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.

Options

A. `15^(@)`
B. `30^(@)`
C. `45^(@)`
D. `60^(@)`

Answer

B

Explanation

`sin theta = (d)/( r)`
`r = (mv)/(qB) = (1)/(B) sqrt((2mV)/(q))`
`sin theta = Bd sqrt((q)/(2mV))`
`= 0.51 xx 0.1 sqrt((1.6 xx 10^(-19))/(2 xx 1.67 xx 10^(-27) xx 500 xx 10^(3)))`
`implies sin theta = (1)/(2) implies theta = 30^(@)`