Question

Two parallel conductors carry current in opposite directions as shown in figure. One conductor carries a current of $10.0A$. Point $C$ is a distance $\frac{d}{2}$ to the right of the $10.0A$ current. If $d=18cm$ and $I$ is adjusted so that the magnetic field at $C$ is zero, the value of the current $I$ is

• A. 10.0 A
• B. 30.0A
• C. 8.0A
• D. 18.0A

Answer 1

Answer: (B)

The magnetic field at $C$ due to first conductor is
$B_1=\frac{{\mu }_0}{2\pi }\frac{I}{3d/2}$
(since, point $C$ is separated by
$d+\frac{d}{2}=\frac{3d}{2}$ from 1st conductor).
The direction of field is perpendicular to the plane of paper and directed outwards.
The magnetic field at $C$ due to second conductor is
$=B_2=\frac{{\mu }_0}{2\pi }\frac{10}{d/2}$
(since, point $C$ is separated by
$\frac{d}{2}$ from 2 nd conductor)
The direction of field is perpendicular to the plane of paper and directed inwards. Since, direction of $B_1$ and $B_2$ at point $C$ is in opposite direction and the magnetic field at $C$ is zero, therefore,
$B_1=B_2\frac{{\mu }_0}{2\pi }\frac{I}{3d/2}=\frac{{\mu }_0}{2\pi }\frac{10}{d/2}$
On solving $I=30.0A$