Derive an expression for the magnetic field produced by a current in a circular arc of a wire using Biot-Savart law.

Answer: At the centre of an arc of radius \(R\) subtending angle \(\phi\) (in radian), \(B = \dfrac{\mu_0 I \phi}{4\pi R}\).

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

By Biot–Savart law,

For every element the field at C has the same direction (perpendicular to the plane of the arc), so the magnitudes add. With \(dl = R\,d\phi\):

Direction: perpendicular to the plane of the arc, given by the right-hand thumb rule. For a full circle \(\phi = 2\pi\) and \(B = \dfrac{\mu_0 I}{2R}\).