If two protons are moving with speed v = 4.5 | vedclass

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(A) The electrostatic force between two protons separated by distance $r$ is given by $F_{E} = \frac{1}{4\pi\epsilon_{0}} \frac{e^{2}}{r^{2}} = \frac{

Vedclass · Accepted Answer

$4.4 	imes 10^{5}$

Vedclass · Answer

(A) The electrostatic force between two protons separated by distance $r$ is given by $F_{E} = \frac{1}{4\pi\epsilon_{0}} \frac{e^{2}}{r^{2}} = \frac{k e^{2}}{r^{2}}$.The magnetic force between two parallel moving charges is given by $F_{M} = \frac{\mu_{0}}{4\pi} \frac{e^{2} v^{2}}{r^{2}}$.Taking the ratio of electrostatic force to magnetic force:$\frac{F_{E}}{F_{M}} = \frac{\frac{1}{4\pi\epsilon_{0}} \frac{e^{2}}{r^{2}}}{\frac{\mu_{0}}{4\pi} \frac{e^{2} v^{2}}{r^{2}}} = \frac{1}{\epsilon_{0} \mu_{0} v^{2}}$.We know that $c^{2} = \frac{1}{\epsilon_{0} \mu_{0}}$,where $c = 3 	imes 10^{8} \, m/s$ is the speed of light.Therefore,the ratio is $\frac{F_{E}}{F_{M}} = \frac{c^{2}}{v^{2}} = \left( \frac{c}{v} ight)^{2}$.Substituting the given values $c = 3 	imes 10^{8} \, m/s$ and $v = 4.5 	imes 10^{5} \, m/s$:$\frac{F_{E}}{F_{M}} = \left( \frac{3 	imes 10^{8}}{4.5 	imes 10^{5}} ight)^{2} = \left( \frac{3000}{4.5} ight)^{2} = \left( \frac{30000}{45} ight)^{2} = \left( \frac{2000}{3} ight)^{2} \approx (666.67)^{2} \approx 4.44 	imes 10^{5}$.

If two protons are moving with speed $v = 4.5 \times 10^{5} \, m/s$ parallel to each other,then the ratio of electrostatic and magnetic force between them is:

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