A charged particle q' is moving with a speed V

cbse
electromagnetism

#1

A charged particle q’ is moving with a speed V perpendicular to the magnetic field of induction B. Find the radius of the path and the time period of the particle.


#2

Let us assume that the field is directed into the page as shown in figure. Then the force experienced by the particle is F = qvB. We know that this force is always directed perpendicular to velocity. Hence the particle moves along a circular path and the magnetic force on a charged particle acts like a centripetal force.
Let r be the radius of the circular path.
We know that centripetal force = m${{v}^{2}}$ / r
q v B = m${{v}^{2}}$ / r
Solving this equation, we get; r = mv / Bq
Time period of the particle; T = 2$\pi$r / v
Substituting r in above equation, we get
T = 2$\pi$m / Bq