Solution to 2008 Problem 36

We use Faraday's Law in integral form
\begin{align*}\int_{\text{closed loop}} \mathbf{E} \cdot d \mathbf{l} = -\frac{d}{dt}\left(\int_{\text{open surface}} \mathbf...
We choose to traverse the loop in a clockwise direction. We also choose the reference direction of the current I to be in the counter-clockwise direction. Then,
\begin{align*}\int_{\text{closed loop}} \mathbf{E} \cdot d \mathbf{l} = V = \frac{Q}{C}\end{align*}
where V is the voltage difference between the upper left-hand corner of the circuit and the lower left-hand corner of the circuit and Q is the charge on the upper plate of the capacitor. Also,
\begin{align*}\frac{d}{dt}\left(\int_{\text{open surface}} \mathbf{B} \cdot d \mathbf{a}\right) = L \frac{dI}{dt} = L \frac{d...
So, Faraday's Law is
\begin{align*}-L \frac{d^2Q}{dt^2} = \frac{Q}{C}\end{align*}
At t = 0, Q = Q_0 and dQ/dt = 0, so the solution to this equation is
\begin{align*}Q(t) = Q_0 \cos \left(\frac{t}{\sqrt{LC}} \right)\end{align*}
\begin{align*}I(t) = \frac{dQ(t)}{dt} = -\frac{Q_0}{\sqrt{LC}} \sin \left(\frac{t}{\sqrt{LC}} \right)\end{align*}
The energy stored in the inductor as a function of time is therefore
\begin{align*}U = \frac{1}{2} L I^2 = \frac{Q_0^2}{2 C}\sin^2 \left(\frac{t}{\sqrt{LC}} \right)\end{align*}
Thus, answer (A) is correct.

return to the 2008 problem list

return to homepage

Please send questions or comments to where X = physgre.