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

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4.Moving Charges and Magnetism

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

A

$0.4 \times {10^{ - 18}}\,N$

B

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

C

$0.8 \times {10^{ - 16}}\,N$

D

$1.6 \times {10^{ - 18}}\,N$

Solution

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

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

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

Standard 12

Physics

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