A charged particle is accelerated through a potential difference of `24 kV` and acquires a speed of `2xx10^(6)m//s`.It is then injected perpendicularly into a magnetic field of strength `0.2 T`.Find the radius of the circle described by it.

Answer

A, B, C

Explanation

`qV=1/2mv^(2)`
`qv_(B)=(mv^(2))/r`
`q/m=v^(2)/(2V)=(mv)/(qB)=v/Bxx(2V)/(v^(2))=(2V)/(Bv)=12cm`