1 \newcommand{\dB}{d\vect{B}}
2 \newcommand{\dl}{d\vect{l}}
4 \begin{problem*}{28.12}
5 Two parallel wires are $5.00\U{cm}$ apart and carry currents in
6 opposite directions, as shown in Fig.~28.37. Find the magnitude and
7 direction of the magnetic field at point $P$ due to the two
8 $1.50\U{mm}$ segments of wire that are opposite each other and
9 $8.00\U{cm}$ from point $P$.
19 real Xsep = sqrt(hypot**2 - (Ysep/2)**2);
23 Distance dtop = Distance((-Xsep,Ysep/2), (0,0), "$8.00\U{cm}$");
24 Distance dbot = Distance((-Xsep,-Ysep/2), (0,0), "$8.00\U{cm}$");
26 dtop.draw(labeloffset=8pt);
27 dbot.draw(labeloffset=-8pt);
31 Wire wbot_seg = Wire((-Xsep-Xseg/2, -Ysep/2), (-Xsep+Xseg/2, -Ysep/2), red);
32 Wire wbot = Wire((-Xsep-Xslush, -Ysep/2), (Xslush,-Ysep/2));
36 TwoTerminal Ibot = current((0,-Ysep/2), 180, "$24.0\U{A}$");
37 Distance dSegbot = Distance((-Xsep-Xseg/2,-Ysep/2), (-Xsep+Xseg/2,-Ysep/2),
38 offset=2mm, "$1.50\U{mm}$");
39 dSegbot.draw(labeloffset=-8pt);
42 Wire wtop_seg = Wire((-Xsep-Xseg/2, Ysep/2), (-Xsep+Xseg/2, Ysep/2), red);
43 Wire wtop = Wire((-Xsep-Xslush, Ysep/2), (Xslush,Ysep/2));
47 TwoTerminal Itop = current((Ibot.end.x,Ysep/2), "$12.0\U{A}$");
48 Distance dSegtop = Distance((-Xsep-Xseg/2,Ysep/2), (-Xsep+Xseg/2,Ysep/2),
49 offset=-2mm, "$1.50\U{mm}$");
50 dSegtop.draw(labeloffset=8pt);
57 Using our right hand rules for current-created \vect{B} fields, we see
58 that both segments create magnetic fields that are into the page at
59 $P$, so the sum will also be directed \ans{into the page}. To find
60 the magnitude, we'll use the Biot-Savart law
62 \dB = \frac{\mu_0}{4\pi}\cdot\frac{I\dl\times\rhat}{r^2}
64 on both segments. All of the components are given except for
65 $\dl\times\rhat$. Drawing a picture for the top segment
74 real Xsep = sqrt(hypot**2 - (Ysep/2)**2);
78 draw((-Xsep,Ysep/2)--(0,0)--(-Xsep,0)--cycle, dashed);
79 label("$8.00$", (-Xsep/2,Ysep/4), NE);
80 label("$2.50$", (-Xsep,Ysep/4), W);
84 Angle theta = Angle((0,Ysep/2), (-Xsep,Ysep/2), (0,0), 10mm, "$\theta$");
86 Angle theta2 = Angle((-Xsep,Ysep/2), (0,0), (-Xsep,0), 10mm, "$\theta$");
89 Vector dL = Vector((-Xsep, Ysep/2), mag=3u, "$\dl$");
90 Vector rhat = Vector((-Xsep, Ysep/2), mag=3u, dir=degrees((Xsep,-Ysep/2)),
96 Using our knowledge of cross products and trigonometry
98 |\dl\times\rhat|=|\dl|\cdot|\rhat|\cdot\sin\theta
99 = 1.50\U{mm}\cdot1\cdot\sin\theta
100 = 1.50\U{mm}\cdot\frac{2.50\U{cm}}{8.00\U{cm}}
101 = 4.6875\E{-4}\U{m} \;.
103 It's also pretty clear that this cross product will have the same
104 magnitude and direction for the bottom segment, although you could
105 work that out in detail if you didn't notice right off.
107 The net magnetic field $B_p$ is then
109 B_p = B_{pt} + B_{pb}
110 = \frac{\mu_0}{4\pi}\cdot\frac{4.6875\E{-4}\U{m}}{(8.00\E{-2}\U{m})^2}
111 \cdot(12.0\U{A}+24.0\U{A})
112 = \ans{264\U{nT}} \;.