Two protons A and B move parallel to the x-ax | vedclass

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(B) The magnetic field $B$ at the position of proton $A$ due to the moving proton $B$ is given by the Biot-Savart law for a point charge:$B = \frac{\m

Vedclass · Accepted Answer

$\frac{v^2}{c^2}$

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(B) The magnetic field $B$ at the position of proton $A$ due to the moving proton $B$ is given by the Biot-Savart law for a point charge:$B = \frac{\mu_0}{4\pi} \frac{e v \sin 90^{\circ}}{d^2} = \frac{\mu_0}{4\pi} \frac{e v}{d^2}$The magnetic force $F_B$ acting on proton $A$ is:$F_B = e v B = e v \left( \frac{\mu_0}{4\pi} \frac{e v}{d^2} \right) = \frac{\mu_0 e^2 v^2}{4\pi d^2}$The electric force $F_e$ acting on proton $A$ due to proton $B$ is given by Coulomb's law:$F_e = \frac{1}{4\pi \varepsilon_0} \frac{e^2}{d^2}$The ratio of magnetic force to electric force is:$\frac{F_B}{F_e} = \frac{\frac{\mu_0 e^2 v^2}{4\pi d^2}}{\frac{e^2}{4\pi \varepsilon_0 d^2}} = \mu_0 \varepsilon_0 v^2$Since the speed of light $c$ is given by $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$,we have $c^2 = \frac{1}{\mu_0 \varepsilon_0}$,or $\mu_0 \varepsilon_0 = \frac{1}{c^2}$.Substituting this into the ratio:$\frac{F_B}{F_e} = \frac{v^2}{c^2}$

Two protons $A$ and $B$ move parallel to the $x$-axis in opposite directions with equal speeds $v$. At the instant shown,the ratio of magnetic force and electric force acting on the proton $A$ is ($c=$ speed of light in vacuum).

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