A proton (mass m) accelerated by a potential difference V flies through a uniform transverse magnetic field B. The field occupies a region of space by width d. If `alpha` be the angle of deviation of proton from initial direction of motion (see figure), the value of sin `alpha` will be

Options

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

Answer

C

Explanation

Energy of proton `=1/2 mv^(2)=qV, v=sqrt((2qV)/m)`
Magnetic force `qvB sin 90^(@)=(mv^(2))/R`
`R=(mv)/(qB), sin alpha= d/R=(dqB)/(mv)=(dqB)/m sqrt(m/(2av)), sin alpha =Bd sqrt(q/(2mv))`