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(C) The magnetic field $B$ produced by a long straight current-carrying wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.Since the curr

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$\frac{\mu_0(I_1+I_2)}{\pi d}$

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(C) The magnetic field $B$ produced by a long straight current-carrying wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.Since the currents $I_1$ and $I_2$ are in opposite directions,by the Right-Hand Thumb Rule,the magnetic fields produced by both wires at the midpoint will point in the same direction.Therefore,the resultant magnetic field is the sum of the individual fields: $B_{net} = B_1 + B_2$.The distance of the midpoint from each wire is $r = \frac{d}{2}$.Substituting the values: $B_{net} = \frac{\mu_0 I_1}{2\pi(d/2)} + \frac{\mu_0 I_2}{2\pi(d/2)}$.Simplifying the expression: $B_{net} = \frac{\mu_0 I_1}{\pi d} + \frac{\mu_0 I_2}{\pi d} = \frac{\mu_0(I_1+I_2)}{\pi d}$.

Two parallel conducting wires of equal length are placed distance $d$ apart and carry currents $I_1$ and $I_2$ respectively in opposite directions. The resultant magnetic field at the midpoint of the distance between both the wires is

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