## Solution to 2008 Problem 70

 From Faraday's Law\begin{align*}\oint \mathbf{E} \cdot d\mathbf{l} = - \frac{d}{dt}\left\{\int \mathbf{B} \cdot d \mathbf{a}\right\}\end{align...If $I$ is the current and Unparseable_latex_formula\Omega$}$ is the resistance of the wire loop, then\begin{align*}I R = \oint \mathbf{E} \cdot d\mathbf{l}\end{align*}$I = dQ/dt$, so\begin{align*}\frac{dQ}{dt} = -\frac{1}{R} \frac{d}{dt}\left\{\int \mathbf{B} \cdot d \mathbf{a}\right\}\end{align*}We integrate both sides with respect to time to obtain the amount of charge that flows past a given point in the wire:\begin{align*}Q = -\left.\frac{1}{R} \left\{\int \mathbf{B} \cdot d \mathbf{a}\right\}\right|_{t = t_0}^{t= t_1}\end{align*}where $t_0$ is some time before the wire loop was pulled out of the magnetic field and $t_1$ is sometime after. At $t = t_1$, there is no magnetic flux through the loop, and at $t= t_0$, the magnetic flux through the loop is $\left(10 \;\mathrm{cm}^2\right) \cdot \left(0.5 \mbox{ T}\right) = 5 \cdot 10^{-4} \;\mathrm{T}\;\mathrm{m}^2$. So,\begin{align*}\left. \left\{\int \mathbf{B} \cdot d \mathbf{a}\right\}\right|_{t = t_0}^{t= t_1} = 0 - 5 \cdot 10^{-4} \;\mat...So,\begin{align*}Q = \frac{5 \cdot 10^{-4} \;\mathrm{T}\;\mathrm{m}^2}{R} = \frac{5 \cdot 10^{-4} \;\mathrm{T}\;\mathrm{m}^2}{5 ...Therefore, answer (A) is correct.