Q. An electron having charge $1.6 \times 10^{- 19}\ C$ and mass $9 \times 10^{- 31}\ kg$ is moving with speed 4$\times 10^{6}\ {ms}^{- 1}$ in a magnetic field of $2 \times 10^{- 1}\ T$ in a circular orbit. The force acting on electron and the radius of the circular orbit will be
A) $18.8\ \times 10^{- 13}\ N,\ 1.1\ \times 10^{- 4}\ m$
B) $12.8\ \times 10^{- 14}\ N,\ 1.1\ \times 10^{- 3}\ m$
C) $12.8\ \times 10^{- 13}\ N,\ 1.1\ \times 10^{- 3}\ m$
D) $1.28\ \times 10^{- 13}\ N,\ 1.1\ \times 10^{- 4}\ m$
Explanation
(d) Force $F = Bev = 2 \times 10^{- 1} \times 1.6 \times 10^{- 19} \times 4 \times 10^{+ 6}\ N$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 12.8 \times 10^{- 14}\ N = 1.28 \times 10^{- 13}\ N$
Now $r = \frac{mv}{Bq} = \frac{9\ \times \ 10^{- 31}\ \times \ 4\ \times 10^{6}}{2\ \times 10^{- 1} \times 1.6\ \times 10^{- 19}}\ m$
$\ \ \ \ \ \ \ \ \ \ \ = 11.25 \times 10^{- 5}\ m = 1.125 \times 10^{- 4}\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.