Question

A proton (mas m) accelerated by a potential difference V files 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:

• A. $Bd\sqrt{\frac{q}{2mV}}$
• B. $\frac{B}{d}\sqrt{\frac{qd}{mV}}$
• C. $qV\sqrt{\frac{Bd}{2m}}$
• D. $\frac{B}{d}\sqrt{\frac{q}{2mV}}$

Answer 1

The correct option is A $Bd\sqrt{\frac{q}{2mV}}$

The proton will move in a circular path as the force due to magnetic field will be always perpendicular to the velocity of proton. The radius of the circle is given by :

$R=\frac{mv}{qB}$...(i)

Where m,v and q are mass, velocity of proton and charge of proton,

The proton is accelerated with potential difference of V, it's KE is given as $K.E.=qV$

$\frac{1}{2}m{v}^2=qV\Rightarrow v=\sqrt{\frac{2qV}{m}}$...(ii)

substituting value of $v$ in equation (i) from equation (ii)

$R=\frac{m\sqrt{\frac{2qV}{m}}}{qB}=\frac{\sqrt{\frac{2mV}{q}}}{B}$...(iii)

from geometry $Rsin\alpha =d\Rightarrow sin\alpha =\frac{d}{R}$

Substituting value of R from equation (iii)

$sin\alpha =dB\sqrt{\frac{q}{2mV}}$.