A proton of mass 'm' moving with a speed v ( $<<$ c, velocity of light in vacuum) completes a circular orbit in time 'T' in a uniform magnetic field. If the speed of the proton is increased to $\sqrt{2}$ v, what will be time needed to complete the circular orbit?

\sqrt{2} T

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\frac{T}{\sqrt{2}}

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\frac{T}{2}

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Solution

The correct option is D T
$T = \frac{2 π m}{q B}$
T is independent of v.

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