Question

Two particles, each of mass $m$ and charge $q$, are attached to the two ends of a light, rigid rod of length $2l$. The rod is rotated about an axis passing through its centre and perpendicular to its length, with constant angular velocity $\omega$. The ratio of magnitudes of magnetic moment of the system and its angular momentum about the centre of the rod is

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

Answer 1

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


The current due to the rotation of charges in a circular path is given by

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

The magnitude of magnetic moment of each particle is given by,

$M=iA=\frac{q\omega }{\pi }\times \pi l^2=q\omega l^2$

The magnitude of angular momentum of each particle is given by,

$L=m{\omega }^{2}r=m{\omega }^{2}l$

Since, the angular momentum of both the particles are in the same direction,

So, $L_{net}=2L=2m{\omega }^{2}l$

$\therefore \frac{M}{L_{net}}=\frac{q}{2m}$

Hence, option $\left(D\right)$ is the correct answer.

Why this Question? Note: This is a very important standard result which must be remembered by students. $\frac{M_{net}}{L_{net}}=\frac{q}{2m}$