Q. A particles of mass $1 \times 10^{- 26}kg$ and charge $1.6 \times 10^{- 19}C$ travelling with a velocity $1.28\ \times \ 10^{6}\ {ms}^{- 1}$ along the positive $X$-axis enters a region in which a uniform electric field E and a uniform magnetic field of induction B are present. If $\mathbf{E} = - 10.24 \times 10^{3}\ \widehat{\mathbf{k}}\ {NC}^{- 1}$ and $B = 8 \times 10^{- 2}\widehat{\mathbf{J}}\ {Wbm}^{- 2},$ the direction of motion of the particles is
Explanation
(a) $m = 1\ \times \ 10^{- 26}kg,\ q = 1.6\ \times \ 10^{- 19}\ C,$
$\ \ \ \ \ \ v = 1.28\ \times \ 10^{6}\ {ms}^{- 1}$
Electric field $\ \ \ \ \ \ \ \ \ \ \ \mathbf{E} = \ - 1024\ \times \ 10^{3}\ \widehat{\mathbf{k}}\ {NC}^{- 1}$
Magnetic field $\ \ \ \ \ \ \ \ \mathbf{B}\ = \ 8\ \times \ 10^{- 2}\ \widehat{\mathbf{j}}\ {Wbm}^{- 2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\left| \mathbf{E} \right|}{\left| \mathbf{B} \right|} = \frac{102.4\ \times \ 10^{3}}{8\ \times 10^{- 2}} = \frac{10.24 \times 10^{6}}{8}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ 1.28\ \times \ 10^{6}$
Hence, $|\mathbf{v}| = \frac{|\mathbf{E}|}{|\mathbf{B}|}$
So, particle will remain undeflected, hence direction of motion of particle is along the positive $X$-axis.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.