Question

A particle of mass m and charge q is moving in a region where uniform, constant electric and magnetic fields E and B are present. E and B are parallel to each other. At time t = 0, the velocity $v_0$ of the particle is perpendicular to E (Assume that its speed is always<< c, the speed of light in vacuum). Find the velocity v of the particle at time t. You must express your answer in terms of t, q, m, the vector $v_0$ , E and B and their magnitudes $v_0$, E and B.

Answer 1

$\hat{j}=\frac{E}{E}or\frac{B}{B}:\hat{i}=\frac{v_0}{v_0}$
$\hat{k}=\frac{v_0\times B_0}{v_{0}B}$
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,
$a_y=\frac{F_e}{m}=\frac{qE}{m}=constant$
Therefore , $v_y=a_{y}t=\frac{qE}{m}.t$ ............(i)
The charged particle under the action of magnetic field
describes a circle in x-z plane (perpendicular to 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 $(F_m)$ will be along positive z-axis
$[Fm=q(v_{0}xB)]$. Let it makes an angle 0 with x-axis at time
t, then
$\therefore v_x=v_{0}cos\omega t=v_{0}cos\left(\frac{qB}{m}t\right)$ ..........(ii)
$v_z=v_{0}sin\omega t=v_{0}sin\left(\frac{qB}{m}t\right)$ ..............(iii)
From Eqs. (i), (ii) and (iii),
$v=v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$
$\therefore v=v_{0}cos\left(\frac{qB}{m}t\right)\left(\frac{v_0}{v_0}\right)+\frac{qB}{m}t\left(\frac{E}{E}\right)$
$+v_{0}sin\left(\frac{qB}{m}t\right)\left(\frac{v_0\times B}{v_{0}B}\right)$
$\therefore v=cos\left(\frac{qB}{m}t\right)(v_0)+\left(\frac{qB}{m}t\right)(E)$
$+v_{0}sin\left(\frac{qB}{m}t\right)\left(\frac{v_0\times B}{v_{0}B}\right)$
NOTE The path of the particle will be a helix of increasing pitch
The axis of the helix will be alongy-axis.