MCQ
An $e^-$ is moving parallel to a long current carrying wire as shown. Force on electron is
- A$0.4 \times {10^{ - 18}}\,N$
- ✓$0.8 \times {10^{ - 18}}\,N$
- C$0.8 \times {10^{ - 16}}\,N$
- D$1.6 \times {10^{ - 18}}\,N$
✓
Answer
Correct option: B.
$0.8 \times {10^{ - 18}}\,N$
b
$\mathrm{F}_{\mathrm{m}}=\mathrm{qvB} \sin \theta=(\mathrm{e})(\mathrm{v})\left(\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}\right) \sin 90^{\circ}$
$\mathrm{F}_{\mathrm{m}}=\mathrm{qvB} \sin \theta=(\mathrm{e})(\mathrm{v})\left(\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}\right) \sin 90^{\circ}$
$\mathrm{F}_{\mathrm{m}}=1.6 \times 10^{-19} \times 10^{5}\left(\frac{2 \times 10^{-7} \times 10}{4 \times 10^{-2}}\right)$
$F_{\mathrm{m}}=0.8 \times 10^{-18}\, \mathrm{N}$
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