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STD 12 - 4. Moving charges and magnetism — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 4. Moving charges and magnetism4 Marks
MCQ
Two protons move parallel to each other, keeping distance $r$ between them, both moving with same  velocity $\vec V\,$. Then the ratio of the electric and magnetic force of interaction between them is
  • ✓
    $c^2 / v^2$
  • B
    $2c^2 / v^2$
  • C
    $c^2 / 2v^2$
  • D
    None
✓

Answer

Correct option: A.
$c^2 / v^2$
a
Electric force $\mathrm{Fe}=\mathrm{kq}_{1} \mathrm{q}_{2} / \mathrm{r}^{2}=\mathrm{kq}^{2} / \mathrm{r}^{2}$

Magnetic force $=\mathrm{qv}\left(\frac{\mu_{0}}{4 \pi} \times \frac{\mathrm{qv}}{r^{2}}\right)$

$\mathrm{F}_{\mathrm{m}}=\mathrm{q}^{2} \mathrm{v}^{2}\left(\frac{\mu_{0}}{4 \pi}\right) \frac{1}{\mathrm{r}^{2}}$

$\frac{F e}{F_{m}}=\frac{k}{v^{2}\left(\frac{\mu_{0}}{4 \pi}\right)}=\frac{1}{v^{2} \varepsilon_{0} \mu_{0}}=\frac{c^{2}}{v^{2}}$

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