Question

Charge q' is uniformly distributed over the surface of an annula, non-conducting disc of inner radius r1 and outer radius r2. The disc is made to rotate about an axis passing through its centre and perpendicular to its plane with a constant frequency f. Magnetic moment of the disc can be expressed as

Answer 1

Answer:

Magnetic moment of the disc can be expressed as

$M = \frac{q\omega(r_1^2 + r_2^2)}{4}$

Explanation:

As we know that the ratio of angular momentum and magnetic moment is constant for a uniformly charged rotating rigid body

now we have

$\frac{M}{L} = \frac{q}{2m}$

here we know that

$L = I\omega$

where we know that

$I = \frac{m(r_1^2 + r_2^2)}{2}\omega$

now from above formula

$M = \frac{m(r_1^2 + r_2^2)}{2}\omega (\frac{q}{2m})$

so we will have

$M = \frac{q\omega(r_1^2 + r_2^2)}{4}$

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Topic : Magnetic Moment

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