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

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The radius of a charged particle moving in a uniform magnetic field is given by $R = \frac{m v}{qB}$ .

The kinetic energy of the particle is $E = \frac{p ^{2}}{2 m}$ , which gives the momentum $p = m v = 2 m E$ .

Substituting the momentum in the radius formula, we get $R = \frac{2 m E}{qB}$ .

This shows that $R \propto E$ .

When the energy becomes $3$ times the original, the new energy is $E^{'} = 3 E$ .

The new radius $R^{'}$ is given by $R^{'} = \frac{2 m ( 3 E )}{qB} = 3 \frac{2 m E}{qB} = 3 R$ .

Answer: $3 R$

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