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 ν→, then

Fe→+Fm→=0→qE→+qv→×B→=0→E→=-v→×B→=B→×v→E=vBsin900=vB ⇒v=EBand, v^=E^×B^v→=vv^=EBE^×B^=E→B×B→Bv→=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 ν→, then

Fe→+Fm→=0→qE→+qv→×B→=0→E→=-v→×B→=B→×v→E=vBsin900=vB ⇒v=EBand, v^=E^×B^v→=vv^=EBE^×B^=E→B×B→Bv→=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 $\vec{ν}$ , then

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

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

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

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

answer is C.

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Detailed Solution

$\vec{F_{e}} + \vec{F_{m}} = \vec{0}$

$q \vec{E} + q (\vec{v} \times \vec{B}) = \vec{0}$

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

$E = vBsin 90^{0} = vB \Rightarrow v = \frac{E}{B}$

and, $\hat{v} = \hat{E} \times \hat{B}$

$\vec{v} = v \hat{v} = \frac{E}{B} (\hat{E} \times \hat{B}) = \frac{\vec{E}}{B} \times \frac{\vec{B}}{B}$

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

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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 ν→, 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 $\vec{ν}$ , then