Question

A particle of mass m and charge q is released from the origin in a region in which the electric field and magnetic field are given by
$\overrightarrow{B}=-B_0\overrightarrow{j}\mathrm{and}\overrightarrow{E}=E_0\overrightarrow{k}$.
Find the speed of the particle as a function of its z-coordinate.

Answer 1

Given:
Mass of the particle = m
Charge of the particle = q
Electric field and magnetic field are given by
$\overrightarrow{B}=-B_0\overrightarrow{j}\mathrm{and}\overrightarrow{E}=E_0\overrightarrow{k}$
Velocity, $v=v_{\mathrm{x}}\hat{i}+v_{\mathrm{y}}\hat{j}+v_{\mathrm{z}}\hat{\mathit{k}}$
So, total force on the particle,
F = q (E + v × B)
$$\begin{gathered} =q\left[E_0\hat{k}-\left(v_{\mathrm{x}}\hat{i}+v_{\mathrm{y}}\hat{j}+v_{\mathrm{z}}\hat{\mathit{k}}\right)\times B_{\mathit{0}}\hat{\mathit{j}}\right] \\ =q{E}_0\hat{k}-v_{\mathrm{x}}{B}_0\hat{k}+v_{\mathrm{z}}{B}_0\hat{i} \end{gathered}$$
Since v_x = 0,
F_z = qE₀
$$\begin{gathered} \mathrm{So},a_{\mathrm{z}}=\frac{q{E}_0}{m} \\ {v}^2=u^2+2as=2\frac{q{E}_0}{m}z \\ \mathrm{So},v=\sqrt{\frac{2q{E}_{0}z}{m}} \end{gathered}$$
Here, z is the distance along the z-direction.