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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
- \( \frac{qB}{m\pi} \)
- \( \frac{qB}{2\pi m} \)
- \( \frac{qB}{2\pi m} \)
- \( \frac{qB}{2\pi E} \)
Correct answer: \( \frac{qB}{2\pi m} \)
Solution
The frequency of rotation of a charged particle in a magnetic field is given by the cyclotron frequency formula, \( f = \frac{qB}{2\pi m} \). The kinetic energy \( E \) does not affect the frequency as it depends only on charge, magnetic field, and mass.
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