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

Applying the Biot-Savart law to every current element of a circular arc, and integrating over the angle subtended, gives $B = \mu_0 I \theta / 4\pi R$.

Consider a circular arc of radius $R$, carrying current $I$, subtending an angle $\theta$ (in radians) at the centre $O$. Consider a small current element $Idl$ on the arc, located at angle $\phi$ from one end.

By the Biot-Savart law, the magnetic field at the centre $O$ due to this element is:

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

Since every element of the arc lies at a fixed distance $R$ from $O$, and the current element $Idl$ is always perpendicular to the radius vector $\hat r$ joining it to the centre ($\sin90°=1$), this simplifies to:

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

All these elemental fields point in the same direction (perpendicular to the plane of the arc, by the right-hand thumb rule), so they add up as scalars:

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