Question

In a cyclotron, if the frequency of the accelerating field is doubled, then the radius of the charged particle moving in a circular path will be

• A. doubled
• B. quadrupled
• C. the same
• D. halved
• E. reduced to one fourth of the original radius

Hint:

Cyclotron frequency depends only on \( B, q, m \).

Answer 1

The Correct Option is C

Step 1: Understanding the Cyclotron Principle:
A cyclotron accelerates charged particles using a high-frequency alternating electric field and a uniform magnetic field. The magnetic field forces the particles to move in a semi-circular path. For acceleration to occur, the frequency of the electric field (\(f\)) must match the particle's revolution frequency, known as the cyclotron frequency (\(f_c\)). This is the resonance condition.
Step 2: Key Formula or Approach:
The cyclotron frequency is given by \(f_c = \frac{qB}{2\pi m}\). It is independent of the particle's speed and radius.
The radius of the particle's path is given by \(r = \frac{mv}{qB}\).
The question implies a change in the operating parameters. If the accelerating frequency \(f\) is doubled, the magnetic field \(B\) must also be doubled to maintain the resonance condition (\(f = f_c \propto B\)).
Step 3: Detailed Explanation:
The question asks what happens to the "radius of the charged particle moving in a circular path". This can be interpreted in two ways, but both lead to the same answer in this context.
Interpreting "radius" as the maximum possible radius of the cyclotron: The maximum radius (\(R_{max}\)) is determined by the physical size of the dees (the semi-circular metal chambers). This is a fixed dimension of the apparatus and does not change when operating parameters like frequency or magnetic field are adjusted.
Interpreting "radius" at a given kinetic energy: This is unlikely to be the intended question, as the energy of the particle changes. However, if we analyze the exit kinetic energy, \(K_{max} = \frac{1}{2}mv_{max}^2 = \frac{(qBR_{max})^2}{2m}\). If we double \(f\) and \(B\), the exit energy becomes \(K'_{max} = \frac{(q(2B)R_{max})^2}{2m} = 4 K_{max}\). The particle exits with much higher energy, but it still exits at the same maximum physical radius \(R_{max}\).
Therefore, the radius of the path, taken to mean the maximum radius of the device, remains the same.
Step 4: Final Answer:
The radius of the circular path, being a physical dimension of the cyclotron, remains the same.