Question

A particle of charge of $q$ and mass $m$ starts moving from the origin under the action of an electric field $\overline{E}=E_0\hat{i}$ and $\overline{B}=B_0\hat{i}$ with a velocity $\overline{v}=v_0\hat{j}.$ The speed of the particle will become $2v_0$ after a time

• A. $t=\frac{2m{v}_0}{qE}$
• B. $t=\frac{2Bq}{m{v}_0}$
• C. $t=\frac{\sqrt{3}m{v}_0}{qE}$
• D. $t=\frac{\sqrt{3Bq}}{m{v}_0}$

Answer 1

The correct option is C $t=\frac{\sqrt{3}m{v}_0}{qE}$

First consider the effect of magnetic field on the particle
Magnetic force on the particle is given by
$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$
The force will be along -z direction and the particle will perform uniform circular motion in y-z plane.



Now, the electric field will try to accelerate the charged particle in the x-direction at each instant.
Acceleration due to electric field $=\frac{qE}{m}$

So,due to the net effect of both forces, particle will be moving in a helical path with increasing pitch.



Since change in magnitude of speed is only in the x-direction (due to electric field) and initial velocity in the x-direction was zero, we can use the first equation of motion to get
At time $t$,
$v_x=u_x+a_{x}t=0+\frac{qE}{m}t=\frac{qE}{m}t$
And there will be no change in the velocity in the y-z plane i.e it will be $v_0$

Net velocity at time $t,v=\sqrt{v_0^2+v_x^2}$
Putting $v=2v_0$,
$v_0^2+v_x^2=4v_0^2$
​​​​​​​$\Rightarrow v_0^2+{\left(\frac{qE}{m}t\right)}^2=4v_0^2$
$\Rightarrow {\left(\frac{qE}{m}t\right)}^2=3v_0^2$
$\Rightarrow \frac{qE}{m}t=\sqrt{3}{v}_0$
$\Rightarrow t=\frac{\sqrt{3}m{v}_0}{qE}$