We need to apply Faraday's Law
in order to find the differential equation for the circuit. We choose a reference current to be in the clockwise direction. Then
$\begin{align*}\int_{\text{closed loop}} \mathbf{E} \cdot d \mathbf{l} = I R - 10 \mbox{ V} = I \cdot 2 \;\Omega - 10 \mbox{ V...$
and
$\begin{align*}-\frac{d}{dt}\left(\int_{\text{open surface}} \mathbf{B} \cdot d \mathbf{a}\right) = - L \frac{d I}{dt} = - 10 ...$
Therefore,
$\begin{align*}I \cdot 2 \;\Omega - 10 \mbox{ V} = - 10 \cdot 10^{-3} \mbox{ F} \cdot \frac{d I}{dt}\end{align*}$
The solution to this differential equation with the initial condition that $I(t) = 0 \mbox{ A}$ is
$\begin{align*}I(t) = 5 \mbox{ A} \left(1 - e^{-t/(5\cdot 10^{-3} \mbox{ s})} \right)\end{align*}$
The voltage across the inductor is not well-defined; because the electric field is nonconservative, the integral of $\mathbf{E} \cdot d \mathbf{l}$ depends on the path that you take. However, the problem clarifies this with the phrase ``as seen on an oscilloscope."

Solution to 2001 Problem 40