1 \begin{problem*}{22.37}
2 Four long, parallel conductors carry equal currents of $I =
3 5.00\U{A}$. Figure P22.37 is an end view of the conductors. The
4 current direction is into the page at points $A$ and $B$ and out of
5 the page at points $C$ and $D$. Calculate the magnitude and direction
6 of the magnetic field at point $P$, located at the center of the
7 square of edge length $a=0.200\U{m}$.
14 dotlabel.bot(btex $P$ etex, origin);
15 draw (-r,r)--(-r,-r)--(r,-r)--(r,r)--cycle dashed evenly;
16 p := draw_Ifletch((-r, r));
17 p := draw_Ifletch((-r,-r));
18 p := draw_Itip( ( r,-r));
19 p := draw_Itip( ( r, r));
20 label.bot(btex 0.200\mbox{ m} etex, (0,-r));
21 label.rt(btex 0.200\mbox{ m} etex, (r,0));
23 label.lft(btex $A$ etex, (-r, r));
24 label.lft(btex $B$ etex, (-r,-r));
25 label.rt( btex $D$ etex, ( r,-r));
26 label.rt( btex $C$ etex, ( r, r));
30 \end{problem*} % problem 22.37
33 First, let us pick a coordinate system by choosing unit vectors.
34 Let \ihat\ be down and to the left,
35 \jhat\ be down and to the right, and
36 \khat\ be straight down.
38 Using the right-hand rule, we determine the direction of the magnetic
39 field at $P$ generated by each wire to be
41 \widehat{B_A} &= \ihat \\
42 \widehat{B_B} &= \jhat \\
43 \widehat{B_C} &= \ihat \\
44 \widehat{B_D} &= \jhat
47 The magnitude of each $B$ is given by
49 B = \frac{\mu_0 I}{2 \pi r}
51 And since the currents have the same magnitude, and each corner is
52 equidistant from the square center, each magnetic field contribution
53 will have the same magnitude. The distance $r$ is given by
55 r = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2}
59 We still have to add our vector fields, which gives
61 \vect{B}_P = \vect{B}_A + \vect{B}_B + \vect{B}_C + \vect{B}_D
63 = 2\frac{\mu_0 I}{2 \pi r}\cdot\sqrt{2}\khat
64 = \frac{\sqrt{2}\mu_0 I}{\pi r}\khat
65 = \frac{2\mu_0 I}{\pi a}\khat