Question

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

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

Answer 1

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

When $\overrightarrow{E}$ and $\overrightarrow{B}$ are perpendicular and velocity has no change then $qE=qvB$ i.e. $v=\frac{E}{B}$. The two forces oppose each other so, v is along $\overrightarrow{E}\times \overrightarrow{B}$ i.e., $\overrightarrow{v}=\frac{\overrightarrow{E}\times \overrightarrow{B}}{B^2}$

Option B