Two protons move parallel to each other,keepi | vedclass

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Question

(A) The electric force $F_e$ between two protons separated by distance $r$ is given by Coulomb's Law: $F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}$

Vedclass · Accepted Answer

$c^2 / v^2$

Vedclass · Answer

(A) The electric force $F_e$ between two protons separated by distance $r$ is given by Coulomb's Law: $F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}$.The magnetic force $F_m$ between two parallel moving charges is given by $F_m = qvB$,where $B$ is the magnetic field produced by one proton at the location of the other: $B = \frac{\mu_0}{4\pi} \frac{qv}{r^2}$.Thus,$F_m = qv \left( \frac{\mu_0}{4\pi} \frac{qv}{r^2} \right) = \frac{\mu_0}{4\pi} \frac{q^2 v^2}{r^2}$.Taking the ratio of electric force to magnetic force:$\frac{F_e}{F_m} = \frac{\frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}}{\frac{\mu_0}{4\pi} \frac{q^2 v^2}{r^2}} = \frac{1}{\epsilon_0 \mu_0 v^2}$.Since the speed of light $c$ is defined as $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$,we have $c^2 = \frac{1}{\mu_0 \epsilon_0}$.Substituting this into the ratio,we get $\frac{F_e}{F_m} = \frac{c^2}{v^2}$.

Two protons move parallel to each other,keeping a distance $r$ between them,both moving with the same velocity $\vec{v}$. Then the ratio of the electric force to the magnetic force of interaction between them is:

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