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 clockwise direction. Then,
$\begin{align}\int_{\text{closed loop}} \mathbf{E} \cdot d \mathbf{l} = - 5 \mbox {V} - I R = - 5 \mbox {V} - I 10 \;\Omega\en...$
By the right-hand corkscrew rule, $d \mathbf{a}$ points into the page. Therefore, $\mathbf{B} \cdot d \mathbf{a}$ is positive. Therefore,
$\begin{align*}-\frac{d}{dt}\left(\int_{\text{open surface}} \mathbf{B} \cdot d \mathbf{a}\right) = - \frac{d}{dt}\left(B \cdo...$
Therefore,
$\begin{align*}- 5 \mbox {V} - I 10 \;\Omega = -1.5 \;\mathrm{T}\;\mathrm{m}^2\mathrm{/}\mathrm{s} \Rightarrow I = -3.5 \mbox{...$
The minus sign indicates that the current is actually in the counterclockwise direction. Therefore, the answer is (B).

Solution to 1996 Problem 2