Question

A cyclotron accelerates a particle of mass \(m\) and charge \(q\) in a magnetic field \(B\). The cyclotron frequency \(\nu\) is

• A. \(\frac{mq}{2\pi B}\)
• B. \(\frac{qB}{2\pi m}\)
• C. \(\frac{mqB}{2\pi}\)
• D. \(2\pi mqB\)
• E. \(\frac{2\pi m}{qB}\)

Hint:

Remember that cyclotron frequency is independent of radius. If you forget the formula, use dimensional analysis: frequency has units \(T^{-1}\). Only option (B) matches the dimensions of frequency.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In a cyclotron, a charged particle moves in a circular path under the influence of a perpendicular magnetic field. The centripetal force is provided by the Lorentz magnetic force.

Step 2: Key Formula or Approach:
1. Magnetic Force = Centripetal Force \(\implies qvB = \frac{mv^2}{r} \implies v = \frac{qBr}{m}\)
2. Frequency \(\nu = \frac{v}{2\pi r}\)

Step 3: Detailed Explanation:
Substitute the expression for velocity \(v\) into the frequency formula:
\[ \nu = \frac{qBr/m}{2\pi r} \]
\[ \nu = \frac{qB}{2\pi m} \]
This frequency is known as the cyclotron frequency or oscillator frequency. It is independent of the radius of the orbit and the speed of the particle.

Step 4: Final Answer:
The cyclotron frequency is \(\frac{qB}{2\pi m}\).