The fractional change in the magnetic field intensity at a distance 'r' from centre on the axis of current carrying coil of radius 'a' to the magnetic field intensity at the centre of the same coil is : (Take r < a)

Solution

Physics sir · 4 steps

Factor the centre field out of the axial field, then expand the bracket binomially for small r/a.

1. Two fields. At the centre, B₀ = (μ₀ I)/2a. On the axis at distance r, B = (μ₀ I a²)/2(a² + r²)^(3/2).
2. Take the ratio. Take a³ out of the denominator. The axial field becomes the centre field times a correction factor.

   B/B₀ = (1 + r²/a²)^(-3/2) ≈ 1 - 3/2r²/a²

3. Binomial expansion. Since r is small compared with a, r²/a² is small, so (1+x)ⁿ ≈ 1 + nx with n = -3/2. Keep only that first term.
4. So. The fractional change is (B₀ - B)/B₀ = 3/2r²/a². The field drops by this fraction as you move along the axis. The answer is option (d).

The idea to keep. Whenever a small distance appears next to a big one, factor out the big one and use (1+x)ⁿ ≈ 1 + nx. The power n becomes the coefficient.

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