Question:hard

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 \(\mu_0 = 4 \pi \times 10^{-7} \, \text{SI unit})\).

Show Hint

Magnetic field due to perpendicular wires: compute fields individually, combine vectorially.
Updated On: Jul 18, 2026
  • \(1 \times 10^{-6} \, \text{T}\)
  • \(2 \times 10^{-6} \, \text{T}\)
  • \(1 \times 10^{-7} \, \text{T}\)
  • \(2 \times 10^{-7} \, \text{T}\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Notice both fields must lie along the same axis.
Both wires and the point $(2,4)$ sit in the same plane, so the field from each infinite wire, which circles the wire, points straight out of or into that plane, along the same axis rather than at an angle to each other. That means the two fields add or subtract along one line, they are never combined as a Pythagorean sum.
Step 2: Find the perpendicular distance from each wire.
For the wire along the x-axis, the perpendicular distance to the point is its y-coordinate, $r_1 = 4\ \text{m}$. For the wire along the y-axis, it is the x-coordinate, $r_2 = 2\ \text{m}$.
Step 3: Compute each field magnitude.
\[ B_1 = \frac{\mu_0 I_1}{2\pi r_1} = \frac{(4\pi\times10^{-7})(8)}{2\pi(4)} = 4\times10^{-7}\ \text{T} \]
\[ B_2 = \frac{\mu_0 I_2}{2\pi r_2} = \frac{(4\pi\times10^{-7})(6)}{2\pi(2)} = 6\times10^{-7}\ \text{T} \]
Step 4: Subtract, since the right-hand rule puts these two fields in opposite directions out of the plane.
\[ B_{\text{net}} = |B_2-B_1| = 6\times10^{-7}-4\times10^{-7} = 2\times10^{-7}\ \text{T} \]
\[ \boxed{2\times10^{-7}\ \text{T}} \]
Was this answer helpful?
0