# Non-conservative EMF around a circuit and delocalization of motional EMF

Around a circuit, the EMF is non-conservative. In fact, we are in a closed loop and "its" work (the EMF is actually the work itself per unit charge) is non-zero (contrarily to conservative forces/fields...). However, in general the EMF (not motional) does not act around the whole circuit, but only on a part of it.

• How do we see that EMF is not conservative?

Edit 17/01/'20: Let me explain a little more my point. EMF is defined around the whole closed circuit. However, if we have a battery, we know the EMF acts within that battery only. The force which moves the charges inside the battery (or whatever) does not generate a vector field in the surroundings, as Newton or Coulomb's forces. If we had a vector field through all the circuit, then we could say 'hey, that's due to a non-conservative force since the work on the closed path is non zero'. But instead we don't have such a field; it may be similar to assuming gravity is working on a closed square path only at one side of such path, while at the other sides gravity is switched off and the travelling particles are moved by another external force. Well, the work due to gravity is non zero all over the path, nevertheless Newton's force is conservative.

• Why motional EMF is not localized on the circuit?
• What do you mean by "EMF is defined around the whole closed circuit." Right away you point out that the EMF exists only in the battery. Aren't those two statements in contradiction? Commented Jan 17, 2020 at 11:45
• EMF is a line integral, but not necessarily around a closed loop. See, for example, equation 2 in this web page EMF is very commonly introduced with respect to Faraday's Law, so one first sees it as an integral around a closed path. The text that goes with it often muddies the issue. Usually it's something like "In Faraday's Law the EMF is the integral of the electric field around a closed loop." That doesn't imply that the definition of EMF includes a closed loop, but I can see how one can be misled. Commented Jan 17, 2020 at 13:48
• More explicitly, see this Wikipedia article Commented Jan 17, 2020 at 13:56
• EMF can be calculated around a loop. It's just not defined around a loop. Commented Jan 17, 2020 at 14:50
• Emf is not a actual force by the way. Commented Jan 24, 2020 at 7:43

However, if we have a battery, we know the EMF acts within that battery only.

Not true.

The force which moves the charges inside the battery (or whatever) does not generate a vector field in the surroundings, as Newton or Coulomb's forces.

Not true.

Here is a fairly realistic simulation I did of the fields in a DC circuit:

As you can see, there are electric fields outside the battery.

Faraday's law tells us that the electric field in a DC circuit is conservative.

• Thanks for the answer! Where do the outside electric field come from? Commented Jan 17, 2020 at 15:26
• @Shootforthemoon: They exist because of the battery, in the sense that if you remove they battery, they won't exist. Some of their sources are on the wires, as you can see by applying Gauss's law and looking carefully at all the little arrowheads. The circuit acts somewhat like a capacitor, which gets charged by the battery.
– user4552
Commented Jan 17, 2020 at 15:44
• Ok, but they exist as a consequence of the electric field from the emf that moves the charges across the circuit, they are not the emf's field itself, right? Commented Jan 17, 2020 at 16:10
• @Shootforthemoon: The distinction you're trying to make is not a physically meaningful distinction.
– user4552
Commented Jan 17, 2020 at 18:10
• Why? Thought the emf acts only within the battery with a (say conservative or not) force, then what happens outside along the circuit is a response to the action of such force Commented Jan 17, 2020 at 18:24

Emf is defined as the work done to take a unit positive charge one complete round around the closed circuit.It can induced by changing the net magnetic flux through the loop or by connecting the circuit with a battery..In the first case it is quite obvious that the emf is non conservative as the induced electric fields due to the changing magnetic flux is circular hence the fields are non conservative.If we connect the circuit with a battery the electric field produced due to Potential difference of the battery will be conservative.However when we say emf of a battery we mean the work done by the battery to transfer a unit positive charge from the positive terminal to the negative terminal of the battery through the circuit and hence the next work done by the battery for a complete round is not zero(Since we neglect the change in potential as the charge passes through the battery)..Hence the emf is considered non conservative

Your example with gravity is a bit misguided. The work due to gravity around any closed path is always zero. If there were a curve with gravity acting on only part of it, and being "switched off" otherwise, it would mean that gravity were nonconservative. An example of such a curve:

A better gravitational analogy is a ball rolling around a closed path in a hilly landscape. The potential is the height above the ground. Naturally, at the start and the end of each lap, the ball will have the same speed. This is because of energy conservation and the fact that the gravitational force field is conservative. Now, if you find that the ball's speed increases each lap, you immediately know that something other than gravity—a nonconservative force field—is giving it an extra push.

The EMF is analogous to this extra kinetic energy per lap, which you can easily measure. But measuring its size doesn't tell you its origin: It can be localised (somebody hitting it with a golf club at a certain point each lap, like a battery) or not (a conveyor belt along the whole path, like a magnetically induced EMF).

• Thank you for the answer! Unfortunately, I don't get how gravity is nonconservative in that case: it would not dissipate energy, and mechanical energy on its path wil be preserved by its action. Also a cycle along its unique side of action would make the work zero. Commented Jan 21, 2020 at 6:20
• The definition of a conservative force field is that it is the gradient of some potential function: $\vec{F} = -\nabla V$. Thus, the work done along a path is the difference $V_1 - V_0$ of the potential at the end and that at the start. In a cycle, the two points are the same, so the work must be zero regardless of the choice of cycle. You could say that gravity in your example is "locally conservative" along "its" side (the waterfall in the picture), but it is not conservative globally. That's why the picture looks completely normal if you just look locally, but impossible globally. Commented Jan 21, 2020 at 11:00

The voltage of a battery is clearly conservative. If you connect the battery to a voltmeter, it will read the same value independent of the path of the wires. If you move or prolong the wires the readout will be the same. This proves that the voltage of EMF is conservative. Now connect a coil to the voltmeter. If you now place a magnet near the coil. The voltage is zero. If you move the coil and /or the magnet around you will read a voltage. If it is an electromagnet it suffices to change the current. In these cases the voltage does depend on the position of the wires , the coil. This shows that the voltage is not conservative. Note that unless your cables are well shielded there may be a voltage on the coil due to quite a variety of sources of static and dynamic magnetic fields, such as Earth, your washing machine, signal transmission through the atmosphere, a lightning bolt, a passing electric tram or even a car etc.