The potential difference between the inductor is the emf of the source when t=0, so does the potential difference vary until the steady state current? What is the voltage-time graph for an inductor when the switch closes?

Hint: You could begin from the expression for the voltage across the inductor. You could then find the value of voltage by substituting for t as 0. From the very expression, you will clearly get the exponential relationship between the quantities. You could accordingly make the required plot.

Complete step-by-step solution:
We know very well that the potential difference between the inductor would be the emf of the source when the time was t=0. We are supposed to find how the potential difference varies until it reaches the steady state current. We are also required to find the voltage-time graph for an inductor when the switch is closed.
Let us recall the voltage across an inductor given by,
${{V}_{L}}={{V}_{B}}{{e}^{-\left( \dfrac{tR}{L} \right)}}$
When the time t=0, ${{V}_{L}}={{V}_{B}}$
We know from the relation that the voltage over the inductor would show an exponential decrease and for the voltage across series would show an exponential increase. So the voltage across the resistor would be,
${{V}_{R}}={{V}_{B}}-{{V}_{L}}$
Also, ${{V}_{R}}=IR$
$\Rightarrow I=\dfrac{{{V}_{R}}}{R}\left( 1-{{e}^{-\left( \dfrac{tR}{L} \right)}} \right)$
 So the required graph would look as the following:

Note: In the expression for voltage, we could take the value of resistance as 0, that is, on substituting R=0, ${{e}^{-\dfrac{Rt}{L}}}=1$. But if we assume an ideal voltage source here, the current would asymptote to infinity for R=0. Hence, we could say that this wouldn’t happen.