Question

A particle of mass m carrying charge $q$ is accelerated by a potential difference $V$. It enters perpendicularly in a region of uniform magnetic field $B$ and executes circular arc of radius $R$, then $q/m$ is equal to :

• A. $\frac{2V}{B^2{R}^2}$
• B. $\frac{V}{2BR}$
• C. $\frac{VB}{2R}$
• D. $\frac{mV}{BR}$

Answer 1

The correct option is A $\frac{2V}{B^2{R}^2}$

particle accelerated by P.d. V
So $qV=\frac{1}{2}m{v}^2$

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

And
$R=\frac{mv}{qB}$

$R=\frac{m}{qB}\sqrt{\frac{2qV}{m}}$

$\frac{q}{m}=\frac{2V}{B^2{R}^2}$