Question

What is the magnetic field strength at the center of the semicircle in $figureP29.54$?

Short answer

The magnetic field strength at the center of the semicircle is$\frac{{\mu }_{o}I}{4R}$.

Step-by-step solution

Given Information.

We need to find the magnetic field strength at the center of the semicircle.

Simplify

With Point P' using the Biot-Savart law the radial and the path length direction is always at a right angle, with this cross product turns into multiplication. Due to the element $\overrightarrow{dl}$of current carrying wire.

${\overrightarrow{B}}_{currentsegment}=\frac{{\mu }_o}{4\pi }\frac{I∆\overrightarrow{s}\times {\displaystyle \stackrel{\wedge }{r}}}{r^2}$

Two wires are parallels to the center of semi-circle, the both wires apply zero magnetic field at the center.

In equation (1) is only good if the length of the line segment is very small when it is compared to the distance from the current element to the point. If not the Biot-Savart law must be used over the entire line segment to calculate the magnetic field and the radius of the circle is close to the length of the segments. The distance over the path $dl$is related to $ds=2\pi dR$

$$\begin{gathered} B=\frac{{\mu }_{o}I}{4\pi R^2}{\int }_0^{1}ds \\ =\frac{{\mu }_{o}I}{4\pi R^2}\left(2\pi R\right) \\ =\frac{{\mu }_{o}I}{2R} \end{gathered}$$

The semi-circle and magnetic field is:

$B_{center}-\frac{1}{2}\left(\frac{{\mu }_{o}I}{2R}\right)=\boxed{\frac{{\mu }_{o}I}{4R}}$