Question

What are the strength and direction of the magnetic field at point P in FIGURE P29.46?

Short answer

The magnetic field at point$P$is$B_P=7.9\times {10}^{-5}T$into the page.

Step-by-step solution

Given information    

We need to find the strength and direction of the magnetic field at point $P$.

Simplify  

Inner radius $a=1cm$and outer radius $b=2cm$. The megnatic field can be determine at point $P$using biot-savart. The path length and radial directions are always at a right angle, so the cross product turns into multiplication. The magnetic field due to the element $\overrightarrow{dl}$of current-carrying wire is given by

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

The approximation works best if the length of the line segment is very small compared to the distance from the element to the point. To calculate the magnetic field, the integral form of the Biot-Savart law must be used over the entire line segment, and since the radius of a circle is near the length of the segments, we integrate over the radius.The distance along the path $dl$is related to the radius by $ds=2\mathrm{\pi dR}$

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

The magnetic field for semicircle is

$B_{arc}=\frac{{\mu }_{o}I}{4R}\left(2\right)$

As the current in the small arc is in the opposite direction from the current in the big arc, the magnetic field due to both segments at point $P$is given by

$$\begin{gathered} B_P=B_a-B_b \\ =\frac{{\mu }_{o}I}{4_a}-\frac{{\mu }_{o}I}{4_b} \\ =\frac{{\mu }_{o}I}{4_{ab}}(b-a)\left(3\right) \end{gathered}$$

Simplify  

Putting values in equation $\left(3\right)$for $a,bandI$to get $B_P$

$$\begin{gathered} B_P=\frac{{\mu }_{o}I}{4_{ab}}(b-a) \\ =\frac{\left(\left(4\times {10}^{-7}\right)\left(T\times m/A\right)\right)\left(5A\right)}{4\left(0.01m\right)\left(0.02m\right)}\left(0.02m-0.01m\right) \\ =7.9\times {10}^{-5}T \end{gathered}$$

After applying right hand rule the magnetic field isinto the page.