Question

A charged particle accelerated through a potential difference of V volts acquires a speed u. The particle is then made to enter perpendicularly in a uniform magnetic field B. The radius of the circular path followed by the charged particle will be proportional to

• A. V/u
• B. u/V
• C. V²/u²
• D. u²/V

Hint:

In questions asking for proportionality, write down the main equations for the physical quantities involved. Then, manipulate these equations to express one quantity in terms of the others. Here, expressing both \(r\) and the options in terms of fundamental quantities like \(m, u, q\) helps to find the correct relationship.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem combines two key concepts: the work-energy theorem for a charged particle in an electric field, and the dynamics of a charged particle in uniform circular motion in a magnetic field. We need to find a relationship between the radius of the circular path and the initial accelerating voltage and final speed.

Step 2: Key Formula or Approach:
1. When a particle of charge \(q\) is accelerated by a potential difference \(V\), the kinetic energy gained is equal to the work done: \(KE = qV\). Also, \(KE = \frac{1}{2}mu^2\). Therefore, \(qV = \frac{1}{2}mu^2\). 2. When this particle enters a magnetic field \(B\) perpendicularly, the magnetic force provides the centripetal force for circular motion: \(quB = \frac{mu^2}{r}\). 3. From this second equation, the radius of the circular path is: \(r = \frac{mu}{qB}\).

Step 3: Detailed Explanation:
We have two main equations: (i) \(V = \frac{mu^2}{2q}\) (ii) \(r = \frac{mu}{qB}\) The question asks for the proportionality of \(r\) in terms of \(V\) and \(u\). We need to combine the equations to find this relationship. Let's look at the expression in option (A), which is \(V/u\). From equation (i), we can write: \[ \frac{V}{u} = \frac{mu^2/2q}{u} = \frac{mu}{2q} \] Now let's compare this with our expression for the radius \(r\) from equation (ii): \[ r = \frac{mu}{qB} \] We can see that both \(r\) and \(V/u\) are directly proportional to the term \(mu/q\). We can write \(r\) in terms of \(V/u\): \[ r = \frac{mu}{qB} = \frac{2}{B} \left( \frac{mu}{2q} \right) = \frac{2}{B} \left( \frac{V}{u} \right) \] Since the magnetic field \(B\) is uniform and constant, we have the proportionality: \[ r \propto \frac{V}{u} \]

Step 4: Final Answer:
The radius of the circular path is proportional to \(V/u\).