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...
\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.

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