91Ó°ÊÓ

We need to Find an expression for the magnetic field strength at the center $(pointP)$of the circular arc.

Wires carrying current produce a magnetic field, and we use the Biot-Savart law to calculate the magnitude and direction of that field at any point where the magnetic field due to the segment of current-carrying wire is given by equation

${\stackrel{→}{B}}_{currentsegment}=\frac{{\mathrm{μ}}_o}{4\mathrm{π}}\frac{I△s\sin \mathrm{θ}}{r^2}\left(1\right)$

At the point $P$, Both the two straight wires are parallel and applying zero magnetic fields at point $P$. at point$P$ the arc wire apply a perpendicular magnetic field, so the magnetic field due to the arc is

${\stackrel{→}{B}}_{arc}=\frac{{\mathrm{μ}}_o}{4\mathrm{π}}\frac{I△s\sin 90°}{r^2}=\frac{{\mathrm{μ}}_o}{4\mathrm{π}}\frac{I△s}{r^2}\left(2\right)$

At very small segments $△s$,$\sin \mathrm{θ}≈\mathrm{θ}$is very small angle and from the given figure, the segment $△s$is

$△s=r\sin \mathrm{θ}=r\mathrm{θ}$

The radius $r$of arc can be substitute the expression $â–³s$into equation $\left(2\right)$to get $B_P$

$$\begin{gathered} B_P=\frac{{\mathrm{μ}}_o}{4\mathrm{Ï€}}\frac{Iâ–³s}{r^2} \\ =\frac{{\mathrm{μ}}_o}{4\mathrm{Ï€}}\frac{I\mathrm{θ}}{r^2} \\ =\frac{{\mathrm{μ}}_{o}I\mathrm{θ}}{4\mathrm{Ï€°ù}} \end{gathered}$$