A charged particle with charge q enters region of constant, uniform and mutually orthogonal fields E→ and B→ , and comes out without any change in magnitude or direction of υ→ , then

F→=q(V→×B→)and F→=qE→ ∴q(V→×B→)=qE→ V→×B→=E→ (V→×B→)×B→=E→×B→ (B→.B→)V→−(B→.B→)V→=E→×B→ V→=E→×B→B2

A charged particle with charge q enters region of constant, uniform and mutually orthogonal fields E→ and B→ , and comes out without any change in magnitude or direction of υ→ , then

F→=q(V→×B→)and F→=qE→ ∴q(V→×B→)=qE→ V→×B→=E→ (V→×B→)×B→=E→×B→ (B→.B→)V→−(B→.B→)V→=E→×B→ V→=E→×B→B2

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A charged particle with charge q enters region of constant, uniform and mutually orthogonal fields $\vec{E}$ and $\vec{B}$ , and comes out without any change in magnitude or direction of $\vec{υ}$ , then

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\vec{υ} = (\vec{E} \times \vec{B}) {/E}^{2}

\vec{υ} = (\vec{E} \times \vec{B}) {/B}^{2}

\vec{υ} = (\vec{B} \times \vec{E}) {/E}^{2}

\vec{υ} = (\vec{B} \times \vec{E}) {/B}^{2}

answer is C.

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Detailed Solution

\vec{F} = q (\vec{V} \times \vec{B}) and \vec{F} = q \vec{E}

$∴ q (\vec{V} \times \vec{B}) = q \vec{E}$

$\vec{V} \times \vec{B} = \vec{E}$

$(\vec{V} \times \vec{B}) \times \vec{B} = \vec{E} \times \vec{B}$

$(\vec{B} . \vec{B}) \vec{V} - (\vec{B} . \vec{B}) \vec{V} = \vec{E} \times \vec{B}$

$\vec{V} = \frac{\vec{E} \times \vec{B}}{B^{2}}$

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A charged particle with charge q enters region of constant, uniform and mutually orthogonal fields E→ and B→ , and comes out without any change in magnitude or direction of υ→ , then

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A charged particle with charge q enters region of constant, uniform and mutually orthogonal fields $\vec{E}$ and $\vec{B}$ , and comes out without any change in magnitude or direction of $\vec{υ}$ , then