# EMF vs Potential difference

Kaushik
• Is emf the work done to move a positive charge from LOWER potential to HIGHER potential to maintain the potential difference or else the charges move from higher potential to lower potential and will reach a point where the potential is the same between the two points and the charge will stop flowing?
• Why is EMF defined as the potential difference when the circuit is open but is always more than the potential difference when the circuit is closed?
• What is the terminal potential difference?

Homework Helper
EMF is a force, that absent other forces, accelerates a positive charge towards lower potential.

https://en.wikipedia.org/wiki/Electromotive_force

Terminal potential difference is probably a reference to a battery. The potential (voltage) is the potential energy per unit charge, between the two terminals.

Mentor
@vanhees71 this is a distinction you mention with some regularity. Maybe you would like to weigh in?

Gold Member
2022 Award
I think the Wikipedia article is not very clear. It is also unfortunate that we call the quantity in question still electromotive "force", because it's from a time in the history of physics, where force had another meaning than today. Today we call a force the vector quantity ##\vec{F}## in Newton's law ##m \vec{a}=\vec{F}##, but in the days from which we still use the expression "electromotive force" the word "force" meant what we nowadays call work (or potential energy, if the force in the modern sense has a potential). E.g., Helmholtz's famous paper where he described what we call "energy conservation" today was titled "Über die Erhaltung der Kraft" ("On the conservation of force").

That said it becomes better understandable, why we define the electromotive force as a line integral along a closed curve which in general can change in time (in SI units):
$$\mathcal{E}=\int_C \mathrm{d} \vec{r} \cdot (\vec{E}+\vec{v} \times \vec{B}).$$
Here ##(\vec{E},\vec{B})## is the electromagnetic field and ##\vec{v}## the velocity field of the line along which we integrate.

The confusion about the electromotive force often comes from a misleading statement of Faraday's Law of induction in AC circuit theory. The problem is that many introductory textbooks start with the Maxwell equations in integral form (for whichever reason, because it's for sure didactically not a good idea since the integral form is way more complicated than the local form in terms of differential equations). One should always start from Maxwell's equations in differential form, which are the fundamental laws of classical electrodynamics. Here the Faraday Law of induction reads (in SI units)
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}.$$
To derive the integral form, one starts with Stokes's theorem, i.e., one integrates the equation along a surface ##F## with the closed boundary curve ##\partial F## (with the usual relative orientation of the curve to the choice of the orientation of the surface-normal vectors along ##F## according to the right-hand rule):
$$\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{E}=-\int_F \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B}.$$
Now one usually likes to express Faraday's Law in terms of the magnetic flux
$$\Phi=\int_F \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$
Now the dilemma is that in the general case, where the surface and its boundary move (i.e., are a function of time). You cannot simply interchange the integration over the surface in the above (correct) integral form of Faraday's Law, but you have to apply the corresonding "Reynolds transport theorem" to the surface integral, which reads
$$\mathrm{d}_t \Phi=\int_F \mathrm{d}^2 \vec{f} \cdot [\partial_t \vec{B} + \vec{v} (\vec{\nabla} \cdot \vec{B}] - \int_{\partial F} \mathrm{d} \vec{r} \cdot (\vec{v} \times \vec{B}).$$
Now using Maxwell's equations (in differential form) for the integrands,
$$\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{E}=-\partial_t \vec{B}$$
leads to (after again using Stokes's theorem)
$$\mathrm{d}_t \Phi = - \int_{\partial F} \mathrm{d} \vec{r} \cdot (\vec{E}+\vec{v} \times \vec{B})=-\mathcal{E},$$
i.e., you have to use the complete form of the electromotive force including the ##\vec{v} \times \vec{B}## term to get the correct integral form of Faraday's Law in the general case with moving surfaces and boundaries. Only then this form is consistent with the definition of the fields due to the force (in the modern sense) on a point charge given by Lorentz's formula ##\vec{F}=q (\vec{E}+\vec{v} \times \vec{B})##. Then it's clear that the electromotive force is the work per charge along a closed path.

Only in the electrostatic case, where ##\vec{B}=0## and ##\vec{\nabla} \times \vec{E}=0## you have a potential for the electric field, ##\vec{E}=-\vec{\nabla} V## and then the EMF is 0 due to the (complete form) of Faraday's Law.

Now it's also clear that there's an important difference between emf vs. voltage (potential difference): In the general case of time-dependent fields and moving charges (i.e., the presence of currents) the electric field is NOT a potential field, i.e., you cannot define a "voltage drop" as a potential difference, and the integral over the closed loop ##\partial F## in Faraday's Law doesn't vanish, and that's why in AC you have to be careful how to connect a "volt meter" if time-dependent magnetic fields are around. The volt meter in such a case doesn't measure "voltage drops" (because there's no voltage, i.e., electrostatic potential to begin with) but it measures "electromotive forces" along the closed path defined by the wires connecting the volt meter to the circuit.

David Lewis, Kaushik, Klystron and 2 others
weirdoguy
EMF is a force

As @vanhees71 pointed out, EMF is not a force. We don't measure force in volts.

vanhees71 and Kaushik
Homework Helper
Gold Member
First, emf is not potential difference. Emf is numerically the work needed to move 1 unit of charge through a potential difference of 1V. But its dimenison in the rationalized mks system is volts (## ML^2T^{-2}Q^{-1} ##), not Newtons (## MLT^{-2} ##). Emf can move ("force") charge against an opposing E field, so in that sense it's a force.

Emf is always the result of converting a different form of energy into electrical energy. This happens in a chemical battery, a coil subject to a time-varying magnetic field, the Seebeck effect in a wire, and many other ways.

As example, a battery generates emf but it also has a potential difference. What happens in a battery is a simultaneous generation of two electric fields. The emf produces a field ## \bf E_m ## forcing charge to move from the cathode to the anode. This charge separation sets up an opposing field ## \bf E_s ## which is electrostatic and can therefore be expressed as voltage. (So the battery has both emf and voltage). The battery reaches equilibrium when the two fields cancel each other (the battery's internal resistance is ignored in this statement.)

The closed-contour integral of the ## \bf E_s## field is always zero. The closed-contour integral of ## \bf E_m ## is always finite in the presence of finite emf. The total electric field is the algebraic sum: ## \bf E = \bf E_m + \bf E_s##.