Question

A charged particle is moving in a uniform magnetic field and enters a lead layer, losing half of its kinetic energy. The radius of curvature of its path becomes:

• A. No change
• B. Reduced by \( \frac{1}{2} \) times of its initial value
• C. Reduced to \( \frac{1}{\sqrt{2}} \) times of its initial value
• D. Reduce to \( \frac{1}{4} \) times of its initial value

Hint:

Radius of circular path \( r \propto v \propto \sqrt{K.E} \). Halving the kinetic energy reduces the radius by \( \frac{1}{\sqrt{2}} \).

Answer 1

The Correct Option is C

The radius of curvature in a magnetic field is given by: \[ r = \frac{mv}{qB} \] Kinetic energy: \[ K.E = \frac{1}{2}mv^2 \Rightarrow v \propto \sqrt{K.E} \] If kinetic energy becomes half: \[ v' = v \cdot \frac{1}{\sqrt{2}} \Rightarrow r' = \frac{mv'}{qB} = r \cdot \frac{1}{\sqrt{2}} \] So, the radius decreases by a factor of \( \frac{1}{\sqrt{2}} \).