Question

A charged particle moving in a uniform magnetic field loses \(4\%\) of its kinetic energy. The radius of curvature of its path changes by:

• A. \(2\%\)
• B. \(4\%\)
• C. \(10\%\)
• D. \(12\%\)

Hint:

In magnetic fields: \[ r \propto \sqrt{K} \] Always halve the percentage change of kinetic energy.

Answer 1

The Correct Option is A

Step 1: Radius of circular path: \[ r = \frac{mv}{qB} \]
Step 2: Kinetic energy: \[ K = \frac{1}{2}mv^2 \Rightarrow v \propto \sqrt{K} \]
Step 3: Small percentage change: \[ \frac{\Delta r}{r} = \frac{1}{2}\frac{\Delta K}{K} \]
Step 4: Given \(\Delta K = 4\%\): \[ \Delta r = 2\% \]