Q. A particle with $10^{- 11}coulomb$ of charge and $10^{- 7}kg$ mass is moving with a velocity of $10^{8}m/s$ along the $y$-axis. A uniform static magnetic field $B = 0.5\ tesla$ is acting along the $x$-direction. The force on the particle is
A) $5 \times 10^{- 11}N\ $along $\widehat{i}$
B) $5 \times 10^{3}N\ $along $\widehat{k}$
C) $5 \times 10^{- 11}N\ $along $- \widehat{j}$
D) $5 \times 10^{- 4}N\ $along $- \widehat{k}$
Explanation
(d) $\overrightarrow{F} = q\left( \overrightarrow{v} \times \overrightarrow{B} \right) = 10^{- 11}(10^{8}\widehat{j} \times 0.5\widehat{i})$
$= 5 \times 10^{- 4}\left( \widehat{j} \times \widehat{i} \right) = 5 \times 10^{- 4}N( - \widehat{k})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.