Question

A particle of mass $1\times {10}^{-26}\text{kg}$ and charge $1.6\times {10}^{-19}\text{C}$ travelling with a velocity $1.28\times {10}^6{\text{ms}}^{-1}$ along the positive $X$-axis enters a region in whch a uniform electric field $\overrightarrow{E}=(-102.4\times {10}^3\hat{k})\frac{N}{C}$ and magnetic field $\overrightarrow{B}=(8\times {10}^{-2})\hat{j}\frac{Wb}{m^2}$ exists. The direction of motion of the particle is along

• A. negative $X$ axis
• B. positive $Y$- axis
• C. positive $Z$- axis
• D. positive $X$- axis

Answer 1

The correct option is D positive $X$- axis

Force due to electric field
$\overrightarrow{F_e}=q\overrightarrow{E}$
$=1.6\times {10}^{-19}\times (-102.4\times {10}^3)\hat{k}$
$=-163.84\times {10}^{-16}\hat{k}\text{(-ve z-axis)}$
Force due to magnetic field
$\overrightarrow{F_m}=q(\overrightarrow{V}\times \overrightarrow{B})$
$=1.6\times {10}^{-19}(1.28\times {10}^6\hat{i}\times 8\times {10}^{-2}\hat{j})$
$=+163.84\times {10}^{-16}\hat{k}\text{(+ve z -axis)}$
$\overrightarrow{F_e}$ is equal in magnitude and opposite in direction to that $\overrightarrow{F_m}.$ Thus, net force on the charge is zero.
$\overrightarrow{F_e}+\overrightarrow{F_m}=0$
Hence, It remain along x-axis.