Q. A charge particle of mass $m$ and charge $q$ enters a region of uniform magnetic field B perpendicular of its velocity v. The particle initially at rest was accelerated by a potential difference $V$ (volts) before it entered the region of magnetic field. What is the diameter of the circular path followed by the charged particle in the region of magnetic field?
A) $\frac{2}{B}\sqrt{\frac{mV}{q}}$
B) $\frac{2}{B}\sqrt{\frac{2mV}{q}}$
C) $B\sqrt{\frac{2mV}{q}}$
D) $\frac{B}{q}\sqrt{\frac{2mV}{B}}$
Explanation
(d) $Bqv = \frac{mv^{2}}{r}\ \Rightarrow r = \frac{mv}{Bq}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
Since particle was initially at rest and gained a velocity $v$ due to a potential difference of $V$ volt. So,
KE of particle $= \frac{1}{2}\ mv^{2} = qV$
$v = \sqrt{\frac{2qV}{m}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
From Eqs. (i) and (ii), we get
$r = \frac{m}{Bq}\sqrt{\frac{2qV}{m}}$
$r = \frac{1}{B}\sqrt{\frac{2mV}{q}}$
$\therefore$ Diameter of the circular path
$d = 2r = \frac{2}{B}\sqrt{\frac{2mV}{q}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.