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A particle of charge q and mass m undergoes circular motion in a uniform magnetic field B. Its frequency of revolution (cyclotron frequency) is:
AqB / (2πm)
B2πm / (qB)
CqBm / (2π)
D2πqB / m
Answer & Solution
Correct answer: A. qB / (2πm)
1. The angular frequency of the circular motion is $\omega = \dfrac{qB}{m}$.
2. Frequency relates to angular frequency by $\omega = 2\pi\nu$.
3. Therefore $\nu_c = \dfrac{\omega}{2\pi} = \dfrac{qB}{2\pi m}$.
4. This is independent of the particle's speed and radius, which is exploited in the cyclotron.
_Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.156_
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