Question

A charged particle q moving in a straight line is accelerated by a potential difference V. It enters a uniform magnetic field B perpendicular to its path. Deduce in terms of V an expression for the radius of the circular path in which it travels.

Answer 1

When charged particle is accelerated through a potential diffe. V, then work done=gain in K.E.
or `qV=1/2mv^2` or `v=sqrt((2qV)/(m))`
The force on charged particle due to perpendicular magnetic field `(=qvB)` provides the centripetal force `(=mv^2//r)` so
`qvB=(mv^2)/(r)`
or `r=(mv)/(qB)=(m)/(qB)=sqrt((2qV)/(m))=sqrt((2mV)/(qB^2))`