Question

An electric field $\left(2x\right)\text{V/m}\hat{i}$ and a magnetic field of constant value $2\text{T}\hat{i}$ exists in a region. If a charged particle of mass $1\text{kg}$ and having charge $1\text{C}$ is released from origin (0,0). what will be its speed when it has travelled $4\text{m}$ along $(+ve)x$ - axis?

• A. $4\sqrt{2}\text{m/s}$
• B. $\sqrt{2}\text{m/s}$
• C. $\frac{1}{\sqrt{2}}\text{m/s}$
• D. $\frac{\sqrt{3}}{2}\text{m/s}$

Answer 1

The correct option is A $4\sqrt{2}\text{m/s}$


Due to presence of electric field in $(+ve)$ $x$ - direction, the charge will accelerate in $(+ve)x$ - direction.

Thus magnetic force acting on charge is,

${\overrightarrow{F}}_m=q(\overrightarrow{v}\times \overrightarrow{B})=0$

$(\because \overrightarrow{B}||\overrightarrow{E}||\overrightarrow{v})$

Here only electric force is the external force acting on the particle.

$F_e=qE=\left(1\right)\left(2x\right)=2x$

From Newton's 2nd law,

$a=\frac{F_{net}}{m}$

or, $a=2x$

or, $v\frac{dv}{dx}=2x$

$vdv=2xdx$

Inegrating both sides with linits $v=0$ at $x=0$ & $v=v$ at $x=4\text{m}$ we get:

${\int }_0^{v}vdv=2{\int }_0^{4}xdx$

$\frac{v^2}{2}=2{\left[\frac{x^2}{2}\right]}_0^4$

$\frac{v^2}{2}=\frac{2}{2}[4^2-0]=\frac{32}{2}$

$v^2=32$

$\therefore v=\sqrt{32}=4\sqrt{2}\text{m/s}$