Question

Ab electron moving with a velocity of $4\times {10}^{-5}m$ $s^{-1}$ along the positive x direction experiences a force of magnitude $3.2\times {10}^{-20}N$ at the point $R$. Find the value of $I$

Answer 1

Magnetic field at $R$ due to both wires $P$ and $Q$ will be downward as shown in figure. Therefore, net field at $R$ will be sum of these two.
$B=B_P+B_Q$

$=\frac{{\mu }_0}{2\pi }\frac{I_P}{5}+\frac{{\mu }_0}{2\pi }\frac{I_Q}{2}=\frac{{\mu }_0}{2\pi }(\frac{2.5}{5}+\frac{I}{2})$
$=\frac{{\mu }_0}{4\pi }(I+1)={10}^{-7}(I+1)$

Net force on the electron:

$F=Bqv\sin {90}^o$
$(3.2\times {10}^{-20})=\left({10}^{-7}\right)(I+1)(1.6\times {10}^{-19})(4\times {10}^5)$
$I+1=5$

$\therefore I=4A$