\begin{problem*}{23.10} % induction
A piece of insulated wire is shaped into a figure eight as shown in
Figure P23.10.  The radius of the upper circle is $r_s = 5.00\U{cm}$
and that of the lower circle is $r_b = 9.00\U{cm}$.  The wire has a
uniform resistance per unit length of $\lambda = 3.00\U{$\Omega$/m}$.
A uniform magnetic field is applied perpendicular to the plane of the
two circles, in the direction shown.  The magnetic field is increasing
at a constant rate of $dB/dt = 2.00\U{T/s}$.  Find the magnitude and
direction of the induced current in the wire.
\end{problem*}

\begin{solution}
Pick a direction for the current to be counterclockwise in the bottom
loop (so clockwise in the top).  Thus, the area vector of the top loop
is antiparallel to \vect{B} and that of the bottom loop is parallel to
\vect{B}.  The magnetic flux is then
\begin{equation}
 \Phi_B = \vect{A}\cdot\vect{B} = (\pi r_s^2 - \pi r_b^2)B \;.
\end{equation}

Using Ampere's law
\begin{equation}
 \varepsilon = - \frac{d\Phi_B}{dt} = \pi(r_b^2 - r_s^2)\frac{dB}{dt} \;.
\end{equation}

The resistance of the entire figure eight is
\begin{equation}
 R = \lambda (2 \pi r_s + 2 \pi r_b) \;.
\end{equation}

Plugging that into Ohm's law yields
\begin{align}
 \varepsilon &= V = I R \\
 I &= \frac{(r_b^2 - r_s^2) \frac{dB}{dt}}{2 \lambda (r_s + r_b)}
    = \ans{25.2\U{mA}} \;.
\end{align}
\end{solution}
