There are two infinitely long straight current carrying conductors and they are held at right angles to each other so that their common ends meet at the origin as shown in the figure given below. The ratio of current in both conductor is 1 : 1. The magnetic field at point P is ____________.

Solution

Physics sir · 4 steps

Treat each arm as a half-infinite wire, check both fields point into the page, then add them.

1. Finite wire formula. A straight wire at perpendicular distance d gives B = (μ₀ I)/(4π d)(sin α₁ + sin α₂). Here α₁ and α₂ are the angles the two ends make with the perpendicular from P. An end at infinity gives sin 90° = 1. Let r = √(x²+y²) be the distance OP.
2. Wire along x-axis. P is a distance y above it. The end at O makes angle θ₁ with the vertical, so sin θ₁ = x/r. This gives B₁ = (μ₀ I)/(4π y)(1 + x/r). The current flows in the -x direction with P above it, so the right-hand rule puts B₁ into the page.
3. Wire along y-axis. P is a distance x from it. The end at O makes angle θ₂ with the horizontal, so sin θ₂ = y/r. This gives B₂ = (μ₀ I)/(4π x)(1 + y/r). The current flows in the +y direction with P to its right, so B₂ is also into the page.

   B = (μ₀ I)/4π[1/y + x/yr + 1/x + y/xr] = (μ₀ I)/(4π xy)[(x+y) + (x²+y²)/r]

4. So. Since (x²+y²)/r = r = √(x²+y²), the total is B = (μ₀ I)/(4π xy)[√(x²+y²) + (x+y)], which is option (a).

The idea to keep. For bent wires, split the wire into straight pieces and use (μ₀ I)/(4π d)(sin α₁+sin α₂) for each. Fix every direction with the right-hand rule before you add or subtract.

Answer from the NTA final answer key. Solution by PrepWiser.