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(A) Let the two wires be located at $x = -d$ and $x = +d$. Both carry current $I$ in the outward direction (positive $z$-axis). The magnetic field due

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(A) Let the two wires be located at $x = -d$ and $x = +d$. Both carry current $I$ in the outward direction (positive $z$-axis). The magnetic field due to a wire at distance $r$ is $B = \frac{\mu_0 I}{2\pi r}$.Using the right-hand rule, the magnetic field due to the left wire at a point $x$ is $B_1 = \frac{\mu_0 I}{2\pi (x + d)}$ (directed upwards for $x > -d$).The magnetic field due to the right wire at a point $x$ is $B_2 = -\frac{\mu_0 I}{2\pi (d - x)}$ (directed downwards for $x The net magnetic field is $B_{net} = B_1 + B_2 = \frac{\mu_0 I}{2\pi} \left( \frac{1}{x+d} - \frac{1}{d-x} \right)$.At the midpoint $x = 0$, $B_{net} = \frac{\mu_0 I}{2\pi} (\frac{1}{d} - \frac{1}{d}) = 0$.For $x For $x > d$, both fields are directed upwards, so $B_{net}$ is positive.Between the wires $(-d < x < d)$, the field changes from negative to positive, passing through zero at $x = 0$. This corresponds to the graph shown in the solution image.

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

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