The magnetic field at the centre of a circular coil of radius r, due to current I flowing through it, is B. The magnetic field at a point along the axis at a distance r/2 from the centre is :

Solution

Physics sir · 4 steps

Do not redo the integral: divide the axial formula by the centre formula and only a cube is left.

1. Two formulas. At the centre, B = (μ₀ I)/2r. At a distance x along the axis, Bₓ = (μ₀ I r²)/2(r² + x²)^(3/2).
2. Take the ratio. Divide the second by the first. The factor (μ₀ I)/2 cancels, and the leftover powers of r combine into a single cube.

   Bₓ/B = r³/(r² + x²)^(3/2) = (r/√(r² + x²))³

3. Substitute the distance. With x = r/2, r² + r²/4 = 5r²/4, so √(r² + x²) = √5/2r. The bracket becomes 2/√5.
4. So. Bₓ = (2/√5)³ B ≈ 0.72B, which is option (c).

The idea to keep. For a circular loop, Bₐₓᵢₛ/B_(centre) = (r/√(r²+x²))³. The field is strongest at the centre, so this ratio is always less than 1.

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