Question

Two long parallel wires carrying currents \(4\text{ A}\) and \(3\text{ A}\) in opposite directions are placed at a distance of \(5\text{ cm}\) from each other. A point \(P\) is at equidistance from both the wires such that the line joining the point \(P\) to the wires are perpendicular to each other. The magnitude of magnetic field at point \(P\) is ( \(\mu_0 = 4\pi \times 10^{-7}\) SI unit )

• A. \(4 \times 10^{-5}\text{ T}\)
• B. \(\sqrt{2} \times 10^{-5}\text{ T}\)
• C. \(2 \times 10^{-5}\text{ T}\)
• D. \(2\sqrt{2} \times 10^{-5}\text{ T}\)

Hint:

If magnetic fields are perpendicular → use Pythagoras.

Answer 1

The Correct Option is D

Concept:
Magnetic field due to long straight wire: \[ B = \frac{\mu_0 I}{2\pi r} \] Step 1: Find distance of point P. Since geometry is perpendicular and equidistant: \[ r = \frac{5}{\sqrt{2}} \text{ cm} = \frac{5}{\sqrt{2}} \times 10^{-2}\text{ m} \]
Step 2: Calculate fields. \[ B_1 = \frac{\mu_0 \cdot 4}{2\pi r}, \quad B_2 = \frac{\mu_0 \cdot 3}{2\pi r} \]
Step 3: Resultant field. Fields are perpendicular: \[ B = \sqrt{B_1^2 + B_2^2} \] \[ = \frac{\mu_0}{2\pi r} \sqrt{4^2 + 3^2} = \frac{\mu_0}{2\pi r} \cdot 5 \]
Step 4: Substitute values. \[ B = \frac{4\pi \times 10^{-7}}{2\pi \cdot \frac{5}{\sqrt{2}} \times 10^{-2}} \cdot 5 \] \[ = \frac{2 \times 10^{-7} \cdot 5}{\frac{5}{\sqrt{2}} \times 10^{-2}} = 2\sqrt{2} \times 10^{-5} \]
Step 5: Conclusion. \[ B = 2\sqrt{2} \times 10^{-5}\text{ T} \]