A small charged ball having mass m and charge q is suspended from a rigid support by means of an inextensible thread of length l. It is made to rotate on a horizontal circular path in a uniform, time independent magnetic field of induction B which is directed upward. The time period of revolution of the ball is `T_0.` If the thread is always stretched, calculate the radius of circular path on which the ball moves.

Answer

`r=[L^(2)-{(mg)/(momega^(2)+-qomegaB)}^(2)]^(1//2)`

Explanation

Charge ball move in the form of conical pendulum.Magnetic force act along radial direction.
`T sin theta=mr omega^(2)+-q.(romega).B`
`Tcos theta=mg`
`tantheta=((mgomega^(2)+-qomegaB)r)/(mg)`
`r/(sqrt(r^(2)+L^(2)))=((momega^(2)+-qomegaB)r)/(mg) rArrr=[L^(2)-{(mg)/(momega^(2)+-qomegaB)}^(2)]^(1//2)`