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A charged particle is moving in a magnetic field at right angles to the field in a circular path of radius $$R$$. If a particle with same charge and twice greater mass were to be moving with half the velocity in the same magnetic field with its velocity at right angles to the field, the radius of its circular path would be
The radius of the path is $$r = \frac{mv}{qB}$$. With mass $$2m$$ and speed $$\frac{v}{2}$$, $$r' = \frac{2m \cdot \frac{v}{2}}{qB} = \frac{mv}{qB} = R$$.
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