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 charged particle moving in a uniform magnetic field loses \(4\%\) of its kinetic energy. The radius of curvature of its path changes by:
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In magnetic fields:
\[
r \propto \sqrt{K}
\]
Always halve the percentage change of kinetic energy.
Updated On: Mar 23, 2026
- \(2\%\)
- \(4\%\)
- \(10\%\)
- \(12\%\)
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The Correct Option is A
Solution and Explanation
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 \(\Delta K = 4\%\): \[ \Delta r = 2\% \]
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