Question

Two long straight parallel wires are a distance $2 d$ apart. They carry steady equal currents flowing out of the plane of the paper. The variation of magnetic field $B$ along the line $xx'$ is given by

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

Answer 2

The magnetic field due to a long straight current carrying wire is given by, $B =\left(\mu_{0} I \right) / 2 \pi r$ or, $B \propto 1 / r$ At the point exactly mid-way between the conductors, the net magnetic field is zero. Using right hand thumb rule, we find that the magnetic field due to left wire will be in $\hat{j}$ direction while due to the right wire is in $(-\hat{j})$ direction. Magnetic field at a distance $x$ from the left wire, lying between the wires. $B=\frac{\mu_{0} I}{2 \pi x} \hat{j}+\frac{\mu_{0} I}{2 \pi(2 d-x)}(-\hat{j})$ or, $B=\frac{\mu_{0} I}{2 \pi}\left(\frac{1}{x}-\frac{1}{2 d-x}\right)$ At $x=d, B=0$ For $x < d, B$ is along $\hat{j}$ For $x > d, B$ is along $-\hat{j}$ On the left side of the left conductor, magnetic fields due to the currents will add up and the net magnetic field will be along $(-\hat{j})$ direction. To the right side of second conductor, the total magnetic field will be along $\hat{j}$ direction.