1 \begin{problem*}{30.32}
2 Four long, parallel conductors carry equal currents of $I=5.00\U{A}$.
3 Figure~P30.32 is an end view of the conductors. The current direction
4 is into the page at points $A$ and $B$ and out of the page at points
5 $C$ an $D$. Calculate \Part{a} the magnitude and \Part{b} direction
6 of the magnetic field at point $P$, located at the center of the
7 square of edge length $l=0.200\U{m}$.
15 Vector Ill = Current((0,0), phi=-90, "$B$");
17 Vector Ilr = Current((d,0), phi=90, "$D$");
19 Vector Iul = Current((0,d), phi=-90, "$A$");
21 Vector Iur = Current((d,d), phi=90, "$C$");
24 Distance dx = Distance((0,0), (d,0), "$l$", offset=18pt);
26 Distance dy = Distance((0,0), (0,d), "$l$", offset=-18pt);
28 dot("$P$", (d,d)/2, S);
34 Using right hand rules, vector addition, and the forumula for magnetic
35 field from a long, straight wire, we can find the magnetic field at
36 $P$. Because each wire carries the same current $I$ and is the same
37 distance $r=l/\sqrt{2}$ from $P$, the only thing that will change
38 between per-wire contributions is the direction of the generated
42 = \vect{B}_{AP} + \vect{B}_{BP} + \vect{B}_{CP} + \vect{B}_{DP}
43 = \frac{\mu_0 I}{2\pi r} \cdot \p({
44 \frac{-\ihat-\jhat}{\sqrt{2}} + \frac{\ihat-\jhat}{\sqrt{2}}
45 + \frac{\ihat-\jhat}{\sqrt{2}} + \frac{-\ihat-\jhat}{\sqrt{2}}
47 = \frac{\mu_0 I}{2\pi\frac{l}{\sqrt{2}}} \cdot \frac{-4\jhat}{\sqrt{2}}
48 = \frac{-2 \mu_0 I}{\pi l}\jhat
49 = \ans{-20.0\jhat\U{$\mu$T}}