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