Question

Two long parallel conductors carry currents in opposite directions as shown. One conductor carries a current of $ 10 \,A $ and the distance between the wires is $ d = 10\, cm $ . Current $ I $ is adjusted, so that the magnetic field at $ P $ is zero. $ P $ is at a distance of $ 5\, cm $ to the right of the $ 10\, A $ current. Value of $ I $ is :

• A. $ 40\,A $
• B. $ 30\,A $
• C. $ 20\,A $
• D. $ 10\,A $

Answer 1

Answer: (B)

From Biot-Savart's law the magnetic field $(B)$ due to a conductor carrying current $I$, at a distance $r_{1}$ is

Magnetic field at $P$ due to current in second conductor is
$B_{2}=\frac{\mu_{0} I_{2}}{2 \pi\left(r_{1}+d\right)}$
From Fleming's right hands rule the fields at $P$ are directed opposite.
$\therefore $ Resultants, field $B_{1}=B_{2}$
$\therefore \, \frac{\mu_{0} I_{1}}{2 \pi r_{1}}=\frac{\mu_{0} I_{2}}{2 \pi\left(r_{1}+d\right)}$
Given, $I_{1}=10\, A, r_{1}=5, \,r_{1}+d=5+10=15 \,cm$
$\therefore \, I_{2}=\frac{I_{1}}{r_{1}} \times\left(r_{1}+d\right)$
$I_{2}=\frac{10}{5} \times 15=30 \,A$