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An electric charge $+ q$ moves with velocity $\overrightarrow V = 3\hat i + 4\hat j + \hat k$ in an electromagnetic field given by : $\overrightarrow E = 3\hat i + \hat j + 2\hat k$ and $\overrightarrow B = \hat i + \hat j - 3\hat k$. The $y-$ component of the force experienced by $+ q$ is :-
$2\,q$
$11\,q$
$5\,q$
$3\,q$
Solution
$\mathrm{F}=\mathrm{F}_{\mathrm{e}}+\mathrm{F}_{\mathrm{m}}$
$ = {\rm{q}}\overrightarrow {\rm{E}} + {\rm{q}}(\overrightarrow {\rm{V}} \times \overrightarrow {\rm{B}} )$
$ \Rightarrow {\rm{q}}[\overrightarrow {\rm{E}} + (\overrightarrow {\rm{V}} \times \overrightarrow {\rm{B}} )]$
$ \Rightarrow {\rm{q}}\left[ {(3\widehat {\rm{i}} + \widehat {\rm{j}} + 2\widehat {\rm{k}})} \right] + \{ (3\widehat {\rm{i}} + 4\widehat {\rm{j}} + \widehat {\rm{k}}) \times (\widehat {\rm{i}} + \widehat {\rm{j}} – 3\widehat {\rm{k}})\} $
$\Rightarrow \mathrm{q}[(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})+(-13 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-\hat{\mathrm{k}})]$
$\mathrm{F}=\mathrm{q}[-10 \hat{\mathrm{i}}+11 \hat{\mathrm{j}}+\hat{\mathrm{k}}]$
so $y$ – component of the force is $=11 \mathrm{\,q}$
