\end{problem*}
\begin{solution}
-\end{solution}
+\Part{a}
+Moving the bar to the right increases the magnetic flux directed into
+the page by increasing the area of magnetic field enclosed by the
+loop. Increasing the flux induces a counter-clockwise current to
+resist the change.
+\begin{align}
+ \Phi_B &= AB = lxB \\
+ |\EMF| &= |-\deriv{t}{\Phi_B}| = lB \deriv{t}{x} = lBv \\
+ 0 &= \sum_\text{loop} V_i = \EMF - IR \\
+ I &= \frac{\EMF}{R} = \frac{lBv}{R} \;.
+\end{align}
+The counter-clockwise current is moving up through the sliding bar, so
+the magnetic force on the bar is directed to the left, with a
+magnitude of
+\begin{equation}
+ F_B = IlB\sin(90\dg) = IlB = \frac{l^2B^2v}{R} \;.
+\end{equation}
+
+For the bar to move to the right at a constant speed, the net force in
+the $\ihat$ direction should be zero. We'll have to apply an external
+$F_\text{app}$ to the right to counter $F_B$.
+\begin{align}
+ 0 &= \sum_i F_{i,x} = F_\text{app} - F_B \\
+ F_\text{app} &= F_B = \frac{l^2B^2v}{R}
+ = \frac{(1.20\U{m}\cdot2.50\U{T})^2\cdot2.00\U{m/s}}{6.00\U{\Ohm}}
+ = \ans{3.00\U{N}} \;.
+\end{align}
+\Part{b}
+The power delivered to a resister is
+\begin{equation}
+ P_R = IV = I^2R = \p({\frac{lBv}{R}})^2 R = \frac{(lBv)^2}{R}
+ = \frac{(1.20\U{m}\cdot2.50\U{T}\cdot2.00\U{m/s})^2}{6.00\U{\Ohm}}
+ = \ans{6.00\U{W}} \;.
+\end{equation}
+
+If you want to look at the problem in terms of energy conservation,
+the external force is putting energy into the system at a rate of
+\begin{equation}
+ P_\text{in} = \deriv{t}{F_\text{app} x} = F_\text{app}\deriv{t}{x}
+ = F_\text{app}v = \frac{l^2B^2v}{R} \cdot v = \frac{(lBv)^2}{R} \;.
+\end{equation}
+All this energy has to go somewhere, and the only place for it to go
+is into resistor heat, so it's no surprise that we get the same
+formula for $P_\text{in}$ that we got for $P_R$.
+\end{solution}
\begin{problem*}{31.30}
A rectangular coil with resistance $R$ has $N$ turns, each of length
$l$ and width $w$ as shown in Figure~P31.30. The coil moves into a
-uniform magnetic fiield $\vect{B}$ with constant velocity $\vect{v}$.
+uniform magnetic field $\vect{B}$ with constant velocity $\vect{v}$.
What are the magnitude and direction of the total magnetic force on
the coil \Part{a} as it enters the magnetic field, \Part{b} as it
moves within the field, and \Part{c} as it leaves the field?
\end{problem*}
\begin{solution}
-\end{solution}
+\Part{a}
+If we define ``into the page'' as the positive direction, the flux
+through the loop will be increasing as the coil enters the field,
+which will induce a current in the counter-clockwise direction
+opposing the changing flux. The right side of the coil and portions
+of the top and bottom sides will be in the field regions, and because
+of the current will be subject to magnetic forces directed to the
+left, down, and up respectively. Because equal portions of the top
+and bottom side will be in the field, there will be no vertical
+component in the net force, which will be directed to the left.
+
+The magnitude of the induced \EMF\ is
+\begin{equation}
+ |\EMF| = |-\deriv{t}{\Phi_B}| = \deriv{t}{AB} = B\deriv{t}{Nxw}
+ = NBw\deriv{t}{x} = NBwv \;.
+\end{equation}
+This leads to a current of
+\begin{equation}
+ I = \frac{\EMF}{R} = \frac{Bwv}{R} \;,
+\end{equation}
+which causes a magnetic force of
+\begin{equation}
+ F_B = NIwB\sin(90\dg) = \ans{\frac{N^2B^2w^2v}{R}} \;.
+\end{equation}
+\Part{b}
+While the coil is completely inside the field, the flux remains
+constant, so there is no induced current and \ans{no magnetic force}.
+
+\Part{c}
+As the coil leaves the field, the flux drops back towards zero,
+indicing a clockwise current trying to keep the flux up. Again, the
+vertical components of the resulting magnetic force cancel out, but
+the upward current in the left side will be subject to a magnetic
+force directed to the left. The magnitude is the same as
+for \Part{a}.
+\end{solution}
\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}
\end{problem*}
\begin{solution}
+Picking up as the positive flux direction, the induced \EMF\ is
+\begin{equation}
+ \EMF = -\deriv{t}{\Phi_B} = -\deriv{t}{NAB} = -NA\deriv{t}{B}
+ = -NA\frac{-2B}{\Delta t} = \frac{2NAB}{\Delta t} \;.
+\end{equation}
+This \EMF\ drives a current through the resistor
+\begin{align}
+ 0 &= \EMF - IR \\
+ I &= \frac{\EMF}{R} = \frac{-2NAB}{R\Delta t} \;.
+\end{align}
+Current is defined as the charge passing through a cross section of
+your circuit per unit time, so the charge entering one end of the
+resistor is
+\begin{equation}
+ \Delta q = \frac{\Delta q}{\Delta t} \Delta t = I \Delta t
+ = \frac{2NAB}{R} = \ans{0.880\U{C}} \;.
+\end{equation}
\end{solution}
-
\end{problem*}
\begin{solution}
-\end{solution}
+\Part{a}
+A clockwise current induces an inward magnetic field inside the loop,
+so the external flux must be increasingly out of the page. Therefore,
+the magnetic field is \ans{increasing}.
+\Part{b}
+The induced \EMF\ must be
+\begin{align}
+ 0 &= \EMF - IR \\
+ \EMF &= IR \;,
+\end{align}
+This is related to the changing field via magnetic flux.
+\begin{align}
+ |\EMF| &= |-\deriv{t}{\Phi_B}| = \deriv{t}{AB} = A\deriv{t}{B}
+ = \pi r^2 \deriv{t}{B} \\
+ \deriv{t}{B} &= \frac{\EMF}{\pi r^2} = \frac{IR}{\pi r^2}
+ = \ans{62.2\U{T/s}} \;.
+\end{align}
+\end{solution}
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