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Consider a circuit with a generator and an inductor as shown -

Circuit 1

Now, the e.m.f. that both the generator and the inductor create are due to changing magnetic fields. So, the electric field in the wires is non-conservative.

So, is Kirchoff's Voltage Law or Loop rule valid here? What is the meaning of "potential difference" across A and B or $v_\mathrm{L}$? Is it the work done to move from A to B through the wire? Is $\mathcal{E} = v_\mathrm{L}$ true or not?

For the difference between e.m.f. and electric potential, see this.

EDIT-

(From comments) I'm trying to understand the $\nabla$ operator but an explanation not involving it would be highly appreciated. There are many students at high-school level(like me) who do not understand $\nabla$ as it is not taught at that level but Kirchoff's law and inductance are taught. It would be of great help if someone gives an explanation at a lower mathematics level(which I believe can be done). The answer can include integration.

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Apoorv Potnis
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1 Answers1

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This is just an answer to the second question.

In Situations like these, Potential-difference always means the terminal voltage between two points of the circuit, that is, the voltage you can measure if you take some multimeter, and connect it's ends to the two points.

As you may know, the electric field $\vec{E}$ (a vector field) can always be written as a sum of a conservative field $\vec{E}_c = - \nabla \Phi$ (which doesn't have rotation) and a source-free field $\vec{E}_i$, with $\nabla \vec{E}_i = 0$. This is the helmholtz decomposition. We can easily identify those components as the ones generated by a potential, and the ones induced by changing magnetic fields:

In the following answer I will make the following assumption: We have a circuit whose wires are 100% conducting, their resistance is 0. This results in no electric field components in the wires, as charges can counterbalance any field infinitely fast (I'm ignoring the mass that electrons are having at this point, If you want, you can use it as the additional assumption that electrons don't have mass).

Additionally we assume, that there are no changing fields inside the multimeter, and inside the wires connecting the multimeter to two abitrary points $A$ and $B$ of your circuit.

The measurement of voltage inside your multimeter (voltage in the sense of a length integral of the electromotive force) is then exactly the difference $\Phi(\vec{r}_A) - \Phi(\vec{r}_B)$: $\int_{\text{closed path}} d\vec{l} \vec{E}_c = 0 = \int_{\text{path from A to B}} d\vec{l} \vec{E}_c + \int_{\text{path from B over Multimeter to A}} d\vec{l} \vec{E}_c \\ = \Phi(\vec{r}_B) - \Phi(\vec{r}_A) + \int_{\text{path over Multimeter}} d\vec{l} \vec{E} \\ = \Phi(\vec{r}_B) - \Phi(\vec{r}_A) - U_{Multimeter} = 0 $

This makes use of the the fact that inside the Multimeter $E = E_c$ because of the absence of changing fields, and the fact that there is no field in the wire connecting the multimeter. In the multimeter, the measured voltage $U_{\text{Multimeter}}$ is the electric force integrated over a certain distance, and that is why $U_{\text{Multimeter}} = \int_{\text{path over Multimeter}}$.

To be very exactly about this, most multimeters (if not all) measure the steady state current that flows through a defined ohmic resistance. This is equal to measuring a field strength over the length of the ohmic resistance, and this is equal to measuring $\int_{\text{path over Multimeter}}$.

Long story short: The Potential difference from $A$ to $B$ is the quantity you can measure when plugging in a multimeter at the two points (That's how you measure it). Furthermore, you have a field called "Potential": $\Phi(\vec{x})$ whose derivatives $ - \nabla \Phi(\vec{x})$ will give you the conservative part of the electric field. The Potential difference is by definition the difference between the values $\Phi(\vec{r}_A)$ and $\Phi(\vec{r}_B$. The Potential difference is not the work done by the electric field along a path from $A$ to $B$, but instead just the work done by the conservative Part of the electric field.

kushal
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Quantumwhisp
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  • I didn't understand your second paragraph but still. When the multimeter is NOT connected, assuming the resistance of the inductor to be zero, would there be accumulation of charges at A and B as they do across a resistor and thus create a potential drop across it? If yes, then there is an electric field in the wire and even if not, there still should be as the generator generates electric field due to changing magnetic flux. Am I right? – Apoorv Potnis Mar 24 '17 at 11:28
  • Also, would it be better to use an ideal potentiometer to measure p.d. as current does not flow through it? – Apoorv Potnis Mar 24 '17 at 11:30
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    Say you have a function f of variables x, y, and z. In order to compute a derivative, you must compute partial derivatives, as in df/dx, df/dy, and df/dz. the delta operator means df/dx + df/dy + df/dz in Cartesian coordinates. If you transform x y z to say r theta phi, the derivatives transform too. –  Mar 24 '17 at 11:54
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    @ApoorvPotnis The conservative part of the potential is created by charges, so yes: You can think of charges in a device that form the potential. A good example is a capacitor: You have charges at the two planes the capacitor consists of, a field (Potential difference) inside, but no field outside. In case of an inductor, just consisting of a conducting wire with a magnetic flux in it, in equilibrium there shoudln't be a field inside the wire: The Charges at the beginning and the end of the inductor create an electric field that cancels out the field created by the change of magnetic flux. – Quantumwhisp Mar 24 '17 at 12:14
  • How did you write $\int_{\mathrm{closed path}}\vec{E},\mathrm{d}\vec{l} = 0$? Shouldn't that be equal to $-\frac{\mathrm{d}\Phi_B}{\mathrm{d}t}$ by Faraday's Law? – Apoorv Potnis Mar 24 '17 at 13:17
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    I wrote $\int_{closedpath} \vec{E}_c d\vec{l} = 0$. It's just the conservative part of the electric field. You can Split the electric field into one conservative part that is generated by the Potential $\Phi$, and a source-free part that is generated by the changing magnetic flux. I wanted to show that it's exactly that part of the electric field that you can measure if you use a multimeter. – Quantumwhisp Mar 24 '17 at 13:22
  • Thanks very much. It cleared a very big problem of mine. – Apoorv Potnis Mar 24 '17 at 18:47