Magnetics Ques 97
- Two particles, each of mass $m$ and charge $q$, are attached to the two ends of a light rigid rod of length $2 R$. 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
(1998, 2M)
(a) $q / 2 m$
(b) $q / m$
(c) $2 q / m$
(d) $q / \pi m$
Show Answer
Answer:
Correct Answer: 97.(a)
Solution:
Formula:
Magnetic Moment Of A Current Carrying Loop:
- Current, $i=($ frequency $)($ charge $)=\frac{\omega}{2 \pi}(2 q)=\frac{q \omega}{\pi}$
Magnetic moment, $M=(i)(A)=\frac{q \omega}{\pi}\left(\pi R^{2}\right)=\left(q \omega R^{2}\right)$
Angular momentum, $L=2 I \omega=2\left(m R^{2}\right) \omega$
$$ \therefore \quad \frac{M}{L}=\frac{q \omega R^{2}}{2\left(m R^{2}\right) \omega}=\frac{q}{2 m} $$
BETA
