\begin{problem*}{31.31}
Two parallel rails with negligable resistance are $10.0\U{cm}$ apart
and are connected by a resistor of resistance $R_3=5.00\U{\Ohm}$.  The
circuit also contains two metal rods having resistances of
$R_1=10.0\U{\Ohm}$ and $R_2=15.0\U{\Ohm}$ sliding along the rails
(Fig.~P31.31).  The rods are pulled away from the resistor at constant
speeds of $v_1=4.00\U{m/s}$ and $v_2=2.00\U{m/s}$, respectively.  A
uniform magnetic field of magnitude $B=0.0100\U{T}$ is applied
perpendicular to the plane of the rails.  Determine the current in
$R_3$.
\begin{center}
% --+-----+-----+--
% x | x x Z x x | x
% <-| x x ZR3 x |->
% v1| x x Z x x |v2
% --+-----+-----+--
%   R1            R2
\begin{asy}
import Mechanics;
import ElectroMag;
import Circ;

real u = 1cm;
real w = 1mm;

MultiTerminal R3 = resistor(dir=90, "$R_3$", draw=false);
real rlen = R3.terminal[1].y - R3.terminal[0].y;
Vector B = BField(phi=-90);
vector_field(R3.center, width=3*u, height=rlen, v=B);
R3.draw();
real yt = R3.terminal[1].y;
real yb = R3.terminal[0].y;
wire((-1.5*u, yt), (1.5*u, yt));
wire((-1.5*u, yb), (1.5*u, yb));
dot(R3.terminal[0]);
dot(R3.terminal[1]);
pair p1 = (-u,(yt+yb)/2);
Vector v1 = Velocity(p1, dir=180, "$v_1$");  v1.draw();
Block R1 = Block(p1, width=w, height=rlen);  R1.draw();
label("$R_1$", (p1.x, yb), align=S);
pair p2 = (-p1.x, p1.y);
Vector v1 = Velocity(p2, "$v_2$");  v1.draw();
Block R2 = Block(p2, width=w, height=rlen);  R2.draw();
label("$R_2$", (p2.x, yb), align=S);
\end{asy}
\end{center}
\end{problem*}

\begin{solution}
The magnetic flux through the loop will increase as the area enclosed
increases, inducing \EMF\ in each rod.
\begin{align}
  \EMF_1 &= BLv_1 = 4.00\U{mV} \\
  \EMF_2 &= BLv_2 = 2.00\U{mV} \;.
\end{align}
The direction of the induced \EMF{}s will try to resist the increasing
flux, so $\EMF_1$ will be directed downward and $\EMF_2$ will be
directed upward.  The situation is then equivalent to the following
circuit:
\begin{center}
\begin{asy}
import Circ;

real u = 1cm;

MultiTerminal V1 = source(dir=-90, type=DC, Label("$\EMF_1$", align=W));
MultiTerminal R1 = resistor(V1.terminal[1], dir=-90, Label("$R_1$", align=W));
MultiTerminal R3 = resistor(dir=90, "$R_3$", draw=false);
R3.centerto(V1.terminal[0], R1.terminal[1], offset=u);  R3.draw();
MultiTerminal V2 = source(R1.terminal[1] + (2*u, 0), dir=90, type=DC,
    Label("$\EMF_2$", align=E));
MultiTerminal R2 = resistor(V2.terminal[1], dir=90, Label("$R_2$", align=E));
wire(V1.terminal[0], R2.terminal[1]);
wire(R1.terminal[1], V2.terminal[0]);
pair top = (R3.center.x, V1.terminal[0].y);
pair bot = (top.x, R1.terminal[1].y);
wire(top, R3.terminal[0]);
wire(bot, R3.terminal[1]);
dot(top);
dot(bot);
\end{asy}
\end{center}
Solve this circuit using the usual Kirchhoff approach.  Label the
current through the three branches $I_1$, $I_2$, and $I_3$ each
pointing up, and you have
\begin{align}
  I_1 + I_2 + I_3 &= 0 \\
  \EMF_1 + I_1 R_1 - I_3 R_3 &= 0 \\
  \EMF_2 - I_2 R_2 + I_3 R_3 &= 0 \;.
\end{align}
This gives three equations with three unknowns.  Solve using your
favorite method.
\begin{align}
 \begin{pmatrix}
  0 \\
  \EMF_1 \\
  \EMF_2
 \end{pmatrix}
  &=
 \begin{pmatrix}
  1\U{\Ohm} & 1\U{\Ohm} & 1\U{\Ohm} \\
  -R_1 & 0 & R_3 \\
  0 & R_2 & -R_3
 \end{pmatrix}
 \begin{pmatrix}
  I_1 \\
  I_2 \\
  I_3
 \end{pmatrix} \\
 \begin{pmatrix}
  I_1 \\
  I_2 \\
  I_3
 \end{pmatrix}
  =&
 \begin{pmatrix}
  1\U{\Ohm} & 1\U{\Ohm} & 1\U{\Ohm} \\
  -R_1 & 0 & R_3 \\
  0 & R_2 & -R_3
 \end{pmatrix}^{-1}
 \begin{pmatrix}
  0 \\
  \EMF_1 \\
  \EMF_2
 \end{pmatrix}
  =
 \begin{pmatrix}
  -327 \\
  182 \\
  145
 \end{pmatrix} \U{$\mu$A}
 \;.
\end{align}
Therefore, \ans{$I_3=145\U{$\mu$A}$ upwards}.
\end{solution}
