Derive an expression for the magnetic field produced at the centre of a circular arc of a wire carrying current I.

Answer: \(B = \dfrac{\mu_0 I\theta}{4\pi R}\), where \(\theta\) is the angle (in radians) subtended by the arc at its centre.

Consider a circular arc of radius \(R\) carrying current \(I\), subtending angle \(\theta\) at its centre C. Take a small element \(d\vec{l}\) of the arc. The position vector from the element to C has magnitude \(R\) and is perpendicular to \(d\vec{l}\), so the angle between them is \(90^\circ\).

By the Biot–Savart law, the field at C due to the element is

By the right-hand rule, the field due to every element of the arc at C is in the same direction (perpendicular to the plane of the arc), so the magnitudes add. With \(dl = R\,d\phi\):

For a complete circular loop \(\theta = 2\pi\), giving \(B = \dfrac{\mu_0 I}{2R}\).