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

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Question

(b)

Magnetic field at $A$ due to $B$

$B=\frac{e v \sin 90^{\circ}}{d^2} 	imes \frac{\mu_0}{4 \pi}$

$B=\frac{e v}{d^2} 	imes \frac{\mu_0}{4

Vedclass · Accepted Answer

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

Vedclass · Answer

(b)

Magnetic field at $A$ due to $B$

$B=\frac{e v \sin 90^{\circ}}{d^2} 	imes \frac{\mu_0}{4 \pi}$

$B=\frac{e v}{d^2} 	imes \frac{\mu_0}{4 \pi}-(i)$

Magnetic force on $A$ is

$F_B=e v B$

From $(i)$, $F_B=e v \frac{(e v)}{d^2} 	imes \frac{\mu_0}{4 \pi}$

$F_e=e t$

$=\frac{e Ke }{ d ^2}$

$\frac{F_B}{F_e}=\frac{e^2 v^2}{d^2} 	imes \frac{d^2}{e^2 k} 	imes \frac{\mu_0}{4 \pi}$

$=\frac{v^2}{\frac{1}{4 \pi \varepsilon_0} 	imes 4 \pi}$

$=\frac{v^2}{\frac{1}{\mu_0} \varepsilon_0}$

$\frac{F_B}{F_e}=\frac{v^2}{c^2}$

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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