A proton accelerated by a potential difference 500 kV flies through a uniform transverse magnetic field 0.1 T.
The field is spread on a region of 1.0 cm thickness. The angle through which the proton gets deviated from its original direction is , (mass of proton `= 1.6 xx 10^(-27) ` kg and charge of proton `=1.6 xx 10^(19) C`)

Options

A. 0.01 rad
B. 0.1 rad
C. 0.05 rad
D. 0.08 rad

Answer

A

Explanation

According to the question

Here first we have to find the radius of circular path (shown in figure ) which is given by relation,
`R= sqrt(2m(KE))/(qB)`
where, KE = qV
Now, putting the values, where mass of proton `m= 1.6 xx 10 ^(-27) kg` ,
`q= 1.6 xx 10^(-19) C ` potential difference V = 500 kV , and magnetic field , B= 0.1 T
putting the given values we get
`R- sqrt(2xx 1.6 xx 10-^(-27) xx 1.6 xx 10^(-19) xx 500 xx 10^3)/(1.6 xx 10^(-19)xx 0.1) `
` rArr R =1 m`
Since are is small so it can be assumed that it can be equal to the thickness of the field.
Hence, `phi + angle ADC = pi` (from figure) ........(i)
where `phi = (arc ) /("radius") = (10^(-2))/1 = 0.01 rad`
` because theta = pi - angle ADC ` (from figure) ........(ii)
From Eqs. (i) and (ii) we get
`phi = theta`
so, `phi = 0.01 rad`