Question

Charge ‘q’ is uniformly distributed over the surface of an annular-non-conducting disc of inner radius $R_1$ and outer radius $R_2$. The disc is made to rotate about an axis passing through its centre and perpendicular to its plane with a constant frequency v (rotations per second). Magnetic moment of the disc can be expressed as :

• A. $\frac{q\pi v(R_2^2-R_1^2)}{2}$
• B. $\frac{q\pi v(R_2^2+R_1^2)}{2}$
• C. $\frac{q\pi v{R}_2^2}{4}$
• D. $\frac{qv(R_2^2-R_1^2)}{4}$

Answer 1

The correct option is B

$\frac{q\pi v(R_2^2+R_1^2)}{2}$

dM=(dI) A =dq v A

$M=\int dM=\int v\pi x^2\times \sigma 2\pi xdx$

$=v\pi \sigma 2\pi {\int }_{R_1}^{R_2}{x}^{3}dx$

$=\frac{2v\sigma {\pi }^2}{4}(R_2^4-R_1^4)$

$=\frac{2v{\pi }^{2}q(R_2^4-R_1^4)}{4\pi (R_2^2-R_1^2)}$

$=\frac{vq\pi (R_2^2+R_1^2)}{2}$