Question

Two particles each of mass m and charge q are attached to the two ends of a light rigid rod of length 2R. The rod is rotated at constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is

• A. $\frac{q}{2m}$
• B. $\frac{q}{m}$
• C. $\frac{2q}{m}$
• D. $\frac{q}{\pi m}$

Answer 1

Answer: (A)

$\frac{q}{2m}$

Let the angular speed be $\omega$.

Thus the current due to the rotation of charges in circular path=$2\times \frac{q}{T}$

$=2\times \frac{q\omega }{2\pi }$

$=\frac{q\omega }{\pi }$

Hence magnetic moment=$M=iA=\frac{q\omega }{\pi }\pi R^2$

$=q\omega R^2$

The angular momentum of the system of charges=$L=2\times m\times {\omega }^{2}R$
Hence $\frac{M}{L}=\frac{q}{2m}$