Field at Centre of Circular Arc

Field at the centre of a circular arc of radius R subtending angle θ (radians) at the centre. For a complete loop (θ = 2π): B = μ₀I/2R.

Setup

A circular arc of radius $R$ carries current $I$ and subtends angle $\theta$ (radians) at centre O.

Derivation

For every element $d\vec{l}$ on the arc:

• Distance to centre: $r = R$ (constant).
• Angle between $d\vec{l}$ (tangential) and $\hat{r}$ (radial): always $90°$.

So $\sin 90° = 1$ everywhere. From Biot-Savart:

$$dB = \frac{\mu_0 I}{4\pi R^2}\,dl$$

All contributions point in the same direction (along the axis). Integrating over arc length $R\theta$:

$$B = \frac{\mu_0 I}{4\pi R^2} \cdot R\theta$$ $$\boxed{B = \frac{\mu_0 I\theta}{4\pi R}}$$

Special Cases

Semicircular arc ($\theta = \pi$):

$$B = \frac{\mu_0 I}{4R}$$

Complete circular loop ($\theta = 2\pi$):

$$\boxed{B = \frac{\mu_0 I}{2R}}$$

$N$ turns:

$$B = \frac{\mu_0 N I}{2R}$$

Direction

Right-hand curl rule: curl fingers in the direction of current; thumb points along $\vec{B}$ at the centre.