Question

Two infinite length wires carry currents 8 A and 6 A respectively and are placed along X and Y axes respectively. Magnetic field at a point P (0,0,d) will be:

• A. \( \frac{7 \mu_0}{\pi d} \)
• B. \( \frac{10 \mu_0}{\pi d} \)
• C. \( \frac{14 \mu_0}{\pi d} \)
• D. \( \frac{5 \mu_0}{\pi d} \)

Hint:

For an infinitely long straight conductor, the magnetic field at a perpendicular distance \( d \) is given by \( B = \frac{\mu_0 I}{2\pi d} \). When two wires are perpendicular, use the Pythagorean Theorem to find the resultant field.

Answer 1

The Correct Option is D

Step 1: Magnetic Field Due to a Long Current-Carrying Wire
The magnetic field at a perpendicular distance \( d \) from an infinitely long straight conductor carrying current \( I \) is given by Ampere’s Law: \[ B = \frac{\mu_0 I}{2\pi d} \] where \( \mu_0 \) is the permeability of free space.
Step 2: Magnetic Fields at Point P Due to Both Wires
- The first wire (along the X-axis) carries current 8 A and contributes a magnetic field:
\[ B_1 = \frac{\mu_0 (8)}{2\pi d} \] - The second wire (along the Y-axis) carries current 6 A and contributes a magnetic field: \[ B_2 = \frac{\mu_0 (6)}{2\pi d} \] Step 3: Resultant Magnetic Field
Since the two magnetic fields are perpendicular to each other, the net magnetic field at P is: \[ B_{\text{net}} = \sqrt{B_1^2 + B_2^2} \] Substituting the values: \[ B_{\text{net}} = \sqrt{\left(\frac{\mu_0 (8)}{2\pi d}\right)^2 + \left(\frac{\mu_0 (6)}{2\pi d}\right)^2} \] \[ B_{\text{net}} = \frac{\mu_0}{2\pi d} \sqrt{8^2 + 6^2} = \frac{\mu_0}{2\pi d} \sqrt{64 + 36} \] \[ B_{\text{net}} = \frac{\mu_0}{2\pi d} \times \sqrt{100} = \frac{\mu_0}{2\pi d} \times 10 \] \[ B_{\text{net}} = \frac{10 \mu_0}{2\pi d} = \frac{5 \mu_0}{\pi d} \] Step 4: Conclusion
Thus, the correct answer is \( \frac{5 \mu_0}{\pi d} \).