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(A) Let the current in both wires be $I$. The magnetic field due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.Fo

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$AB$

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(A) Let the current in both wires be $I$. The magnetic field due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.For any point on the line $AB$ (which makes an angle of $45^\circ$ with the axes),the perpendicular distance from the $X$-axis and the $Y$-axis is the same,say $r$.The magnetic field due to the wire along the $X$-axis at a point $(x, y)$ on $AB$ is directed into the plane (using the right-hand thumb rule),while the magnetic field due to the wire along the $Y$-axis is directed out of the plane.Since the distances are equal and the currents are equal,the magnitudes of the magnetic fields are equal: $B_X = B_Y = \frac{\mu_0 I}{2\pi r}$.Because the directions are opposite,the net magnetic field $B_{net} = B_X - B_Y = 0$ at every point on the line $AB$.

Two very thin metallic wires placed along $X$ and $Y$-axis carry equal currents as shown in the figure. $AB$ and $CD$ are lines at $45^\circ$ with the axes,with the origin of the axes at $O$. The magnetic field will be zero on the line:

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