Step 1: Understanding the Concept:
The magnetic field produced by a long straight wire forms concentric circles around the wire.
The net magnetic field at a point due to multiple wires is the vector sum of the individual magnetic fields produced by each wire.
Step 2: Key Formula or Approach:
The magnitude of the magnetic field due to a long straight wire at a distance \( r \) is \( B = \frac{\mu_0 I}{2\pi r} \).
We will use vector addition to find the resultant field at point \( P \).
Step 3: Detailed Explanation:
Let the two wires be at positions A and B. The distance between them is \( d = AB = 5\text{ cm} \).
Point \( P \) is equidistant from both wires, so \( AP = BP = r \).
We are given that the lines \( AP \) and \( BP \) are perpendicular to each other, forming a right-angled triangle \( \triangle APB \) with the right angle at \( P \).
Using Pythagoras theorem on \( \triangle APB \):
\[ AP^2 + BP^2 = AB^2 \]
\[ r^2 + r^2 = 5^2 \]
\[ 2r^2 = 25 \]
\[ r^2 = \frac{25}{2} \implies r = \frac{5}{\sqrt{2}}\text{ cm} = \frac{5}{\sqrt{2}} \times 10^{-2}\text{ m} \]
Let's find the magnetic fields at \( P \).
Magnetic field due to wire A (\( I_1 = 4\text{ A} \)): \( B_1 = \frac{\mu_0 I_1}{2\pi r} \).
The direction of \( \vec{B}_1 \) is perpendicular to the line \( AP \) and lies in the plane of the triangle.
Magnetic field due to wire B (\( I_2 = 3\text{ A} \)): \( B_2 = \frac{\mu_0 I_2}{2\pi r} \).
The direction of \( \vec{B}_2 \) is perpendicular to the line \( BP \).
Since \( AP \perp BP \), the magnetic field vectors \( \vec{B}_1 \) and \( \vec{B}_2 \) are also perpendicular to each other.
The magnitude of the resultant magnetic field \( B \) is:
\[ B = \sqrt{B_1^2 + B_2^2} = \sqrt{\left(\frac{\mu_0 I_1}{2\pi r}\right)^2 + \left(\frac{\mu_0 I_2}{2\pi r}\right)^2} \]
\[ B = \frac{\mu_0}{2\pi r} \sqrt{I_1^2 + I_2^2} \]
Substitute the known values:
\[ B = \frac{4\pi \times 10^{-7}}{2\pi \times \frac{5}{\sqrt{2}} \times 10^{-2}} \sqrt{4^2 + 3^2} \]
\[ B = \frac{2 \times 10^{-7}}{\frac{5}{\sqrt{2}} \times 10^{-2}} \times \sqrt{16 + 9} \]
\[ B = \frac{2\sqrt{2} \times 10^{-5}}{5} \times \sqrt{25} \]
\[ B = \frac{2\sqrt{2} \times 10^{-5}}{5} \times 5 \]
\[ B = 2\sqrt{2} \times 10^{-5}\text{ T} \]
The direction of currents being opposite doesn't change the fact that the two field vectors are orthogonal in this specific geometry.
Step 4: Final Answer:
The magnitude of magnetic field at point \( P \) is \( 2\sqrt{2} \times 10^{-5}\text{ T} \).