Charged particle with charge q enters a region of constant, uniform and mutually orthogonal fields v

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4.Moving Charges and Magnetism

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

A

$\overrightarrow {\;v} $=$\;\frac{{\left( {\vec BX\vec E} \right)}}{{{E^2}}}$

B

$\overrightarrow {\;v} = \frac{{\left( {\vec EX\vec B} \right)}}{{{B^2}}}$

C

$\overrightarrow {\;v} $=$\;\frac{{\left( {\vec BX\vec E} \right)}}{{{B^2}}}$

D

$\;\overrightarrow {\;v} = \frac{{\left( {\vec EX\vec B} \right)}}{{{E^2}}}$

(AIEEE-2007)

Solution

Here, $\vec{E}$ and $\vec{B}$ are perpendicular to each other and the velocity $\vec{v}$ does not change; therefore

$q E=q v B \Rightarrow v=\frac{E}{B}$

Also,

$\left|\frac{\vec{E} \times \vec{B}}{B^{2}}\right|=\frac{E B \sin \theta}{B^{2}}=\frac{E B \sin 90^{\circ}}{B^{2}}=\frac{E}{B}=|\vec{v}|=v$

Standard 12

Physics

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