Q. An electron enters the space between the plates of a charged capacitor as shown. The charge density on the plate is $\sigma.$ Electric intensity in the space between the plates is $E$.A uniform magnetic field $B$also exists in the space perpendicular to the direction of $E.$ The electron moves perpendicular to both E and B without any change in direction. The time taken by the electron to travel a distance $l$ in the space is
A) $\frac{\sigma l}{\varepsilon_{0}B}$
B) $\frac{\sigma B}{\varepsilon_{0}l}$
C) $\frac{\varepsilon_{0}lB}{\sigma}$
D) $\frac{\varepsilon_{0}l}{\sigma B}$
Explanation
(c) Here, magnetic force =electrostatic force
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ qvB = qE$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = \frac{E}{B} = \frac{\sigma}{\varepsilon_{0}B}$
The time taken by electron to travel a distance $l$ in that space
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t = \frac{1}{v} = \frac{l}{\frac{\sigma}{\varepsilon_{0}B}} = \frac{\varepsilon_{0}lB}{\sigma}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.