Q. An electron (mass $= 9.1 \times 10^{- 31}kg$, charge $= 1.6 \times 10^{- 18}C$) experience no deflection if subjected to an electric field of $3.2 \times 10^{5}V/m$, and a magnetic field of $2.0 \times 10^{- 3}Wb/m^{2}$. Both the fields are normal to the path of electron and to each other. If the electric field is removed, then the electron will revolve in an orbit of radius
A) 45 $m$
B) 4.5 $m$
C) 0.45 $m$
D) 0.045
Explanation
(c) For no deflection is mutually perpendicular electric and magnetic field $v = \frac{E}{B} = \frac{3.2 \times 10^{5}}{2 \times 10^{- 3}} = 1.6 \times 10^{8}\ m/s$.
If electric field is removed then due to only magnetic field radius of the path described by electron
$r = \frac{mv}{qB} = \frac{9.1 \times 10^{- 31} \times 1.6 \times 10^{8}}{1.6 \times 10^{- 19} \times 2 \times 10^{- 3}} = 0.45\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.