1 \begin{problem*}{31.31}
2 Two parallel rails with negligable resistance are $10.0\U{cm}$ apart
3 and are connected by a resistor of resistance $R_3=5.00\U{\Ohm}$. The
4 circuit also contains two metal rods having resistances of
5 $R_1=10.0\U{\Ohm}$ and $R_2=15.0\U{\Ohm}$ sliding along the rails
6 (Fig.~P31.31). The rods are pulled away from the resistor at constant
7 speeds of $v_1=4.00\U{m/s}$ and $v_2=2.00\U{m/s}$, respectively. A
8 uniform magnetic field of magnitude $B=0.0100\U{T}$ is applied
9 perpendicular to the plane of the rails. Determine the current in
26 MultiTerminal R3 = resistor(dir=90, "$R_3$", draw=false);
27 real rlen = R3.terminal[1].y - R3.terminal[0].y;
28 Vector B = BField(phi=-90);
29 vector_field(R3.center, width=3*u, height=rlen, v=B);
31 real yt = R3.terminal[1].y;
32 real yb = R3.terminal[0].y;
33 wire((-1.5*u, yt), (1.5*u, yt));
34 wire((-1.5*u, yb), (1.5*u, yb));
37 pair p1 = (-u,(yt+yb)/2);
38 Vector v1 = Velocity(p1, dir=180, "$v_1$"); v1.draw();
39 Block R1 = Block(p1, width=w, height=rlen); R1.draw();
40 label("$R_1$", (p1.x, yb), align=S);
41 pair p2 = (-p1.x, p1.y);
42 Vector v1 = Velocity(p2, "$v_2$"); v1.draw();
43 Block R2 = Block(p2, width=w, height=rlen); R2.draw();
44 label("$R_2$", (p2.x, yb), align=S);
50 The magnetic flux through the loop will increase as the area enclosed
51 increases, inducing \EMF\ in each rod.
53 \EMF_1 &= BLv_1 = 4.00\U{mV} \\
54 \EMF_2 &= BLv_2 = 2.00\U{mV} \;.
56 The direction of the induced \EMF{}s will try to resist the increasing
57 flux, so $\EMF_1$ will be directed downward and $\EMF_2$ will be
58 directed upward. The situation is then equivalent to the following
66 MultiTerminal V1 = source(dir=-90, type=DC, Label("$\EMF_1$", align=W));
67 MultiTerminal R1 = resistor(V1.terminal[1], dir=-90, Label("$R_1$", align=W));
68 MultiTerminal R3 = resistor(dir=90, "$R_3$", draw=false);
69 R3.centerto(V1.terminal[0], R1.terminal[1], offset=u); R3.draw();
70 MultiTerminal V2 = source(R1.terminal[1] + (2*u, 0), dir=90, type=DC,
71 Label("$\EMF_2$", align=E));
72 MultiTerminal R2 = resistor(V2.terminal[1], dir=90, Label("$R_2$", align=E));
73 wire(V1.terminal[0], R2.terminal[1]);
74 wire(R1.terminal[1], V2.terminal[0]);
75 pair top = (R3.center.x, V1.terminal[0].y);
76 pair bot = (top.x, R1.terminal[1].y);
77 wire(top, R3.terminal[0]);
78 wire(bot, R3.terminal[1]);
83 Solve this circuit using the usual Kirchhoff approach. Label the
84 current through the three branches $I_1$, $I_2$, and $I_3$ each
85 pointing up, and you have
87 I_1 + I_2 + I_3 &= 0 \\
88 \EMF_1 + I_1 R_1 - I_3 R_3 &= 0 \\
89 \EMF_2 - I_2 R_2 + I_3 R_3 &= 0 \;.
91 This gives three equations with three unknowns. Solve using your
101 1\U{\Ohm} & 1\U{\Ohm} & 1\U{\Ohm} \\
117 1\U{\Ohm} & 1\U{\Ohm} & 1\U{\Ohm} \\
131 \end{pmatrix} \U{$\mu$A}
134 Therefore, \ans{$I_3=145\U{$\mu$A}$ upwards}.