JEE Main 2026 · 28 January · morning · Single correct · +4 / −1

The magnetic field at the centre of a current carrying circular loop of radius R is 16μ T. The magnetic field at a distance x=√3R on its axis from the centre is
____ μT.

  1. (a) 8
  2. (b) 2
  3. (c) 4
  4. (d) 2√2
Answer and solution

Answer (b) 2

Solution

Physics sir · 3 steps

Do not redo Biot-Savart: take the ratio of the axial field to the centre field and most terms cancel.

  1. Two formulas. At the centre, B₀ = (μ₀ I)/2R. On the axis at distance x, B = (μ₀ I R²)/2(R² + x²)^(3/2). Dividing the second by the first cancels μ₀ and I.

    B/B₀ = R³/(R² + x²)^(3/2)

  2. Put in x. With x = √3R, R² + x² = R² + 3R² = 4R². Then (4R²)^(3/2) = 8R³, so B/B₀ = R³/8R³ = 1/8.
  3. So. B = 16/8 = 2 μT, which is option (b).

The idea to keep. For a point on a loop's axis, B/B₀ = (R/√(R²+x²))³. Find the distance from the point to the rim, take the ratio R to that distance, and cube it.

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

Study Moving Charges & Magnetism

Asked about 200 times in JEE Main since 2020. This question is from 2026.