Question

An electron of mass $m$ is accelerated through a potential difference of $V$ and then it enters a magnetic field of induction $B$ normal to the lines. Then, the radius of the circular path is

• A. $\sqrt{\frac{2eV}{m}}$
• B. $\sqrt{\frac{2Vm}{e{B}^2}}$
• C. $\sqrt{\frac{2Vm}{eB}}$
• D. $\sqrt{\frac{2Vm}{e^{2}B}}$

Answer 1

The correct option is B $\sqrt{\frac{2Vm}{e{B}^2}}$

Let $F_m$ be the force due to magnetic field.
Thus, $F_m$ = $qvB$
Magnetic force provides the centripetal force to the particle moving in circular motion.
$⟹qvB=\frac{m{v}^2}{r}$
$⟹r=\frac{mv}{qB}$----(i)

We know that
$P=\sqrt{2mK}=mv$ (where $K$ is the kinetic energy)

Also, as electron is accelerated through potential difference of $V$, work done on electron will be change in kinetic energy.
$\Rightarrow K=qV$
$\Rightarrow mv=\sqrt{2mqV}$-----(ii)

Putting (ii) in (i), we get

$r=\frac{\sqrt{2meV}}{eB}$ ( $\because q=e$)
$⟹r=\sqrt{\frac{2mV}{e{B}^2}}$