Q. A particle of charge $- 16 \times 10^{- 18}coulomb$ moving with velocity $10\ ms^{- 1}$ along the $x$-axis enters a region where a magnetic field of induction $B$ is along the $y$-axis, and an electric field of magnitude $10^{4}V/m$ is along the negative $z$-axis. If the charged particle continues moving along the $x$-axis, the magnitude of $B$ is
A) $10^{- 3}Wb/m^{2}$
B) $10^{3}Wb/m^{2}$
C) $10^{5}Wb/m^{2}$
D) $10^{16}Wb/m^{2}$
Explanation
(b) Since particle is moving undeflected
So $qE = qvB \Rightarrow B = E/v = \frac{10^{4}}{10} = 10^{3}Wb/m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.