1 % Commented out because these are defined in problem28.12.tex
2 %\newcommand{\dB}{d\vect{B}}
3 %\newcommand{\dl}{d\vect{l}}
5 \begin{problem*}{28.30}
6 Calculate the magnitude and direction of the magnetic field at point
7 $P$ due to the current in the semicircular section of wire shown in
8 Fig.~28.46. (\emph{Hint:} Does the current in the long, straight
9 section of the wire produce any field at $P$?)
13 import Circ; // TODO: wire(path);
18 Distance R = Distance((0,0), (r*Cos(45),r*Sin(45)), "$r$");
20 draw((-2r,0)--(-r,0){N}..(0,r){E}..{S}(r,0)--(2r,0));
21 TwoTerminal Il = current((-1.5r,0), "I");
22 TwoTerminal Ir = current((1.5r,0), "I");
31 \dB = \frac{\mu_0}{4\pi}\cdot\frac{I\dl\times\rhat}{r^2}
33 gives the magnetic field from an infinitesimal chunk of current.
34 The straight sections of wire do not contribute any magnetic field at $P$, because
36 |\dl\times\rhat| = |\dl|\cdot|\rhat|\sin\theta = 0 \;,
38 since $\theta=0\dg$ to the left and $180\dg$ to the right.
40 For all portions of wire along the semicircle, $\theta=90\dg$, and
41 $\dl\times\rhat$ points into the page. The net magnetic field at $P$
44 B = \int_s \frac{\mu_0}{4\pi}\cdot\frac{I\dl\times\rhat}{r^2}
45 = \frac{\mu_0}{4\pi}\cdot\frac{I}{r^2} \int_s |\dl|\cdot|\rhat|\sin(90\dg)
46 = \frac{\mu_0 I}{4\pi r^2} \int_s |\dl|
47 = \frac{\mu_0 I}{4\pi r^2} \cdot \pi r
48 = \ans{\frac{\mu_0I}{4r}} \;.
50 where $\int_s |\dl|$ is just the length of the semicircle, which is
51 half the $2\pi r$ circumference of the entire circle.