Physics · Moving Charges And Magnetism · NEET
At the centre, x = 0, so the denominator is 2(R²)^(3/2) = 2R³ and B is largest: B = μ₀I/2R. Move to any point on the axis (x not zero) and the denominator (R² + x²)^(3/2) grows, so B falls. The centre is the strongest point on the axis; every other axial point is weaker. This is why NEET options that make the axial field bigger than the centre value are always wrong.
For x >> R, ignore R² next to x² in the bracket: (R² + x²)^(3/2) ≈ (x²)^(3/2) = x³. So B ≈ μ₀ I R² / (2 x³). Writing area A = πR² and magnetic moment m = IA, this becomes B = μ₀ (2m) / (4π x³) = μ₀ · 2m / (4π x³). The loop behaves exactly like a magnetic dipole on its axis, field falling as 1/x³. This dipole link is a favourite NEET idea.
Take a small element I dl on the loop. By Biot-Savart, dB = (μ₀/4π) I dl / (R²+x²) and it is perpendicular to the line joining the element to the point. As you go around the ring, each dB has a component along the axis and a component perpendicular to the axis. By symmetry, every element has a diametrically opposite partner whose perpendicular component points the opposite way, so they cancel in pairs. Only the axial components add, all pointing the same way along the axis, giving the net field.
Curl the fingers of your right hand along the current direction in the loop; your thumb points along the axis in the direction of B. So on one side of the loop the field points away from it (like a north pole) and on the other side into it (south pole). The field is always parallel to the axis at axial points, never tilted.
Each element gives dB = (μ₀/4π) I dl /(R²+x²) since dl is perpendicular to the position vector r = √(R²+x²). Keep only the axial part: multiply by sinφ = R/√(R²+x²). Then B = (μ₀ I R /4π (R²+x²)^(3/2)) ∮ dl. The loop length ∮ dl = 2πR, giving B = μ₀ I R² / [2 (R² + x²)^(3/2)]. For N turns, multiply by N.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
B = μ₀ I R² / [2 (R² + x²)^(3/2)], where R is the loop radius, x is the distance of the axial point from the centre, and I is the current. For N turns, B = μ₀ N I R² / [2 (R² + x²)^(3/2)].
Put x = 0: B = μ₀ I R² / [2 R³] = μ₀ I / 2R. This is the standard centre-of-loop field, and it is the maximum value on the whole axis.
At the centre, x = 0. The field is strongest there and decreases smoothly as you move out along the axis in either direction.
For x >> R it varies as 1/x³: B ≈ μ₀ I R² / (2 x³) = μ₀ · 2m / (4π x³), where m = I·πR² is the magnetic moment. The loop then behaves like a magnetic dipole.
Yes. At every point on the axis the net field points along the axis (direction given by the right-hand rule). The perpendicular components of all current elements cancel by symmetry, leaving only the axial component.
