Question

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

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

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, $I_1=10A,r_1=5,r_1+d=5+10=15cm$
$\therefore I_2=\frac{I_1}{r_1}\times \left(r_1+d\right)$
$I_2=\frac{10}{5}\times 15=30A$