Question

Consider a hypothetic spherical body. The body is cut into two parts about the diameter. One of hemispherical portion has mass distribution $m$ while the other portion has identical charge distribution $q$. The body is rotated about the axis with constant speed $\omega$. Then, the ratio of magnetic moment to angular momentum is

• A. $\frac{q}{2m}$
• B. $>\frac{q}{2m}$
• C. $<\frac{q}{2m}$
• D. cannot be calculated

Answer 1

The correct option is A $\frac{q}{2m}$

Magnetic moment $\mu =iA=\frac{q\omega }{2\pi }\times \pi r^2$

Angular momentum $L=m\omega r^2$

$\frac{\mu }{L}=\frac{\frac{q\omega }{2\pi }\times \pi r^2}{m\omega r^2}=\frac{q}{2m}$

$i$ is the current due to the motion of the charge$,A$ is the area of the loop$,\omega$ is the angular velocity