Q. A particle of charge $- 16 \times 10^{- 18}$ C 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 $10^{4}\ {Vm}^{- 1}$ is along the negative z-axis. If the charged particle continues moving along the x-axis, the magnitude of B is
Explanation
(a) The force on a particle is
So, $\mathbf{F} = q\left( \mathbf{E} + \mathbf{v} \times \mathbf{B} \right)$
or $\mathbf{F} = \mathbf{F}_{e} + \mathbf{F}_{m}$
$\mathbf{F}_{e} = q\mathbf{E}$
$\ = -$16$\times 10^{- 18} \times 10^{4}\left( - \widehat{k} \right)$
=16$\times 10^{- 14}\widehat{k}$
and $\mathbf{F}_{m}$=$- 16 \times 10^{- 18}(10\widehat{i} \times B\widehat{j})$
=$- 16 \times 10^{- 17} \times B( + \widehat{k})$
=$- 16 \times 10^{- 17}B \times \widehat{k}$
Since, particle will continue to move along +$\ x$-axis, so resultant force is equal to 0.
$\mathbf{F}_{e} + \mathbf{F}_{m} = 0$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 16 \times 10^{- 14} = 16 \times 10^{- 17}B$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B$ =$\ \frac{16 \times 10^{- 14}}{\ 16 \times 10^{- 17}} = 10^{3}$
$B = 10^{3}Wb$-$m^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.