question_answer

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 a will be:            [JEE ONLINE 10-04-2015]

A) \[{{q}^{V}}\sqrt{\frac{Bd}{2m}}\]

B) \[Bd\sqrt{\frac{q}{2mV}}\]

C) \[\frac{B}{2}\sqrt{\frac{qd}{mV}}\]

D) \[\frac{B}{d}\sqrt{\frac{q}{2mV}}\]

Correct Answer: B

Solution :

[b] Energy of proton \[=\frac{1}{2}m{{v}^{2}}=qV\]

\[v=\sqrt{\frac{2qV}{m}}\]

magnetic force \[=q\left( \vec{v}\times \vec{B} \right)=\frac{m{{v}^{2}}}{R}\]

\[R=\frac{mv}{qB}\]

\[\sin \alpha =\frac{d}{R}=\frac{dqB}{mv}=\frac{dqB}{m}\sqrt{\frac{m}{2qV}}\]

\[\sin \alpha =Bd\sqrt{\frac{q}{2mV}}\]