Question

A point charge particle of charge q and mass m is moving in circular path of radius r with constant speed v in a uniform magnetic field due to magnetic force. If the speed is doubled to 2 v then what happens to the period of revolution, T?

• A. T increases by a factor of 2
• B. T increases by a factor of 4
• C. T stays the same
• D. T decreases by a factor of 2
• E. T decreases by a factor of 4

Answer 1

The correct option is C T stays the same

The force due to magnetic field on the charged particle provides the centripetal acceleration to the charge.

Thus $qvB=\frac{m{v}^2}{r}$

$⟹\frac{r}{v}=\frac{m}{qB}$

The time period of revolution is $T=\frac{2\pi r}{v}$

$=\frac{2\pi m}{qB}$

which is independent of the speed of the particle.

Thus when the velocity becomes twice, the radius also becomes twice to keep the time period of revolution the same.