Question

A charged particle is moving in a uniform magnetic field in a circular path of radius 'R'. When the energy of the particle becomes 3 times the original, the new radius will be

• A. \(R\)
• B. \(3R\)
• C. \(\sqrt{3}R\)
• D. \(R/3\)

Hint:

Radius is proportional to momentum, so it goes as the root of kinetic energy.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A charged particle in a uniform magnetic field moves in a circle of radius \(R = \dfrac{mv}{qB} = \dfrac{p}{qB}\).

Step 2: Express in terms of energy:
\[ p = \sqrt{2mE} \Rightarrow R = \frac{\sqrt{2mE}}{qB} \propto\sqrt E \]

Step 3: New radius:
\[ \frac{R'}{R} = \sqrt{\frac{3E}{E}} = \sqrt3 \Rightarrow R' = \sqrt3\,R \]

Step 4: Check:
Option (C). Option (B) 3R would be right if \(R\) were proportional to \(E\), and (D) \(R/3\) would be right if it went inversely.

Final Answer:
R is proportional to root of E, so R becomes root 3 times R. \[ \boxed{\text{(C) }\sqrt3\,R} \]