Question

An electron of mass $\text{m}$ and electric charge $e$ is accelerated by a potential difference of $V$ and then it enters in a magnetic field $B$ perpendicular to the magnetic field lines. Radius of the circular path formed by electron is

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

Answer 1

Answer: (B)

$Bev=\frac{mv^{2}}{r} \, or \, r=\frac{mv}{Be}$ as $mv=\sqrt{2 mT}$ $\left(\right.T \, = \, KE\left.\right)$ . So $r=\frac{\sqrt{2 mT}}{Be}$
As the electron has been accelerated from rest through a potential difference of $V$ volt, then $T=eV$
$\therefore \, r=\sqrt{\frac{2 mVe}{B^{2} e^{2}}}=\sqrt{\frac{2 mV}{B^{2} e}}$