Answer:
The value of circular velocity or frequency is \(\dfrac{qB}{m}\).
Explanation:
The charged particle is moving in a circular path.
Let the charge of the particle is \(q\). The radius of the circular path is \(R\). The velocity of the particle is \(v\) and it is moving in a magnetic field is \(B\).
Now, the magnetic force experienced by the charged particle is,
\(F=qvB\) (assuming the magnetic field is perpendicular to the direction of motion)
Now, the centripetal force experienced by the particle will be,
\(F_c=\dfrac{mv^2}{R}\)
Velocity of the particle can be written as (in the form of angular velocity) \(v=R\omega\).
Now, equating the magnetic and centripetal force to get the value of circular velocity,
\(F=F_c\\qvB=\dfrac{mv^2}{R}\\qR\omega B=\dfrac{m(R\omega)^2}{R}\\qR\omega B=\dfrac{mR^2\omega^2}{R}\\\omega=\dfrac{qB}{m}\)
So, the value of circular velocity or frequency is \(\dfrac{qB}{m}\).
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