Question

Find an expression for the magnetic field strength at the center (point P) of the circular arc in $FIGUREP29.45.$

$FIGUREP29.45$ localid="1649251364548" $FIGUREP29.46$

Short answer

$B_p=\frac{{\mu }_{o}I\theta }{4\mathrm{\pi r}}$

Step-by-step solution

Step.1. Given information 

We should find an expression for the magnetic field strength at the center (point P) of the circular arc in $FIGUREP29.45.$

Step.2 solution 

A current-carrying wire produces a magnetic field the Biot-Savart law enables us to calculate the magnetic field at any point where the magnetic field due to the segment ${∆}_s$of current-carryinjg wire is given by equation localid="1649251373932" $\left(29.6\right)$

localid="1649251379150" ${\overrightarrow{B}}_{currentsegment}=\frac{{\mu }_o}{4\pi }\frac{I{∆}_s\sin \theta }{r^2}$ (1)

The two straight wires are parallel to the point P, so both wires apply zero magnetic fields at point P in this case, the magnetic field at point P is due to the arc wire. The arc wire apply a perpendicular magnetic field at point P, so the magnetic field due to the arc is

localid="1649251382747" ${\overrightarrow{B}}_{arc}=\frac{{\mu }_o}{4\mathrm{\pi }}\frac{I{∆}_s\sin {90}^{°}}{r^2}=\frac{{\mu }_o}{4\pi }\frac{I{∆}_s}{r^2}$ (2)

At very small segments ${∆}_s$the angle is very small where localid="1649251386407" $\sin \theta \approx \theta$and from the given figure, segment ${∆}_s$is

localid="1649251390330" ${∆}_s=r\sin \theta =r\theta$

Where r is the radius of the arc. Now, we substitute the expression localid="1649250937705" ${∆}_s$into equation localid="1649250957972" $\left(2\right)$to get localid="1649250990626" $B_p$

localid="1649251393907" $$\begin{gathered} B_p=\frac{{\mu }_o}{4\mathrm{\pi }}\frac{I{∆}_s}{r^2} \\ =\frac{{\mu }_o}{4\pi }\frac{Ir\theta }{r^2} \\ =\frac{{\mu }_{o}I\theta }{4\pi r} \end{gathered}$$