The magnetic field at the centre of a wire loop formed by two semicircular wires of radii R₁=2 π m and R₂=4 π m, carrying current I=4 A as per figure given below is α × 10⁻⁷ T. The value of α is ________. (Centre O is common for all segments)

Solution

Physics sir · 4 steps

Both arcs circle O in the same sense, so their fields add. The straight pieces give nothing at O.

1. Straight segments. The four straight pieces lie on a line through O. For them dl⃗ is parallel to r⃗, so dl⃗ × r⃗ = 0 and their field at O is zero.
2. Direction of each arc. Trace the loop. Going over the top along R₂ from left to right is clockwise about O. Coming back from right to left under O along R₁ is also clockwise. So both semicircles give fields at O in the same direction.

   B = (μ₀ I)/4R₁ + (μ₀ I)/4R₂ = (μ₀ I)/4(1/R₁ + 1/R₂)

3. Put in numbers. (μ₀ I)/4 = (4π × 10⁻⁷ × 4)/4 = 4π × 10⁻⁷. The bracket is 1/2π + 1/4π = 3/4π m⁻¹. Multiplying gives B = 3 × 10⁻⁷ T.
4. So. α = 3.

The common slip: 1. Subtracting the two arc fields gives 4π × 10⁻⁷ × 1/4π = 1 × 10⁻⁷ T, so 1. Both arcs run clockwise about O, so their fields add.

The idea to keep. A semicircle gives (μ₀ I)/4R at its centre, and a straight wire pointing at the centre gives zero. Trace the current around the loop to decide whether the arc fields add or subtract.

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