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.
$$\overrightarrow v = \overrightarrow B \times \frac{{\overrightarrow E }}{{{E^2}}}$$
B.
$$\overrightarrow v = \overrightarrow E \times \frac{{\overrightarrow B }}{{{B^2}}}$$
C.
$$\overrightarrow v = \overrightarrow B \times \frac{{\overrightarrow E }}{{{B^2}}}$$
D.
$$\overrightarrow v = \overrightarrow E \times \frac{{\overrightarrow B }}{{{E^2}}}$$
Answer :
$$\overrightarrow v = \overrightarrow E \times \frac{{\overrightarrow B }}{{{B^2}}}$$
Solution :
Here, $$\overrightarrow E $$ and $$\overrightarrow B $$ are perpendicular to each other and the velocity $$\overrightarrow v $$ does not change; therefore
$$qE = qvB \Rightarrow v = \frac{E}{B}$$
Also,
$$\left| {\frac{{\overrightarrow E \times \overrightarrow B }}{{{B^2}}}} \right| = \frac{{E\,B\sin \theta }}{{{B^2}}} = \frac{{E\,B\sin {{90}^ \circ }}}{{{B^2}}} = \frac{E}{B} = \left| {\overrightarrow v } \right| = v$$