is :


A
14qωR2
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B
12qωR
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C
qωR
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D
12qωR2
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Solution

The correct option is B $$\frac{1}{4}q\omega R^{2}$$
Charge per unit area$$(\sigma )=\dfrac{q}{\pi R^2}$$
Charge on ring of width $$dr$$$$=\sigma \cdot2\pi rdr$$
Magnetic moment of this ring of width dr
$$d\mu=i\cdot A$$
$$\displaystyle \int d\mu=\int_{0}^{R}T\cdot2\pi rdr\dfrac{w}{2\pi}\cdot\pi r^2$$$$=\mu=\dfrac{qwR^2}{4}$$

52158_24018_ans_3320e3f0c78142e08141c8cbeef2f8ce.png

Physics
NCERT
Standard XII

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