A loop ABCDA , carrying current I=12 A, is placed in a plane, consists of two semi-circular segments of radius R₁=6 π m and R₂=4 π m. The magnitude of the resultant magnetic field at center O is k × 10⁻⁷ T. The value of k is_________.
( Given μ₀=4 π × 10⁻⁷ Tm A⁻¹ )

Solution

Physics sir · 4 steps

The two arcs carry current in opposite senses around O, so their fields subtract, not add.

1. Straight parts. AB and CD lie on a line that passes through O. For every element there, dl⃗ is parallel to r⃗, so dl⃗ × r⃗ = 0 and they give no field at O.

   B_(semicircle) = (μ₀ I)/4R

2. Each arc. Outer arc: B₁ = (4π × 10⁻⁷ × 12)/(4 × 6π) = 2 × 10⁻⁷ T. Inner arc: B₂ = (4π × 10⁻⁷ × 12)/(4 × 4π) = 3 × 10⁻⁷ T.
3. Directions. Current goes from D over the top to A on the outer arc, which is anticlockwise about O, so B₁ points out of the page. On the inner arc it goes from B over the top to C, which is clockwise, so B₂ points into the page.
4. So. B = B₂ - B₁ = 3 × 10⁻⁷ - 2 × 10⁻⁷ = 1 × 10⁻⁷ T, into the page. Hence k = 1.

The common slip: 5. Adding the two arc fields instead of subtracting them gives 2 + 3 = 5. The arcs carry current in opposite senses around O.

The idea to keep. For arcs around a common centre, find each field with (μ₀ I θ)/(4π R), then check the sense of the current around the centre. Opposite senses subtract. Straight wires through the centre add nothing.

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