Question

PQ is a long straight conductor carrying a current of $3$A as shown in Figure. An electron moves with a velocity of $2\times {10}^{7}m{s}^{-1}$ parallel to it. Find the force acting on the electron.

Answer 1

Magnetic field at a perpendicular distance $d$ due to long straight conductor is given by $B=\frac{{\mu }_{o}I}{2\pi d}$

Given : $I=3A$, $d=0.6m$

We know ${\mu }_o=4\pi \times {10}^{-7}$ $Tm{A}^{-1}$

$⟹$ $B=\frac{2\times {10}^{-7}\times 3}{\left(0.6\right)}={10}^{-6}T$

Speed of the electron $v=2\times {10}^{7}m/s$

$\therefore$ Force acting on the electron $|F|=|q|vB$

$\therefore$ Force acting on the electron $|F|=(1.6\times {10}^{-19})(2\times {10}^7)\left({10}^{-6}\right)=3.2\times {10}^{-18}N$