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) $\frac{q B}{2 \pi E}$

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Answer:

Correct Answer: 18. (b)

Solution:

  1. (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

$v=\frac{\omega}{2 \pi} \Rightarrow v=\frac{q B}{2 \pi m}$