A current of 30 A each flows in opposite directions in two conducting wires, placed parallel to each other at a distance of 8 cm. The magnetic field at the mid point between the two wires is __________ $$\mu$$T. ($$\dfrac{\mu_0}{4\pi} = 10^{-7}$$ N/A$$^2$$)

Distance of the midpoint from each wire, $$ r = \frac{d}{2} = 4 \text{ cm} = 0.04 \text{ m} $$

Magnetic constant, $$ \frac{\mu_0}{4\pi} = 10^{-7} \text{ T}\cdot\text{m/A} \implies \frac{\mu_0}{2\pi} = 2 \times 10^{-7} \text{ T}\cdot\text{m/A} $$

The magnitude of the magnetic field produced by a long straight current-carrying wire at a distance $$ r $$ is given by:

$$ B = \frac{\mu_0 I}{2\pi r} $$

Using Right-Hand Grip Rule,as currents in the two wires are flowing in opposite directions, the magnetic fields produced by both wires at the midpoint between them will point in the \textbf{same direction}.

Therefore, the net magnetic field ($$ B_{net} $$) at the midpoint is the sum of the magnitudes of the magnetic fields produced by each wire:

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

$$ B_{net} = \frac{\mu_0 I}{2\pi r} + \frac{\mu_0 I}{2\pi r} $$

$$ B_{net} = 2 \left( \frac{\mu_0 I}{2\pi r} \right) = \frac{\mu_0 I}{\pi r} $$

$$ B_{net} = \frac{4\pi \times 10^{-7} \times 30}{\pi \times 0.04} $$

$$ B_{net} = \frac{4 \times 10^{-7} \times 30}{0.04} $$

$$ B_{net} = \frac{120 \times 10^{-7}}{4 \times 10^{-2}} $$

$$ B_{net} = 30 \times 10^{-5} \text{ T} $$

$$ B_{net} = 30 \times 10^{-5} \times 10^6 \ \mu\text{T} $$

$$ B_{net} = 30 \times 10^1 \ \mu\text{T} $$

$$ B_{net} = 300 \ \mu\text{T} $$