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I was trying to solve this question: enter image description here

I understand that if the current is increasing leftward, then via Lenz's Law, the direction of the induced EMF will be to the right. Given that I now know the direction of EMF, how can I figure out which point is at a higher potential?

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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.

enter image description here

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 $a$.

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First keep in mind that current flow by convention is the flow of positive charge, that is, from higher to lower potential.. So what does that tell you about the polarity across the inductor, that is, which side is the positive (higher potential) side? You also know the voltage across an inductor will change in the direction so as to oppose a change in current. So what does that tell you about how the voltage is changing, I.e. is the positive side getting more positive or less positive?

Hope this helps.

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  • $\begingroup$ Since current is flowing to the left, I think the right side must have a positive potential (higher potential). Since voltage across the inductor opposes the change in current, I think (?) the positive side is getting MORE positive since the EMF forces positive charges towards the right? $\endgroup$
    – LamGyro
    Commented Apr 29, 2019 at 6:17
  • $\begingroup$ @JamesBond, the polarity of the voltage across an inductor does not depend on the direction of the current through so the reasoning in the first sentence of your comment above isn't correct. It's the sign of the time rate of change of the current that's important here. $\endgroup$
    – Hal Hollis
    Commented Apr 30, 2019 at 17:34

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