Question:
Two long wires carrying currents \(8 \, \text{A}\) along x-axis and \(6 \, \text{A}\) along y-axis. Find the magnetic field at the point \((2\hat{i} + 4\hat{j})\). (Take \(\mu_0 = 4 \pi \times 10^{-7} \, \text{SI unit})\).
Two long wires carrying currents \(8 \, \text{A}\) along x-axis and \(6 \, \text{A}\) along y-axis. Find the magnetic field at the point \((2\hat{i} + 4\hat{j})\). (Take \(\mu_0 = 4 \pi \times 10^{-7} \, \text{SI unit})\).
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Magnetic field due to perpendicular wires: compute fields individually, combine vectorially.
Updated On: Jul 18, 2026
- \(1 \times 10^{-6} \, \text{T}\)
- \(2 \times 10^{-6} \, \text{T}\)
- \(1 \times 10^{-7} \, \text{T}\)
- \(2 \times 10^{-7} \, \text{T}\)
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The Correct Option is D
Solution and Explanation
Step 1: Recall magnetic field due to long straight wire.
\[ B = \frac{\mu_0 I}{2 \pi r} \]
Direction given by right-hand rule.
Step 2: Identify distances.
Point coordinates \((2, 4)\), wire along x-axis (y = 0), distance \(r_y = 4\) m. Wire along y-axis (x = 0), distance \(r_x = 2\) m.
Step 3: Compute magnetic fields.
\[ B_x = \frac{\mu_0 I_x}{2 \pi r_x} = \frac{4 \pi \times 10^{-7} \cdot 8}{2 \pi \cdot 2} = 8 \times 10^{-7} \, \text{T} \]
\[ B_y = \frac{\mu_0 I_y}{2 \pi r_y} = \frac{4 \pi \times 10^{-7} \cdot 6}{2 \pi \cdot 4} = 3 \times 10^{-7} \, \text{T} \]
Step 4: Combine perpendicular components.
\[ B = \sqrt{B_x^2 + B_y^2} = \sqrt{(8 \times 10^{-7})^2 + (3 \times 10^{-7})^2} \approx 2 \times 10^{-7} \, \text{T} \]
Step 5: Check direction.
Use right-hand rule, components perpendicular, magnitude as computed.
Step 6: Final conclusion.
Hence, the magnetic field at the point is:
\[ \boxed{2 \times 10^{-7} \, \text{T}} \]
\[ B = \frac{\mu_0 I}{2 \pi r} \]
Direction given by right-hand rule.
Step 2: Identify distances.
Point coordinates \((2, 4)\), wire along x-axis (y = 0), distance \(r_y = 4\) m. Wire along y-axis (x = 0), distance \(r_x = 2\) m.
Step 3: Compute magnetic fields.
\[ B_x = \frac{\mu_0 I_x}{2 \pi r_x} = \frac{4 \pi \times 10^{-7} \cdot 8}{2 \pi \cdot 2} = 8 \times 10^{-7} \, \text{T} \]
\[ B_y = \frac{\mu_0 I_y}{2 \pi r_y} = \frac{4 \pi \times 10^{-7} \cdot 6}{2 \pi \cdot 4} = 3 \times 10^{-7} \, \text{T} \]
Step 4: Combine perpendicular components.
\[ B = \sqrt{B_x^2 + B_y^2} = \sqrt{(8 \times 10^{-7})^2 + (3 \times 10^{-7})^2} \approx 2 \times 10^{-7} \, \text{T} \]
Step 5: Check direction.
Use right-hand rule, components perpendicular, magnitude as computed.
Step 6: Final conclusion.
Hence, the magnetic field at the point is:
\[ \boxed{2 \times 10^{-7} \, \text{T}} \]
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