Question

If T is time period and V is maximum speed of a charged particle in cyclotron, then

• A. T \(\propto\) V
• B. T \(\propto\) V\(^2\)
• C. T \(\propto\) 1/V
• D. T \(\propto\) 1/V\(^2\)
• E. None of these

Hint:

A core principle of the cyclotron is that the revolution period is constant. This allows the use of a fixed-frequency AC voltage to accelerate the particles. Remember the formula \(T = \frac{2\pi m}{qB}\); it clearly shows no dependence on velocity or radius. This independence breaks down at very high (relativistic) speeds, which is a limitation of the classic cyclotron.

Answer 1

Answer: (E)

Note: In a standard, non-relativistic cyclotron, the time period of revolution of a charged particle is independent of its speed and the radius of its orbit. This is the fundamental principle upon which the cyclotron operates. Therefore, none of the given proportionality options are correct. We will proceed by deriving the correct relationship.
Step 1: Understanding the Concept:
A cyclotron accelerates charged particles using a constant magnetic field and an oscillating electric field. The magnetic field forces the particle to move in a circular path, and the electric field accelerates it each time it crosses the gap between the two "dee" electrodes. The question asks for the relationship between the time period of one revolution (T) and the particle's speed (V).
Step 2: Key Formula or Approach:
The magnetic force (\(F_B\)) on the charged particle provides the necessary centripetal force (\(F_c\)) to keep it in a circular path.
\[ F_B = F_c \] \[ qvB = \frac{mv^2}{r} \] The time period (T) is the time taken to complete one full circle, which is the circumference divided by the speed.
\[ T = \frac{2\pi r}{v} \] Step 3: Detailed Explanation:
From the force balance equation, we can find an expression for the radius \(r\):
\[ qvB = \frac{mv^2}{r} \implies r = \frac{mv}{qB} \] Now, substitute this expression for the radius \(r\) into the time period equation:
\[ T = \frac{2\pi}{v} \left( \frac{mv}{qB} \right) \] The speed \(v\) cancels out from the numerator and the denominator:
\[ T = \frac{2\pi m}{qB} \] This result shows that the time period \(T\) depends only on the mass (\(m\)) and charge (\(q\)) of the particle, and the strength of the magnetic field (\(B\)). It is independent of the particle's speed (\(v\)) and the radius of its orbit (\(r\)).
Step 4: Final Answer:
The time period \(T\) is independent of the speed \(V\). The correct relationship is \(T \propto V^0\). Since this is not among the options, the question is flawed.