# Loop of wire in changing magnetic field, open circuit vs. short vs. intermediate

Assume you have a loop of wire in a time varying magnetic field. The loop has only one turn and you can either leave the ends disconnected (open circuit), short them, or place a resistor across the ends.

I understand if the ends are disconnected the voltage is given by the Maxwell Faraday equation:

And if the ends are shorted (assuming zero resistance) there is no voltage across the loop so the magnetic flux through the loop can't change (the coil current makes its own magnetic field that cancels out any changes in the external one.

However, what if there is a resistance placed across the coil so the voltage isn't zero, and the current isn't zero (Now it is V=IR). How can you calculate how much current will flow?

Thanks!

I'd use $$\mathscr E-L\frac{dI}{dt}=RI$$ In which $$\mathscr E$$ is the emf due to the changing external magnetic field, as given by the equation you have cited, $$L$$ is the inductance of the loop and $$R$$ is its resistance plus any external resistance. The equation can be solved for $$I$$ if you know how $$\mathscr E$$ varies with time.

$$L$$ is not particularly easy to calculate for a loop, but approximate formulae are easily found online.

• Does this also work if the resistance is R=0 (the ends are shorted)? I guess that would imply E = L dI/Dt Sep 19, 2022 at 16:00
• Yes, I don't see why not. Sep 19, 2022 at 16:13
• I think the only thing that intuitively is confusing about this is that the EMF calculation only depends on the flux through the interior surface of the loop, while the inductance takes into account all of the magnetic field in 3D space. Sep 19, 2022 at 17:29
• But the inductance is defined as the flux linked with the loop due to current in the loop, per unit current. The flux integral involved is a surface integral over any surface bounded by the loop. In other words an integral of exactly the same form as your integral for flux due to the external source. So I don't see the difference that you see. In both cases the fields do extend over space (and the field lines are closed loops), but the flux calculation is essentially of the same type. Sep 19, 2022 at 17:44
• Page 11-6 of this resource shows such a calculation: web.mit.edu/viz/EM/visualizations/coursenotes/modules/… Sep 19, 2022 at 19:40

I disagree with your analysis. Irrelevant of the circuit being closed or opened, there is always a voltage difference.

The voltage difference between any two points is the opposite of the circulation of the electric field between these two points.

$$\triangle V=V_{B}-V_{A}=- \int_a^b \overrightarrow{E}. \overrightarrow{dl}$$

Alike a normal battery, the circuit can be opened and yet have a voltage difference between its points (terminal). If the circuit is closed, the voltage difference drives the current.

For historical reasons, Faraday's equation is widely used in elementary electromagnetic theory lectures. It is unfortunate because, being non local, it fails to explain the true nature of the electric field.

From Maxwell's equations we have:

$$\begin{cases}\overrightarrow{E}=- \frac{\partial \overrightarrow{A} }{\partial t}\\ \overrightarrow{B} = \overrightarrow{ \nabla } \times \overrightarrow{A}\\ \overrightarrow{ \nabla }.\overrightarrow{A}=0 \end{cases}$$

Where $$\overrightarrow{A}$$ is the vector potential.To simplify, let's use:

$$\overrightarrow{A}(t) = B(t).x.\overrightarrow{e_{y}}$$

$$B(t)$$ is any function of time t.

We get for the fields:

$$\begin{cases} \overrightarrow{B}=B(t) \overrightarrow{e_{z}} \\ \overrightarrow{E}=- \frac{dB}{dt}x \overrightarrow{e_{y}} \end{cases}$$

The existence of these two fields is independent of the presence or not of a loop. As soon as the the magnetic field varies in time, there is an electric field as well. This is the main drawback back of Faraday's law. Students have the wrong impression the time variation of the magnetic field induces an electric field into the loop. Again, the electric field exists everywhere in space where the magnetic field is time dependent.

Let's now see what are the effects of this electric field on a circuit. To simplify let's consider a plane square loop.

From the expression of the fields, we see that $$\overrightarrow{A}(t)$$ is along the y axis, positively oriented for x positive and negatively for x negative.

And the electric field is in the opposite direction:

We can now calculate the potential difference between any two points $$M_{1}$$ and $$M_{2}$$ in the circuit.

First case: $$M_{1}$$ and $$M_{2}$$ on AB or DC

The electric field is perpendicular to the displacement, so its circulation is null. $$\overrightarrow{E}. \overrightarrow{dl}=0 ~ \Longrightarrow \triangle V=- \int_{M_{1}}^{M_{2}} \overrightarrow{E}. \overrightarrow{dl}=0$$

Had we considered a finite size cylindrical wire, the electric field would have displaced the electrons towards the surface. This effect is called the Hall effect. The voltage difference that we have calculated is independent of the circuit being closed or opened. It is generated by the external time varying magnetic field.

Second case: $$M_{1}$$ and $$M_{2}$$ on BC

Let's call "a", the abscissa of the segment [BC]. we have: $$M_{1}(a,y_{1})$$ and $$M_{2}(a,y_{2})$$

According to the formula above: $$\overrightarrow{E}=- \frac{dB}{dt} a \overrightarrow{e_{y}}$$

We thus get for the potential difference between the two points:

$$\triangle V=V_{1}-V_{2}=- \int_{M_{1}}^{M_{2}} \overrightarrow{E}. \overrightarrow{dl}=\frac{dB}{dt} a.(y_{1}-y_{2})$$

The special case where $$M_{1}$$ and $$M_{2}$$ are in B and C leads to classical formula obtained with Faraday's law:

$$\triangle V=V_{B}-V_{C}=\frac{dB}{dt} a^{2}$$

Third case: $$M_{1}$$ and $$M_{2}$$ on DA

The only difference with the previous case is the sign of "a".

$$\triangle V=V_{A}-V_{D}=-\frac{dB}{dt} a^{2}$$

1. If the loop is opened, let's say between C and D, the electrons move rapidly and accumulate on the opened end of the loop. Equilibrium is reached when the electrostatic field cancels out the electromotive field. One has to take into account this charge redistribution to calculate the potential

2. If the loop is closed: As explained in other answers, the e.m.f drives a current i(t) that is limited by both the resistance and the self induction.

$$V_{B}-V_{C}= \frac{dB}{dt}a^{2}= r~i+L \frac{di}{dt}$$

The formula is still valid if r=0.

• What about for closed superconducting loop where r=0? Is there still a voltage around the loop? I mean, I guess there must be in order for the electrons to start flowing. Sep 22, 2022 at 19:20
• I know little about superconductors. I just know that 3D superconductors expel the magnetic field, but I have no idea about the wire like ones. Sep 22, 2022 at 22:21