If we have a conducting loop of resistance R in a region of varying external magnetic field, how can we determine the current through the loop?

First, if we consider Ohm's Law, then we get that $$\epsilon = -\frac{d\phi}{dt} = IR$$ However, if we treat it like an inductor, we get $$\epsilon = -\frac{d\phi}{dt} = -L\frac{dI}{dt}$$

I am confused as to which is true and why. Is Ohm's Law even valid here? I found several videos and books that use Ohm's Law which really confused me as I thought Ohm's Law wasn't valid in the case of varying electric fields as it only holds in a condition of steady state.

  • $\begingroup$ This was tagged to be a homework question... I don't know why but it's clearly got nothing to do with any exercise problems or anything like that $\endgroup$ Commented Aug 13, 2023 at 9:18
  • $\begingroup$ Please define/explain your symbols. What is $\epsilon$, what is $\phi$ etc. $\endgroup$
    – kricheli
    Commented Aug 13, 2023 at 9:33
  • $\begingroup$ If you have a conductor with L and R you handle it just as R in row with L and use the complex resitance $\endgroup$
    – trula
    Commented Aug 13, 2023 at 9:38
  • $\begingroup$ You should specify what generates the variable magnetic field. Is is the conductor itself (in which case you are talking about self inductance with an ESR given), or is it an externally generated magnetic field (from another coil, or from a moving magnet nearby, so that self-inductance could be neglected)? $\endgroup$
    – Peltio
    Commented Aug 13, 2023 at 10:08
  • $\begingroup$ I would guess the questions assume that $d\Phi/dt$ is constant so the EMF generated round the loop is constant. Then it is just a matter of equating the constant EMF to the voltage drop in the resistance of the loop IR. You are correct that if $d^2\Phi/dt^2 \ne 0$ the EMF will be changing with time and we would have to consider the inductance of the loop. However for a simple loop the inductance is probably negligible compared to the resistance so you can treat it as purely resistive. $\endgroup$ Commented Aug 13, 2023 at 10:25

1 Answer 1


Your first equation, which I shall write as $$\mathscr E=-\frac{d\Phi}{dt},$$ is correct, as long as we interpret $\Phi$ as the total magnetic flux through the loop. Thus $$\Phi=\Phi_{ext}+LI.$$ As John Rennie has remarked, the self-inductance term (the second term on the right) will no doubt be negligible compared with the external field (the first term on the right). Whether or not this is the case can be assessed by substituting for $\Phi$ from the second equation into the first, giving $$\mathscr E=-\frac{d\Phi_{ext}}{dt}-L\frac{d I}{dt}\ \ \ \ \ \ \text{that is} \ \ \ \ \ \mathscr E=-\frac{d\Phi_{ext}}{dt}-\frac LR\frac{d \mathscr E}{dt}$$ A copper ring of diameter 50 mm and wire radius 0.5 mm has an inductance of $2.94\times 10^{-7}$ H and a resistance of $3.44\times 10^{-3}\Omega$, so you can estimate the rate of change of emf at which the inductance term would be significant.

  • $\begingroup$ Nitpick, the last equation does not follow from only the penultimate one, but we need to assume Kirchhoff's second circuital law $\mathscr{E} = RI$ as well. $\endgroup$ Commented Aug 13, 2023 at 22:44

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