An electric charge + q moves with velocity ov | vedclass

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Question

$\mathrm{F}=\mathrm{F}_{\mathrm{e}}+\mathrm{F}_{\mathrm{m}}$

$ = {m{q}}\overrightarrow {m{E}}  + {m{q}}(\overrightarrow {m{V}}

Vedclass · Accepted Answer

$11\,q$

Vedclass · Answer

$\mathrm{F}=\mathrm{F}_{\mathrm{e}}+\mathrm{F}_{\mathrm{m}}$

$ = {m{q}}\overrightarrow {m{E}}  + {m{q}}(\overrightarrow {m{V}} 	imes \overrightarrow {m{B}} )$

$ \Rightarrow {m{q}}[\overrightarrow {m{E}}  + (\overrightarrow {m{V}} 	imes  \overrightarrow {m{B}} )]$

$ \Rightarrow {m{q}}\left[ {(3\widehat {m{i}} + \widehat {m{j}} + 2\widehat {m{k}})} ight] + \{ (3\widehat {m{i}} + 4\widehat {m{j}} + \widehat {m{k}}) 	imes (\widehat {m{i}} + \widehat {m{j}} - 3\widehat {m{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 $=11 \mathrm{\,q}$

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 :-

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