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A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform
magnetic field. What should be the energy of an α -particle to describe a circle of the same
radius in the same field?

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2 MeV

4 MeV

0.5 MeV

1 MeV

answer is B.

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Detailed Solution

For proton $r = \frac{\sqrt{2 m (KE)}}{qB}$

$q \propto \sqrt{m (KE)} H e n c e \frac{e}{2 e} = \sqrt{\frac{(m_{p}) (1MeV)}{({4m}_{p}) (KE)}} \frac{1}{4} = \frac{1MeV}{4KE} K E = 1 M e V$

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