1 \begin{problem*}{28.23}
2 Four long, parallel power lines each carry $100\U{A}$ currents. A
3 cross-sectional diagram of these lines if a square, $20.0\U{cm}$ on
4 each side. For each of the three cases shown in Fig.~28.41, calculate
5 the magnetic field at the center of the square.
23 label("(a)", (x,-d), S);
24 Vs.push(Vector((x-d,-d), phi=-90, ipen));
25 Vs.push(Vector((x+d,-d), phi=-90, ipen));
26 Vs.push(Vector((x+d, d), phi=-90, ipen));
27 Vs.push(Vector((x-d, d), phi=-90, ipen));
30 label("(b)", (x,-d), S);
31 Vs.push(Vector((x-d,-d), phi= 90, ipen));
32 Vs.push(Vector((x+d,-d), phi=-90, ipen));
33 Vs.push(Vector((x+d, d), phi= 90, ipen));
34 Vs.push(Vector((x-d, d), phi=-90, ipen));
37 label("(c)", (x,-d), S);
38 Vs.push(Vector((x-d,-d), phi= 90, ipen));
39 Vs.push(Vector((x+d,-d), phi= 90, ipen));
40 Vs.push(Vector((x+d, d), phi=-90, ipen));
41 Vs.push(Vector((x-d, d), phi=-90, ipen));
43 for (int i=0; i<Vs.length; i+=1)
50 Note that the magnitude of magnetic field at the center of the square
51 from any corner wire will be
53 B_w = \frac{\mu_0 I}{2\pi \frac{a}{\sqrt{2}}} \;,
55 where $a=20.0\U{cm}$ is the side length of the square, and
56 $r=a/\sqrt{2}=a\cos(45\dg)$ is the distance from the corner of the
59 Using our right-hand rules to determine the direction of the magnetic
79 label("(a)", (x,-d), S);
80 Vs.push(Vector((x-d,-d), phi=-90, iLL));
81 Vs.push(Vector((x,0), dir=-45, iLL));
82 Vs.push(Vector((x+d,-d), phi=-90, iLR));
83 Vs.push(Vector((x,0), dir=45, iLR));
84 Vs.push(Vector((x+d, d), phi=-90, iUR));
85 Vs.push(Vector((x,0), dir=135, iUR));
86 Vs.push(Vector((x-d, d), phi=-90, iUL));
87 Vs.push(Vector((x,0), dir=-135, iUL));
90 label("(b)", (x,-d), S);
91 Vs.push(Vector((x-d,-d), phi= 90, iLL));
92 Vs.push(Vector((x,0), dir=135, iLL));
93 Vs.push(Vector((x+d,-d), phi=-90, iLR));
94 Vs.push(Vector((x,0), dir=45, iLR));
95 Vs.push(Vector((x+d, d), phi= 90, iUR));
96 Vs.push(Vector((x,0), dir=-45, iUR));
97 Vs.push(Vector((x-d, d), phi=-90, iUL));
98 Vs.push(Vector((x,0), dir=-135, iUL));
101 label("(c)", (x,-d), S);
102 Vs.push(Vector((x-d,-d), phi= 90, iLL));
103 Vs.push(Vector((x,+dy), dir=135, iLL));
104 Vs.push(Vector((x+d,-d), phi= 90, iLR));
105 Vs.push(Vector((x,-dy), dir=-135, iLR));
106 Vs.push(Vector((x+d, d), phi=-90, iUR));
107 Vs.push(Vector((x,0), dir=135, iUR));
108 Vs.push(Vector((x-d, d), phi=-90, iUL));
109 Vs.push(Vector((x,0), dir=-135, iUL));
110 Vs.push(Vector((x,0), dir=-180)); // Net B field.
111 Vs[Vs.length-1].mag *= 4*Cos(45);
113 for (int i=0; i<Vs.length; i+=1)
118 For both \Part{a} and \Part{b}, the magnetic fields cancel out and the
119 net magnetic field at the center of the square is \ans{zero}. We can
120 also see this by symmetry, without using the right-hand rule.
121 The \Part{a} configuration is symmetric to $90\dg$ rotations about the
122 square center, so the magnetic field at the center must also be
123 symmetric to $90\dg$ rotations about the square center. A vector with
124 non-zero length is only symmetric to $360\dg$ rotations, so there
125 cannot be any magnetic field in the center of \Part{a}. Similarly,
126 the \Part{b} configuration is symmetric to $180\dg$ rotations, so it
127 cannot have magnetic field at the center. \Part{c} is only symmetric
128 to $360\dg$ rotations, so it can have a magnetic field at the center.
130 The magnitude of the magnetic field for \Part{c} is given by the sum
131 of the horizontal components of the corner magnetic fields, since the
132 vertical components cancel out.
134 B_c = 4 B_w \cos(45\dg)
135 = 4 \cdot \frac{\mu_0 I}{2\pi \frac{a}{\sqrt{2}}} \cdot \frac{1}{\sqrt{2}}
136 = \ans{\frac{2\mu_0 I}{\pi a}} \;.