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I was reading the chapter about self-induction and inductance from "Physics for scientists and Engineers with Modern Physics, vol. 2, Eight edition - Serway | Jewett" and I can't understand the last sentence of the following quote:

Consider a circuit consisting of a switch, a resistor, and a source of emf. When the switch is thrown to its closed position, the current does not immediately jump from zero to its maximum value e/R. Faraday’s law of electromagnetic induction can be used to describe this effect as follows. As the current increases with time, the magnetic flux through the circuit loop due to this current also increases with time. This increasing flux creates an induced emf in the circuit. The direction of the induced emf is such that it would cause an induced current in the loop (if the loop did not already carry a current), which would establish a magnetic field opposing the change in the original magnetic field.

Why does it say "it would cause an induced current in the loop, if the loop did not already carry a current"? What is the reason of using the conditional form?

I thought that an induced emf generates an induced current (which won't be the same as the one generated from the physical source, indeed), so I'm quite confused about this paragraph.

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The author is considering two scenarios.

  1. There is just a loop of wire without any emf source; if you change magnetic field through this loop you will see BOTH induced emf and current in the loop.
  2. There is a loop of wire connected to an emf source through switch; when you close the switch, the changing current(clockwise) produces a changing magnetic field through the loop. This changing magnetic field produces an induced emf in the loop. Because of this induced emf, you will NOT see an induced current in counter-clock-wise direction because the current is already being established in the clock-wise direction by the emf source.
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  • $\begingroup$ Does the fact that we don't see the induced current in the loop mean that it doesn't exist? I'm not sure if I understood.. Basically with the emf source we see the induced emf but we don't see the induced current? $\endgroup$
    – aghin
    Commented Feb 16, 2022 at 18:23
  • $\begingroup$ If the emf source produced clockwise current, then an ammeter would say it was clockwise current. The "induced emf" produces counterclockwise current, but you would not detect this current in the ammeter as counter clockwise. You would just see a decrease in current if the loop was large. An increase in current if the loop was small. $\endgroup$
    – user291258
    Commented Feb 16, 2022 at 18:32
  • $\begingroup$ Think of it as two sources trying to drive the current through resistor. The emf source will always win because it is the original clockwise current due to emf source that is the reason for the changing magnetic field. $\endgroup$
    – user291258
    Commented Feb 16, 2022 at 18:36
  • $\begingroup$ The induced emf due to changing magnetic field opposes the original current due to emf source, but this induced emf cannot exceed the original emf. So you will never see a current in the counter clockwise direction $\endgroup$
    – user291258
    Commented Feb 16, 2022 at 18:37
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    $\begingroup$ Thank you for your comments! I now understand the whole thing better. Regarding the exceeding of the induced emf, why can't it be greater than a physical one? I found this article on quora that gives some example when we have greater induced emf rather than an applied one quora.com/… . $\endgroup$
    – aghin
    Commented Feb 16, 2022 at 19:51

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