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Field on Axis of Circular Loop

Field at axial distance x from centre of a loop of radius R. At x = 0 reduces to μ₀I/2R. For x ≫ R falls as μ₀IR²/2x³ — the magnetic dipole field.
Class 12
Derivation

Setup

A circular loop of radius RR carries current II. Point P is on the axis at distance xx from the centre.

Geometry

For any element I dl⃗I\,d\vec{l} on the loop:

  • Distance to P: r=R2+x2r = \sqrt{R^2 + x^2} (constant for all elements).
  • Angle between dl⃗d\vec{l} (tangential) and r^\hat{r}: always 90°90°.
dB=μ0I4πdlR2+x2dB = \frac{\mu_0 I}{4\pi} \frac{dl}{R^2 + x^2}

Resolving Components

dB⃗d\vec{B} makes angle α\alpha with the axis, where sin⁡α=R/R2+x2\sin\alpha = R/\sqrt{R^2+x^2}.

By symmetry, the perpendicular components from diametrically opposite elements cancel. Only axial components survive:

dBx=dB⋅RR2+x2dB_x = dB \cdot \frac{R}{\sqrt{R^2 + x^2}}

Integration

B=μ0I4π(R2+x2)⋅RR2+x2⋅2πRB = \frac{\mu_0 I}{4\pi(R^2+x^2)} \cdot \frac{R}{\sqrt{R^2+x^2}} \cdot 2\pi R B=μ0IR22 (R2+x2)3/2\boxed{B = \frac{\mu_0 I R^2}{2\,(R^2 + x^2)^{3/2}}}

Special Cases

At the centre (x=0x = 0):

B=μ0I2RB = \frac{\mu_0 I}{2R}

Far field (x≫Rx \gg R):

B≈μ0IR22x3=μ04π2mx3,m=IπR2B \approx \frac{\mu_0 I R^2}{2x^3} = \frac{\mu_0}{4\pi}\frac{2m}{x^3}, \quad m = I\pi R^2

The loop behaves as a magnetic dipole at large distances.

Note
The field is maximum at $x = 0$ and decreases monotonically along the axis. At $x = R/2$, $B = (8/5\sqrt{5})\cdot\mu_0 I/2R$ — a commonly asked result.