Question

A charged particle with charge $q$ enters a region of constant, uniform and mutually orthogonal fields $\overrightarrow{E}$ and $\overrightarrow{B}$ , with a velocity $\overrightarrow{v}$ perpendicular to both $\overrightarrow{E}$ and $\overrightarrow{B}$, and comes out without any change in magnitude or direction of $\overrightarrow{v}$. Then

• A. $\overrightarrow{v}=\frac{\overrightarrow{E}\times \overrightarrow{B}}{B^2}$
• B. $\overrightarrow{v}=\frac{\overrightarrow{E}\times \overrightarrow{E}}{B^2}$
• C. $\overrightarrow{v}=\frac{\overrightarrow{E}\times \overrightarrow{B}}{E^2}$
• D. $\overrightarrow{v}=\frac{\overrightarrow{B}\times \overrightarrow{B}}{E^2}$

Answer 1

The correct option is A $\overrightarrow{v}=\frac{\overrightarrow{E}\times \overrightarrow{B}}{B^2}$

Since there is no change in magnitude and direction of velocity of the charged particle, so force due to magnetic field is equal to force due to electric field.
Therefore,
$qE=qvB$
$\Rightarrow v=\frac{E}{B}$
So magnitude of velocity vector is $\frac{E}{B}$
Given that velocity $\overrightarrow{v}$ perpendicular to both $\overrightarrow{E}$ and $\overrightarrow{B}$, so $\overrightarrow{v}$ is along $(\overrightarrow{E}\times \overrightarrow{B})$
Now, $\overrightarrow{v}=\frac{E}{B}\times \frac{\overrightarrow{E}\times \overrightarrow{B}}{|\overrightarrow{E}\times \overrightarrow{B}|}=\frac{E}{B}\times \frac{\overrightarrow{E}\times \overrightarrow{B}}{EB\sin {90}^{\circ }}=\frac{\overrightarrow{E}\times \overrightarrow{B}}{B^2}$