Question

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

• A. $\sqrt{\frac{q{E}_{0}z}{m}}$
• B. $\sqrt{\frac{2(qv{B}_0+q{E}_0)2z}{m}}$
• C. $\sqrt{\frac{(-qv{B}_0+q{E}_0)2z}{m}}$
• D. $\sqrt{\frac{2q{E}_{0}z}{m}}$

Answer 1

The correct option is D $\sqrt{\frac{2q{E}_{0}z}{m}}$

Given:
$\overrightarrow{B}=B_0\hat{j}$
$\overrightarrow{E}=E_0\hat{k}$

The particle have initial velocity zero, so electric field will be responsible for movement of the particle.

Magnetic force on the particle does not increase the speed of the particle since ${\overrightarrow{F}}_{mag.}\overrightarrow{v}=0(<{90}^{\circ })$

$\Rightarrow {\overrightarrow{F}}_{electric}=q{E}_0\hat{k}$

If $\overrightarrow{a}$ is the acceleration along the ${\overrightarrow{F}}_{electric}$.

$\Rightarrow \overrightarrow{a}=\frac{q{E}_0}{m}\hat{k}$

Since acceleration is constant, applying the one-dimensional kinematic equation:

$v^2-u^2=2a_{z}z$

Initial speed $u=0$,

$\therefore v^2=\frac{2q{E}_{0}z}{m}$

$\Rightarrow v=\sqrt{\frac{2q{E}_{0}z}{m}}$

Hence, option (d) is the correct answer.