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(B) The distance between the two wires is $d = 3 	ext{ m}$. The midpoint is at a distance $r = d/2 = 1.5 	ext{ m}$ from each wire.For a long straigh

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

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(B) The distance between the two wires is $d = 3 	ext{ m}$. The midpoint is at a distance $r = d/2 = 1.5 	ext{ m}$ from each wire.For a long straight wire, the magnetic field is given by $B = \frac{\mu_0 I}{2\pi r}$.For the first wire $(I_1 = 3 	ext{ A})$, the magnetic field at the midpoint is $B_1 = \frac{\mu_0 	imes 3}{2\pi 	imes 1.5} = \frac{3\mu_0}{3\pi} = \frac{\mu_0}{\pi} = \frac{2\mu_0}{2\pi}$.For the second wire $(I_2 = 4.5 	ext{ A})$, the magnetic field at the midpoint is $B_2 = \frac{\mu_0 	imes 4.5}{2\pi 	imes 1.5} = \frac{4.5\mu_0}{3\pi} = \frac{1.5\mu_0}{\pi} = \frac{3\mu_0}{2\pi}$.Since the currents are in opposite directions, the magnetic fields at the midpoint due to both wires will be in the same direction (by the Right-Hand Thumb Rule).Therefore, the resultant magnetic field is $B_{net} = B_1 + B_2 = \frac{2\mu_0}{2\pi} + \frac{3\mu_0}{2\pi} = \frac{5\mu_0}{2\pi}$.

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