Magnetics Ques 96
- A particle of charge $q$ and mass $m$ moves in a circular orbit of radius $r$ with angular speed $\omega$. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on
(2000, 2M)
(a) $\omega$ and $q$
(b) $\omega, q$ and $m$
(c) $q$ and $m$
(d) $\omega$ and $m$
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Answer:
Correct Answer: 96.(c)
Solution:
Formula:
Magnetic Moment Of A Current Carrying Loop:
- Ratio of magnetic moment and angular momentum is given by
$$ \frac{M}{L}=\frac{q}{2 m} $$
which is a function of $q$ and $m$ only. This can be derived as follows,
$$ \begin{aligned} M & =i A=(q f) \cdot\left(\pi r^{2}\right) \\ & =(q) \frac{\omega}{2 \pi}\left(\pi r^{2}\right)=\frac{q \omega r^{2}}{2} \end{aligned} $$
and
$$ L=I \omega=\left(m r^{2} \omega\right) $$
$$ \therefore \quad \frac{M}{L}=\frac{q \frac{\omega r^{2}}{2}}{m r^{2} \omega}=\frac{q}{2 m} $$
BETA
