Question

A particle of mass m and charge q is moving in a region where uniform, constant electric and magnetic fields $\overrightarrow{E}$ and $\overrightarrow{B}$ are present. $\overrightarrow{E}$ and $\overrightarrow{B}$ are parallel to each other. At time $t=0$, the velocity ${\overrightarrow{v}}_0$ of the particle is perpendicular to $\overrightarrow{E}$ (Assume that its speed is always < < c, the speed of light in vacuum). If the velocity $\overrightarrow{v}$ of the particle at time t is $\overrightarrow{v}=X\times cos\left(\frac{qB}{m}t\right)\left({\overrightarrow{v}}_0\right)+\left(\frac{q}{m}t\right)\left(\overrightarrow{E}\right)+X\times sin\left(\frac{qB}{m}t\right)\left(\frac{{\overrightarrow{v}}_0\times \overrightarrow{B}}{B}\right)$. Find X?

Answer 1

$\hat{j}=\frac{\overrightarrow{E}}{E}$ or $\frac{\overrightarrow{B}}{B};\hat{i}=\frac{{\overrightarrow{v}}_0}{v_0};\hat{k}=\frac{{\overrightarrow{v}}_0\times {\overrightarrow{B}}_0}{v_{0}B}$

Force due to electric field will be along y-axis. Magnetic force will not affect the motion of charged particle in the direction of electric field (or y-axis) so.
$a_y=\frac{F_e}{m}=\frac{qE}{m},v_y=a_{y}t$$=\frac{qE}{m}T$ (i)

The charged particle under the action of magnetic field describes a circle in x-z plane (perpendicular to $\overrightarrow{B}$) with:

$T=\frac{2\pi m}{Bq}$ or $\omega =\frac{2\pi }{T}=\frac{qB}{m}$

Initially $(t=0)$ velocity was along x-axis. Therefore, magnetic force $\left({\overrightarrow{F}}_m\right)$ will be along positive z-axis $[{\overrightarrow{F}}_m=q({\overrightarrow{v}}_0\times \overrightarrow{B}\left)\right]$.

Let it this force makes an angle $\theta$ with x-axis at time t, then $\theta =\omega t$.
$\therefore v_x=v_{0}cos\omega t=v_{0}cos\left(\frac{qB}{m}t\right)$ (ii)
$v_z=v_{0}sin\omega t=v_{0}sin\left(\frac{qB}{m}t\right)$ (iii)

From Eqs. (i), (ii), and (iii)
$\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$
$\therefore \overrightarrow{v}=v_{0}cos\left(\frac{qB}{m}t\right)\left(\frac{{\overrightarrow{v}}_0}{v_0}\right)+\frac{qE}{m}t\left(\frac{\overrightarrow{E}}{E}\right)+v_{0}sin\left(\frac{qB}{m}t\right)\left(\frac{{\overrightarrow{v}}_0\times \overrightarrow{B}}{v_{0}B}\right)$
Or $\overrightarrow{v}=cos\left(\frac{qB}{m}t\right)\left({\overrightarrow{v}}_0\right)+\left(\frac{q}{m}t\right)\left(\overrightarrow{E}\right)+sin\left(\frac{qB}{m}t\right)\left(\frac{{\overrightarrow{v}}_0\times \overrightarrow{B}}{B}\right)$
The path of the particle will be a helix of increasing pitch. The axis of the helix will be along y-axis.