1. Magnetic field due to one long straight conductor

   For a long straight wire carrying current $I$, the magnetic field at distance $r$ is

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

2. Given data

   • Distance between wires $= 8\text{ cm} = 0.08\text{ m}$
   • Midpoint is at distance $$r = \frac{0.08}{2} = 0.04\text{ m}$$
   • Resultant magnetic field at midpoint $$B_{\text{net}} = 300\,\mu T = 300 \times 10^{-6}\text{ T}$$

3. Direction of magnetic fields at midpoint

   • If currents are in the same direction, the magnetic fields at the midpoint due to the two wires are in opposite directions, so they cancel.
   • If currents are in opposite directions, the magnetic fields at the midpoint are in the same direction, so they add.

   Since the net field is nonzero, the currents must be in opposite directions.

4. Field due to one wire at midpoint

   $$B_1 = \frac{\mu_0 I}{2\pi r}$$

   Since the two fields add,

   $$B_{\text{net}} = 2B_1 = 2\cdot \frac{\mu_0 I}{2\pi r}$$

   So,

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

5. Substitute values

   Using $\mu_0 = 4\pi \times 10^{-7}$,

   $$300 \times 10^{-6} = \frac{4\pi \times 10^{-7} \cdot I}{\pi \cdot 0.04}$$

   Cancel $\pi$:

   $$300 \times 10^{-6} = \frac{4 \times 10^{-7} I}{0.04}$$ $$300 \times 10^{-6} = 100 \times 10^{-7} I = 10^{-5} I$$

   Therefore,

   $$I = \frac{300 \times 10^{-6}}{10^{-5}} = 30\text{ A}$$

6. Check options

   • A: $30$ A in same direction $\rightarrow$ field at midpoint would cancel, wrong.
   • B: $30$ A in opposite direction $\rightarrow$ correct magnitude and correct direction condition.
   • C: $60$ A in opposite direction $\rightarrow$ gives $600\,\mu T$, wrong.
   • D: $300$ A in opposite direction $\rightarrow$ much too large, wrong.

Final answer: B. $30$ A in the opposite direction