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    $\left( Let \mu_{0}\epsilon_{0}=\frac{1}{9 \times 10^{16}} s^{2}/m^{2}\right)$

A) $1.0 \times 10^{-6}$

B) $1.5 \times 10^{-7}$

C) $3.0 \times 10^{-6}$

D) $2.5 \times 10^{-7}$


Option D


  Given, two charge particles having same charge, q move with same speed

 i.e,  $v_{1}=v_{2}=150 km/s =1.5 \times 10^{5}  m/s$

 Electric force between two moving charge  particle is same as they are static,

 i.e,    $|F_{2}|=\frac{1}{4 \pi \epsilon_{0}}.\frac{q^{2}}{r^{2}}$ ..........(i)

 Magnetic force between two moving charge particles is given by

$F_{1}=\frac{\mu_{0}}{4 \pi }.\frac{q_{1}.q_{2} v_{1} v_{2}}{r^{2}}$

                                                                         $[\because q_{1}=q_{2}=q$ and $v_{1}=v_{2}=v]$

 $|F_{1}|=\frac{\mu_{0}}{4 \pi}\frac{q^{2} v^{2}}{r^{2}}$.........(ii)

 From   Eq.(i) and (ii) , we get

$\therefore$    $\frac{|F_{1}|}{|F_{2}|}=\frac{\frac{\mu_{0}}{4 \pi}\frac{q^{2} v^{2}}{r^{2}}}{\frac{1}{4 \pi\epsilon_{0}}\frac{q^{2} }{r^{2}}}= \epsilon_{0}\mu_{0}v^{2}$

      $=\frac{1}{9 \times 10^{16}}\times (1.5 \times 10^{5})^{2}=2.5 \times 10^{-7}$

 Hence, $\frac{F_{1}}{F_{2}}$ is $2.5 \times 10^{-7}$