Magnetic Field Between Two Parallel Current Carrying Wires

To find the magnetic field at the midpoint between two long parallel current-carrying wires, we use the formula for the magnetic field due to an infinitely long straight wire at a perpendicular distance $r$:

$B = \frac{\mu_0 I}{2\pi r} = \left(\frac{\mu_0}{4\pi}\right) \frac{2I}{r}$

Given Data:

• Current in each wire, $I = 30\text{ A}$
• Separation between the two wires, $d = 8\text{ cm} = 8 \times 10^{-2}\text{ m}$
• Distance of the midpoint from each wire, $r = \frac{d}{2} = 4\text{ cm} = 4 \times 10^{-2}\text{ m}$
• Constant $\frac{\mu_0}{4\pi} = 10^{-7}\text{ N/A}^2$

Direction of Magnetic Field: Since the currents in the two parallel wires flow in opposite directions, according to the Right-Hand Thumb Rule, the magnetic fields produced by both wires at the midpoint point in the same direction. Therefore, the net magnetic field $B_{\text{net}}$ is the sum of the magnitudes of the magnetic fields due to each wire:

$B_{\text{net}} = B_1 + B_2$

Since both wires carry the same magnitude of current and are at equal distances from the midpoint, $B_1 = B_2$:

$B_1 = \left(\frac{\mu_0}{4\pi}\right) \frac{2I}{r} = 10^{-7} \times \frac{2 \times 30}{4 \times 10^{-2}}\text{ T}$

$B_1 = 10^{-7} \times 1500 = 1.5 \times 10^{-4}\text{ T} = 150\,\mu\text{T}$

Now, calculating the net magnetic field at the midpoint:

$B_{\text{net}} = 2 \times B_1 = 2 \times 150\,\mu\text{T} = 300\,\mu\text{T}$

Thus, the magnitude of the magnetic field at the midpoint between the two wires is $300\,\mu\text{T}$.

Correct Option: B