Physics - Magnetism Question with Solution | TestHub

Given, speed of electrons, $v=5\times {10}^5 \mathrm{m} {\mathrm{s}}^{-1}$

Electrostatic force between the electrons,

$F_e=k·\frac{e^2}{r^2} ...\left(\mathrm{i}\right)$

Magnetic field produced due to moving electron with velocity $v$,

$B=\frac{{\mu }_0}{4\pi }·\frac{ev}{r^2} ...\left(\mathrm{ii}\right)$

Magnetic force between the electrons,

$F_m=Bev=\frac{{\mu }_0}{4\pi }·\frac{ev}{r^2}·ev$   (from Eq. (ii))

$F_m=\frac{{\mu }_0}{4\pi }·\frac{e^2{v}^2}{r^2}$

$\therefore$ From Eqs. (i) and (iii), we have

$\frac{F_e}{F_m}=\frac{k{e}^2/r^2}{\frac{{\mu }_0{e}^2{v}^2}{4\pi ·r^2}}=\frac{k·4\pi }{{\mu }_0{v}^2}$

$=\frac{9\times {10}^9\times 4\times 3.14}{4\pi \times {10}^{-7}{\left(5\times {10}^5\right)}^2}=3.6\times {10}^5$