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. `B/2 sqrt((qd)/(mV))`
B. `B/d sqrt(d/(2 mV))`
C. `B d sqrt(q/(2 mV))`
D. `qV sqrt((Bd)/(2m))`

Answer

c

Explanation

If R is the radius of circle described by the particle then angle of deviation can be written as follows:
` sin alpha = d//R ` ….(i)
Radius of the circle inside magnetic field can be written as: `R = (mv)/(qB)`
Particle is accelerated through a potential difference V.
` qV = 1/2 mv^(2) rArr mv^(2) = 2mqV rArr mv = sqrt(2 mqV) `
We can write the radius as follows:
`R = (mv)/(qB) = sqrt(2mqV)/(qB) = 1/B sqrt((2mV)/q)`
Substituting in equation (i) we get the following:
` sin alpha = Bd sqrt(q/(2mV))`
Hence option ( c) is correct.