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: Speed first:
Energy $\tfrac12mv^2$ triples, so $v$ becomes $\sqrt3\,v$.

Step 2: Radius:
$R = mv/qB$ is proportional to $v$.

Step 3: Result:
$R' = \sqrt3\,R$, option (C).

Final Answer:
The new radius is root 3 times R. \[ \boxed{\text{(C) }\sqrt3\,R} \]