Question

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

• A. $\frac{B}{2}\sqrt{\frac{ed}{mV}}$
• B. $eV\sqrt{\frac{Bd}{2m}}$
• C. $\frac{B}{2}\sqrt{\frac{e}{2mV}}$
• D. $Bd\sqrt{\frac{e}{2mV}}$

Answer 1

The correct option is D $Bd\sqrt{\frac{e}{2mV}}$

D. $Bd\sqrt{\frac{e}{2mV}}$

lets,

$m=$ mass of proton

$V=$ potential difference

$v=$ velocity of proton

$e=$ charge of proton

$d=$ The field occupies a region of space of width.

$R=$ radius of circle

$\alpha =$ angle of deviation.

Energy of proton,

$E=\frac{1}{2}m{v}^2=eV$

$v=\sqrt{\frac{2eV}{m}}$

The magnetic force,

$F=e(\overrightarrow{v}\times \overrightarrow{B})=\frac{m{v}^2}{R}$

$R=\frac{mv}{eB}$

$sin\alpha =\frac{d}{R}=\frac{deB}{mv}=\frac{deB}{m}\sqrt{\frac{m}{2qV}}$

$sin\alpha =Bd\sqrt{\frac{e}{2mV}}$