Answer:
The frequency of revolution of the second particle is f
Explanation:
When an object is moving under the magnetic field on a circular [path, the centripetal force is balanced by the magnetic force as :
[tex]\dfrac{mv^2}{r}=qvB[/tex]
Since,
[tex]v=r\omega[/tex]
and
[tex]\omega=2\pi f[/tex]
On rearranging the above equation, the formula for the frequency pf revolution is given by :
[tex]f=\dfrac{qB}{2\pi m}[/tex]
It is clear that the frequency is independent of the velocity. So, the frequency of revolution of the second particle remains the same i.e. f
