A charged particle is moving in a uniform magnetic field in a circular path of radius ' R '. When the kinetic energy of the particle is increased to three times, then the new radius will be

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

Since kinetic energy is $K = \frac{p ^{2}}{2 m}$ , the momentum is $p = 2 m K$ .

Substituting $p = m v$ into the radius formula gives $R = \frac{2 m K}{qB}$ .

This shows that $R \propto K$ .

When the kinetic energy is increased to three times its initial value, $K^{'} = 3 K$ .

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

Answer: $3 \cdot R$

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