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 such a way so that when the loop is in the position shown in the figure, the traversal appears clockwise when viewed as in the figure. We also choose the reference direction of the current $I$ to be in the same direction as the traversal. Then
$\begin{align*}\int_{\text{closed loop}} \mathbf{E} \cdot d \mathbf{l} = I R = I 9 \;\Omega\end{align*}$
By the right-hand corkscrew rule, $d \mathbf{a}$ points in the $- \mathbf{n}$ direction. Therefore, $\mathbf{B} \cdot d \mathbf{a} = B_0 \hat{\mathbf{x}} \cdot \hat{\mathbf{n}} d a = B_0 \cos \theta da$ where $\theta$ is the angle between $\hat{ \mathbf{x}}$ and the normal vector $\hat{ \mathbf{n}}$ . We are given that

$\begin{align*}\frac{d \theta}{dt} = \omega\end{align*}$
and that $\theta(t = 0) = \pi/2$ . Therefore, $\theta = \omega t + \pi/2$ . Therefore,

$\begin{align*}-\frac{d}{dt}\left(\int_{\text{open surface}} \mathbf{B} \cdot d \mathbf{a}\right) &= - \frac{d}{dt}\left(1...$
Therefore,
$\begin{align*}I 9 \;\Omega = 0.225 \pi \cos \omega t \mbox{ V} \Rightarrow I = 0.025 \pi \cos \omega t \mbox{ A} = 25 \pi \co...$
Note that the reference direction that we chose was actually the correct, because when $\omega t \approx - \frac{\pi}{4}$ , so that the coil is in the orientation that is shown in the figure, then the current I is positive. Therefore, answer (E) is correct.

Solution to 2001 Problem 86