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(C) The magnetic field $B$ due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.For a point $(x, y)$ in the first q

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$y=\left(\frac{I_1}{I_2}\right) x$

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(C) The magnetic field $B$ due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.For a point $(x, y)$ in the first quadrant,the magnetic field due to the wire along the $x$-axis (carrying current $I_1$) is $B_1 = \frac{\mu_0 I_1}{2 \pi y}$ (directed out of the plane).The magnetic field due to the wire along the $y$-axis (carrying current $I_2$) is $B_2 = \frac{\mu_0 I_2}{2 \pi x}$ (directed into the plane).For the net magnetic field to be zero,the magnitudes must be equal: $B_1 = B_2$.$\frac{\mu_0 I_1}{2 \pi y} = \frac{\mu_0 I_2}{2 \pi x}$.Simplifying this,we get $\frac{I_1}{y} = \frac{I_2}{x}$,which implies $y = \left(\frac{I_1}{I_2}\right) x$.

Two long straight wires are placed along the $x$-axis and $y$-axis. They carry currents $I_1$ and $I_2$ respectively. The equation of the locus of points with zero magnetic induction in the magnetic field produced by them is:

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