Question

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 :

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

Answer 1

Answer: (C)

$Bd\sqrt{\frac{q}{2mV}}$

Answer 2