Question

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

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

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.
$B=\frac{{\mu }_{0}I}{2\pi x}\hat{j}+\frac{{\mu }_{0}I}{2\pi (2d-x)}(-\hat{j})$
or, $B=\frac{{\mu }_{0}I}{2\pi }\left(\frac{1}{x}-\frac{1}{2d-x}\right)$
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.