\begin{problem*}{23.1} % induction
A flat loop of wire consisting of a single turn of cross-sectional
area $A=8.00\U{cm$^2$}$ is perpendicular to a magnetic field that
increases uniformly in magnitude from $B_i = 0.500\U{T}$ to $B_f =
2.50\U{T}$ in $1.00\U{s}$.  What is the resulting induced current if
the loop has a resistance of $R = 2.00\Omega$.
\end{problem*}

\begin{solution}
By Faraday's law
\begin{equation}
 \varepsilon = - \frac{d\Phi_B}{dt}
             = - \frac{(2.0\U{T})\cdot(8.00\E{-4}\U{m$^2$})}{1.00\U{s}}
	     = - 1.6\U{mV} \;.
\end{equation}

By Ohm's law
\begin{align}
 \varepsilon &= V = IR \\
 I &= \frac{\varepsilon}{R} = \frac{-1.6\U{mV}}{2.00\Omega}
   = \ans{-0.80\U{mA}} \;.
\end{align}
\end{solution}

