Moving Charges and Magnetism - Result Question 18
18. A charged particle of charge $q$ and mass $m$ enters perpendicularly in a magnetic field $\vec{B}$. Kinetic energy of the particle is E; then frequency of rotation is
[2001]
(a) $\frac{q B}{m \pi}$
(b) $\frac{q B}{2 \pi m}$
(c) $\frac{q B E}{2 \pi m}$
(d) $\frac{q B}{2 \pi E}$
Show Answer
Answer:
Correct Answer: 18. (b)
Solution:
- (b) For circular path in magnetic field, $mr \omega^{2}=q v B$
$\Rightarrow \omega^{2}=\frac{q v B}{m r} \quad$ As $v=r \omega$
$\therefore \omega^{2}=\frac{q(r \omega) B}{m r} \Rightarrow \omega=\frac{q B}{m}$
$\therefore$ If $v$ is frequency of rotation, then
$v=\frac{\omega}{2 \pi} \Rightarrow v=\frac{q B}{2 \pi m}$
