Question

A charged particle enters a region of constant, uniform and mutually orthogonal field $\overrightarrow{E}$ & $\overrightarrow{B}$, with velocity $\overrightarrow{U}$ perpendicular to both $\overrightarrow{E}$ & $\overrightarrow{B}$ and Comes out without any charge in magnitude or direction of $\overrightarrow{U}$. Then ,

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

Answer 1

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


As per the information, electric field $\left(\overrightarrow{E}\right)$ and magnetic field $\left(\overrightarrow{B}\right)$, both are perpendicular to each other as well as to the velocity of the charged particle.

If there is no charge in the magnitude & direction of the velocity of charge $q$, then

$|f_E|=|f_B|$

$\Rightarrow qE=qUB$

$\Rightarrow U=\frac{E}{B}$

The given situation is shown in the figure below:
For the above situation, $\overrightarrow{B}$ must be into the plane $⨂$

$\therefore \overrightarrow{U}=\frac{\overrightarrow{E}\times \overrightarrow{B}}{B^2}=\frac{EB\sin {90}^{\circ }}{B^2}\Rightarrow \begin{array}{c}U=\frac{E}{B}\end{array}$

Hence, option $\left(A\right)$ is the correct answer.