JEE MAIN - Physics (2007 - No. 14)
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
$$\overrightarrow v = \overrightarrow B \times \overrightarrow E /{E^2}$$
$$\overrightarrow v = \overrightarrow E \times \overrightarrow B /{B^2}$$
$$\overrightarrow v = \overrightarrow B \times \overrightarrow E /{B^2}$$
$$\overrightarrow v = \overrightarrow E \times \overrightarrow B /{E^2}$$
Explanation
Here, $$\overrightarrow E $$ and $$\overrightarrow B $$ are perpendicular to each other and the velocity $$\overrightarrow v $$ does not change; therefore
$$qE = qvB \Rightarrow v = {E \over B}$$
Also, $$\left| {{{\overrightarrow E \times \overrightarrow B } \over {{B^2}}}} \right| = {{E\,\,B\sin \theta } \over {{B^2}}}$$
$$ = {{E\,\,B\sin {{90}^ \circ }} \over {{B^2}}} = {E \over B} = \left| {\overrightarrow v } \right| = v$$
$$qE = qvB \Rightarrow v = {E \over B}$$
Also, $$\left| {{{\overrightarrow E \times \overrightarrow B } \over {{B^2}}}} \right| = {{E\,\,B\sin \theta } \over {{B^2}}}$$
$$ = {{E\,\,B\sin {{90}^ \circ }} \over {{B^2}}} = {E \over B} = \left| {\overrightarrow v } \right| = v$$
Comments (0)
