For a resistor you know that the (conventional) current flows from the higher potential to the lower potential but that is not necessarily true for an inductor.
To answer your question it is perhaps easier not to consider the inductor in isolation but rather as part of a complete circuit as shown below.
In the left hand diagram the current in the circuit is increasing and the potential of node $a$ is zero and that of node $b$ is $\mathcal E = iR$.
Since $\mathcal E > iR$ the potential of node $b$ is higher than that of node $a$.
In the right hand circuit the current through the inductor is decreasing and the potential of node $a$ is again zero but the potential of node $b$ is $-iR$.
In this case, the potential of node $b$ is lower than that of node $a$.