Question

A proton accelarted by a potential differnce `V = 500 kV` fieles through a unifrom transverse magnetic filed the induction `B = 0.54 T`. The field occupies a region of space `d = 10 cm` in thickness (Fig). Find the angle `alpha` through which the proton deviates from the initial direction of its motion.

Answer 1

From the figure,
`sin alpha = (d)/(R) = (d q B)/(m v)`
As radius of the arc `R = (m v)/(q B)`, where `v` is the velocity of the particle, when it enteres into the filed. Form initial condition of the problem,
`qV = (1)/(2) mv^(2)` or, `v = sqrt((2qV)/(m))`
Hence, `sin alpha = (dqB)/(m sqrt((2qV)/(m))) = dB sqrt((q)/(2 mV))`
and `alpha = sin^(-1) (dB sqrt((q)/(2 mV))) = 30^(@)`, on putting the values.