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(D) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.Given the distan

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$\frac{5 \mu_0}{2 \pi}$

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(D) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.Given the distance between the wires is $d = 3 \ m$,the midpoint is at a distance $r = 1.5 \ m$ from each wire.For the first wire,$I_1 = 3 \ A$. The magnetic field is $B_1 = \frac{\mu_0 	imes 3}{2 \pi 	imes 1.5} = \frac{2 \mu_0}{2 \pi}$.For the second wire,$I_2 = 4.5 \ A$. The magnetic field is $B_2 = \frac{\mu_0 	imes 4.5}{2 \pi 	imes 1.5} = \frac{3 \mu_0}{2 \pi}$.Since the currents are in opposite directions,by the right-hand rule,the magnetic fields produced by both wires at the midpoint point in the same direction.Therefore,the net magnetic field is $B = B_1 + B_2 = \frac{2 \mu_0}{2 \pi} + \frac{3 \mu_0}{2 \pi} = \frac{5 \mu_0}{2 \pi}$.

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