An e^- is moving parallel to a long current c | vedclass

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Question

$\mathrm{F}_{\mathrm{m}}=\mathrm{qvB} \sin 	heta=(\mathrm{e})(\mathrm{v})\left(\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}ight) \sin 90^{\circ}$

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Vedclass · Accepted Answer

$0.8 	imes {10^{ - 18}}\,N$

Vedclass · Answer

$\mathrm{F}_{\mathrm{m}}=\mathrm{qvB} \sin 	heta=(\mathrm{e})(\mathrm{v})\left(\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}ight) \sin 90^{\circ}$

$\mathrm{F}_{\mathrm{m}}=1.6 	imes 10^{-19} 	imes 10^{5}\left(\frac{2 	imes 10^{-7} 	imes 10}{4 	imes 10^{-2}}ight)$

$F_{\mathrm{m}}=0.8 	imes 10^{-18}\, \mathrm{N}$

An $e^-$ is moving parallel to a long current carrying wire as shown. Force on electron is