Q. A charge of $2.0\ \mu C$ moves with a speed of $3.0 \times 10^{6}\ ms^{- 1}$ along $+ ve\ X$-axis. A magnetic field of strength $\overrightarrow{B} = - 0.2\ \widehat{k}\ tesla$ exists in space. What is the magnetic force $({\overrightarrow{F}}_{m})$ on the charge
A) $F_{m} = 1.2\ N$ along $+ ve\ x -$ direction
B) $F_{m} = 1.2\ N$ along $- ve\ x -$ direction
C) $F_{m} = 1.2\ N$ along $+ ve\ y -$ direction
D) $F_{m} = 1.2\ N$ along $- ve\ y -$ direction
Explanation
(c) By using ${\overrightarrow{F}}_{m} = q(\overrightarrow{v} \times \overrightarrow{B})$
$\Rightarrow {\overrightarrow{F}}_{m} = 2 \times 10^{- 6}\left\{ 3 \times 10^{6}\widehat{i} \times ( - 0.2)\widehat{k} \right\} = - 1.2\left( \widehat{i} \times \widehat{k} \right) = + 1.2\widehat{j}$
$i.e.,$ 1.2 $N$ in positive $y$ direction
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.