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(D) Given two particles with equal charge $q$ moving with speed $v = 150 \ km/s = 1.5 	imes 10^5 \ m/s$.The electric force $|F_2|$ between two charge

Vedclass · Accepted Answer

$2.5 	imes 10^{-7}$

Vedclass · Answer

(D) Given two particles with equal charge $q$ moving with speed $v = 150 \ km/s = 1.5 	imes 10^5 \ m/s$.The electric force $|F_2|$ between two charges at distance $r$ is given by Coulomb's law:$|F_2| = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{r^2}$ $(i)$The magnetic force $|F_1|$ between two moving charges is given by:$|F_1| = \frac{\mu_0}{4 \pi} \cdot \frac{q^2 v^2}{r^2}$ (ii)Dividing equation (ii) by equation $(i)$:$\frac{|F_1|}{|F_2|} = \frac{\frac{\mu_0}{4 \pi} \cdot \frac{q^2 v^2}{r^2}}{\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{r^2}} = \mu_0 \varepsilon_0 v^2$Substituting the given values $\mu_0 \varepsilon_0 = \frac{1}{9 	imes 10^{16}} \ s^2/m^2$ and $v = 1.5 	imes 10^5 \ m/s$:$\frac{|F_1|}{|F_2|} = \frac{1}{9 	imes 10^{16}} 	imes (1.5 	imes 10^5)^2 = \frac{2.25 	imes 10^{10}}{9 	imes 10^{16}} = 0.25 	imes 10^{-6} = 2.5 	imes 10^{-7}$.

List-$I$	List-$II$
$a$. Fleming's left-hand rule	$e$. Direction of induced current
$b$. Fleming's right-hand rule	$f$. South pole
$c$. Clockwise current	$g$. North pole
$d$. Anticlockwise current	$h$. Direction of force

Two particles carrying equal charges move parallel to each other with the speed $150 \ km/s$. If $F_1$ and $F_2$ are magnetic and electric forces between two charged particles, then $\frac{|F_1|}{|F_2|}$ is (Let $\mu_0 \varepsilon_0 = \frac{1}{9 \times 10^{16}} \ s^2/m^2$)

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List-$I$ List-$II$
$a$. Fleming's left-hand rule $e$. Direction of induced current
$b$. Fleming's right-hand rule $f$. South pole
$c$. Clockwise current $g$. North pole
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