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A particle of mass $m$ and charge $q$ is thrown from origin at $t = 0$ with velocity $2\hat{i}$ + $3\hat{j}$ + $4\hat{k}$ units in a region with uniform magnetic field $\vec B$ = $2\hat{i}$ units. After time $t =\frac{{\pi m}}{{qB}}$ , an electric field is switched on such that particle moves on a straight line with constant speed. $\vec E$ may be
$5\hat k - \,10\hat j \ units$
$-6\hat k - \,9\hat j \ units$
$-6\hat k + \,8\hat j \ units$
$6\hat k + \,8\hat j \ units$
Solution
In time $\frac{\mathrm{T}}{2}=\frac{\pi \mathrm{m}}{\mathrm{qB}}$
$\mathrm{V}_{\perp}$ gets reversed
$\overrightarrow {\rm{v}} = 2\widehat {\rm{i}} – 3\widehat {\rm{j}} – 4\widehat {\rm{k}}$
$\mathrm{q}(\overrightarrow{\mathrm{E}}+\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=0$
$\overrightarrow {\rm{E}} = – (\overrightarrow {\rm{v}} \times \overrightarrow {\rm{B}} ) = – (2\widehat {\rm{i}} – 3\widehat {\rm{j}} – 4\widehat {\rm{k}}) \times 2\widehat {\rm{i}} = – 6\widehat {\rm{k}} + 8\widehat {\rm{j}}$
