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: Conceptual Foundation:
This problem integrates two core principles: the work-energy theorem for charged particles in electric fields and the dynamics of charged particles in uniform circular motion within magnetic fields. The objective is to establish a relationship between the radius of the circular trajectory and the initial accelerating voltage along with the final velocity.

Step 2: Essential Formulas and Methodology:
1. Acceleration of a charged particle (\(q\)) by a potential difference (\(V\)) results in kinetic energy gain equivalent to the work done: \(KE = qV\). Since \(KE = \frac{1}{2}mu^2\), it follows that \(qV = \frac{1}{2}mu^2\).2. Upon entering a magnetic field (\(B\)) perpendicularly, the magnetic force acts as the centripetal force for circular motion: \(quB = \frac{mu^2}{r}\).3. Rearranging the second equation yields the radius of the circular path: \(r = \frac{mu}{qB}\).

Step 3: Detailed Derivation:
We have established two primary equations:(i) \(V = \frac{mu^2}{2q}\)(ii) \(r = \frac{mu}{qB}\)The problem requires determining the proportionality of \(r\) with respect to \(V\) and \(u\). Combining the equations is necessary to derive this relationship. Consider the expression \(V/u\), as presented in option (A).From equation (i), we can derive:\[ \frac{V}{u} = \frac{mu^2/2q}{u} = \frac{mu}{2q} \]Comparing this with the expression for radius \(r\) from equation (ii):\[ r = \frac{mu}{qB} \]Both \(r\) and \(V/u\) are directly proportional to the term \(mu/q\). We can express \(r\) in terms of \(V/u\) as follows:\[ r = \frac{mu}{qB} = \frac{2}{B} \left( \frac{mu}{2q} \right) = \frac{2}{B} \left( \frac{V}{u} \right) \]Given that the magnetic field \(B\) is uniform and constant, the proportionality is established as:\[ r \propto \frac{V}{u} \]

Step 4: Conclusion:
The radius of the circular trajectory is directly proportional to the ratio \(V/u\).