1 \begin{problem*}{28.60}
2 Figure~28.54 shows an end view of two long, parallel wires
3 perpendicular to the $xy$-plane, each carrying a current $I$ but in
4 opposite directions. \Part{a} Copy the diagram, and draw vectors to
5 show the \vect{B} field of each wire and the net \vect{B} field at a
6 point $P$. \Part{b} Derive the expression for the magnitude of
7 \vect{B} at any point on the $x$-axis in terms of the $x$-coordinate
8 of the point. What is the direction of \vect{B}? \Part{c} Graph the
9 magnitude of \vect{B} at points on the $x$-axis. \Part{d} At what
10 value of $x$ is the magnitude of \vect{B} a maximum? \Part{e} What is
11 the magnitude of \vect{B} when $x\gg a$?
28 Vector Vt = Vector(Scale((0,a)), phi=90, ipen);
29 Vector Vb = Vector(Scale((0,-a)), phi=-90, ipen);
33 dot("P", Scale((x,0)), N);
35 label("$a$", align=W, Scale((0,a/2)));
36 label("$a$", align=W, Scale((0,-a/2)));
37 label("$x$", align=S, Scale((x/2,0)));
59 Vs.push(Vector(Scale((0,a)), phi=90, ipen));
60 Vs.push(Vector(Scale((0,-a)), phi=-90, ipen));
61 Vs.push(BField((x,0), dir=degrees((x,-a))+90)); // From top wire
62 Vs.push(BField((x,0), dir=degrees((x,a))-90)); // From bottom wire
63 Vs.push(BField((x,0))); // Net
65 for (int i=0; i<Vs.length; i+=1) {
69 dot("P", Scale((x,0)), N);
71 label("$a$", align=W, Scale((0,a/2)));
72 label("$a$", align=W, Scale((0,-a/2)));
73 label("$x$", align=S, Scale((x/2,0)));
78 The magnitude of magnetic field from each wire at $P$ is
80 B_w = \frac{\mu_0 I}{2\pi r}
81 = \frac{\mu_0 I}{2\pi\sqrt{x^2+a^2}} \;.
83 The net magnetic field is the sum of the horizontal components, since
84 the vertical components cancel.
87 = 2\cdot\frac{\mu_0 I}{2\pi\sqrt{x^2+a^2}}\cdot\frac{a}{\sqrt{x^2+a^2}}
88 = \ans{\frac{\mu_0 I a}{\pi(x^2+a^2)}} \;,
90 where $\cos\theta=a/r$ comes from
100 Vector Bt = BField((x,0), dir=degrees((x,-a))+90); // From top wire
101 Angle theta1 = Angle((x,0)+dir(0), (x,0), (x,0)+dir(Bt.dir), "$\theta$");
102 Angle theta2 = Angle((0,0), (0,a), (x,0), "$\theta$");
106 draw((0,0)--(0,a)--(x,0)--cycle);
108 label("$r$", (x/2,a/2), NE);
109 label("$a$", (0,a/2), W);
117 size(5cm, 2cm, IgnoreAspect);
121 return 1.0/(x**2 + 1);
126 draw(graph(B, -2*a, 2*a, n=333), red);
127 label("$a$", align=S, Scale((a,0)));
128 label("$-a$", align=S, Scale((-a,0)));
133 The magnitude of \vect{B} has a maximum at $\ans{x=0}$.
136 For $x\gg a$, $x^2+a^2\approx x^2$, so
138 B = \frac{\mu_0 I a}{\pi(x^2+a^2)}
139 \approx \ans{\frac{\mu_0 I a}{\pi x^2}} \;.