\end{problem*}
\begin{solution}
-\end{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}