Question

In a certain region, uniform electron field $\overrightarrow{E}=E_0(-\hat{k})$ and magnetic field $\overrightarrow{B}=B_0\hat{k}$ are present. At time $t=0$, a particle of mass $m$ and charge $q$ is given a velocity $v_0\hat{j}+v_0\hat{k}$. The time, when the speed of particle is minimum will be:

• A. $\frac{2m{v}_0}{q{E}_0}$
• B. $\frac{m{v}_0}{q{E}_0}$
• C. $\frac{m{v}_0}{2q{E}_0}$
• D. $\frac{3m{v}_0}{q{E}_0}$

Answer 1

The correct option is B $\frac{m{v}_0}{q{E}_0}$

Magnetic force experienced by the particle is given by

$\overrightarrow{F_m}=q\left[\right(v_0\hat{j}+v_0\hat{k})\times B_0\hat{k}]=q{v}_0{B}_0\hat{i}$

So, magnetic force along $y\&z-$ axis will be zero.

Acceleration due to electric field in $-vez-$ direction will be

$\therefore a_z=-\frac{q{E}_0}{m}$

The component of velocity in $+vez-$ direction will be opposed by electric field, and it will eventually become zero $(v_z=0)$, at that time the particle will have minimum speed.

Given that, $\overrightarrow{v_i}=v_0\hat{j}+v_0\hat{k}$

And at the condition of minimum speed, velocity along $z-$ axis, $v_z=v_0\hat{k}$ become zero due to electric force, but $v_0\hat{j}$ will have same magnitude but different direction due to magnetic force.

So, $v_f=\sqrt{v_0^2+0^2}=v_0$

$\therefore v_{min}=v_0$

Let the $v_{min}$ is at time $t$, and using first equation of motion along $z-$ direction

$v_z=u_z+a_{z}t$

$\Rightarrow 0=v_0-\left(\frac{q{E}_0}{m}\right)t$

$\therefore t=\frac{m{v}_0}{q{E}_0}$

Hence, option $\left(b\right)$ is correct.

Why this question ? Tip: In vector form we can write velocity in this question as $\overrightarrow{v}=v_y\hat{j}+v_z\hat{k}$. The $y-$ component of velocity will change only its direction and not the speed due to action of magnetic force ${\overrightarrow{F}}_m$. Concept : ${\overrightarrow{v}}_{min}$ will be achieved only if $v_z=0$.