I am trying to explain this problem:

A circular conducting loop composed of N turns of wire has a radius of r and a total resistance of R. Perpendicular to the plane of the loop is a magnetic field of strength B. At what rate (in T/s) must this field change, if the induced current that flows in the loop is to be I?

So Faraday's law tells me that voltage is induced in a coil by a changing magnetic field.

emf = -N*B*A/t

And then I can solve for B/t and get

B/t = -emf/(N*A)

Easy, but what about the current? Everywhere around the internets, it's telling me to use Ohm's law to get

B/t = -R*I/(N*A)

And that certainly seems like what I'm supposed to do based on what I'm given in the problem. But that doesn't make sense to me because it's not a resistor; it's a coil, more like an inductor. Does it not store energy when current passes through it? Why don't I have to use the crazy formulae for voltage/current through an inductor with all the derivatives and such? Does the fact that it's closed have anything to do with it?

Please explain to me how is Ohm's law valid here.



1 Answer 1


But that doesn't make sense to me because it's not a resistor; it's a coil, more like an inductor.

It's not an ideal resistor - since ideal resistors have only resistance - and it's not an ideal inductor -since ideal inductors have only inductance.

If this were a loop of ideal conductor, which has zero resistance, a constant current could exist in the loop without an emf generating, time changing magnetic field linking the loop since there is no dissipation of energy.

However, when there is resistance in the loop, sustaining a current $I$ requires a non-zero emf since the resistance dissipates energy.

When the voltage across the resistance (given by Ohm's law) and the emf generated by the time changing magnetic field are of the same magnitude, the current is constant with time.

  • $\begingroup$ So in real life, the inductance of the coil might matter as well? (I'm thinking of the magnetic energy stored in the coil due to the current.) However, we can't calculate it based on the problem because they didn't give the length of the coil, so it's understood that it does not affect the current calculation significantly? $\endgroup$
    – user42630
    Commented Mar 17, 2014 at 2:57

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