Question

Find an expression for the magnetic field strength at the center $(pointP)$of the circular arc in FIGURE P29.45.

Short answer

Expression for the magnetic field strength at the center $(pointP)$of the circular arc is$B_P=\frac{{\mu }_{o}I\theta }{4\mathrm{\pi r}}$

Step-by-step solution

Given information   

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

Simplify  

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

${\overrightarrow{B}}_{currentsegment}=\frac{{\mu }_o}{4\mathrm{\pi }}\frac{I△s\sin \theta }{r^2}\left(1\right)$

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

${\overrightarrow{B}}_{arc}=\frac{{\mu }_o}{4\mathrm{\pi }}\frac{I△s\sin 90°}{r^2}=\frac{{\mu }_o}{4\mathrm{\pi }}\frac{I△s}{r^2}\left(2\right)$

At very small segments $△s$,$\sin \theta \approx \theta$is very small angle and from the given figure, the segment $△s$is

$△s=r\sin \theta =r\theta$

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

$$\begin{gathered} B_P=\frac{{\mu }_o}{4\mathrm{\pi }}\frac{I△s}{r^2} \\ =\frac{{\mu }_o}{4\mathrm{\pi }}\frac{I\theta }{r^2} \\ =\frac{{\mu }_{o}I\theta }{4\mathrm{\pi r}} \end{gathered}$$