Obtain expression for the radius of the circular path and its frequency, of a charged particle entering into an uniform magnetic field.

Explanation

Let B represent the uniform magnetic applied perpendicular to the plane of the paper and pointing inwards.
Let q represents a +ve charge entering the field with a velocity v at right angles to the field. Let F represent the magnetic force on the charged particle, which acts perpendicular to the direction of velocity. No work is done by the force.
Magnetic force `vec(F)=qvec(v) times vec(B)`. This force acts as a centripetal force and produces a circular motion (as per the Fleming's Left Hand Rule).

i.e., `" "(mv^(2))/r=qvBsin90^(@)`
i.e., `" "r=(mv)/(qB)`
Therefore, radius of the circular path `r propto m/q` for a given uniform speed and magnetic field.
Period of revolution `T=(2pir)/v=(2pim)/qB " and frequency " v=l/T=(qB)/(2pim)`
In the presence of the component of velocity along B, the path described by the charged particle will be helix and whose pitch `p=v_(h)T, v_(h)` is the horizontal component of velocity along B.