1 \begin{problem*}{23.1} % induction
2 A flat loop of wire consisting of a single turn of cross-sectional
3 area $A=8.00\U{cm$^2$}$ is perpendicular to a magnetic field that
4 increases uniformly in magnitude from $B_i = 0.500\U{T}$ to $B_f =
5 2.50\U{T}$ in $1.00\U{s}$. What is the resulting induced current if
6 the loop has a resistance of $R = 2.00\Omega$.
12 \varepsilon = - \frac{d\Phi_B}{dt}
13 = - \frac{(2.0\U{T})\cdot(8.00\E{-4}\U{m$^2$})}{1.00\U{s}}
19 \varepsilon &= V = IR \\
20 I &= \frac{\varepsilon}{R} = \frac{-1.6\U{mV}}{2.00\Omega}
21 = \ans{-0.80\U{mA}} \;.