A charge moves in a circular path perpendicular to a magnetic field. The time period of the revolution is…
A charge moves in a circular path perpendicular to a magnetic field. The time period of the revolution is independent of
strength of charge.
magnetic field.
mass of the charge.
velocity of the charge.
Solution
The time period of a charged particle undergoing circular motion under uniform perpendicular magnetic field is given by $T=\frac{2 \pi m}{q B}$
So, it is independent of velocity.