# Closed loop integral of Electric Field dotted with dL

$$\oint\mathbf{E}\cdot d\mathbf{l} =-\frac{d\Phi}{dt} = -\frac{d}{dt}\int_S\mathbf{B}\cdot d\mathbf{S}$$

Integral of an Electric field dotted with $$d\mathbf{l}$$ gives the value of the potential, but the work done across a closed loop should be $$0$$. How come this equation makes sense?

• Is it because the answer is a function of time and as we are going along the circuit, the value of potential will change and go to 0 when we are done with the loop? Commented Aug 7, 2022 at 13:53
• The work done in a closed loop is only necessarily $0$ when the force can be derived from a potential. What you've found is that not all forces arise from a potential. A classic example is friction. It most certainly takes work to move an object experiencing friction in a closed loop. This is not relevant for you yet, but I can't resist: something a bit strange is going on in the electromagnetic case. As the name suggests, electricity is just one part of what we should think of as a unified electro-magnetic force. And that force, it turns out, actually does have a potential. Commented Aug 8, 2022 at 19:50
• This potential, however, is not a scalar as you are probably used to; it is a $1$-form. The electromagnetic field is best thought of as a $2$-form. In the same way we get the field by taking the differential of the potential (up to a sign of course), we get the electromagnetic field $2$-form from taking the differential of the electromagnetic potential $1$-form. And instead of integrals around loops always evaluating to $0$, we find that integrals of the $2$-form on surfaces (think flux) always evaluate to $0$. Commented Aug 8, 2022 at 19:59

You are studying time dependent phenomena where electrostatic thinking must be abandoned.

In electrostatics, $$W=\oint \ q\vec{E}\cdot{d\vec{l}} =0,$$ which locally translates into: $$\vec{E}=- \vec{\nabla} \phi.$$ But when the fields are time dependent, the circulation of the electric field is no longer null, $$W=\oint \ q\vec{E}\cdot{d\vec{l}} \neq 0.$$ Locally, it can be shown the electric field is no longer a pure gradient, $$\vec{E}=-\vec{\nabla}\phi-\frac{\partial\vec{A}}{\partial t},$$ where $$\vec{A}$$ is the vector potential.

Electrostatics:

$$\oint \vec{E} \cdot \vec{dl} = 0$$

comes from Coulomb's law

$$\vec{E} = \int \frac{\rho(r')\hat r}{4\pi\epsilon_0 |\vec{r}-\vec{r}'|^2}d^3r'$$

$$\nabla × \vec{E} = 0$$

$$\vec{E} = -\nabla V$$

Computing the closed line integral directly, we obtain:

$$\oint \vec{E} \cdot \vec{dl} = 0$$

Electrodynamics:

Faraday's law states that the closed line integral is actually equal to the negative of the rate of change of magnetic flux:

$$\oint \vec{E} \cdot \vec{dl} = - \iint \frac{\partial \vec{B}}{\partial t} \cdot \vec{da}$$

Firstly, your equation is not correct in general; the time derivative should be on the inside of the integral, this is actually important since your equation in its present form says that motional EMF is caused by the electric field, which is not true. You can see this because if $$\vec{da}$$ is time-dependent, you'd get different answers depending on whether or not the time derivative is in the inside or outside

$$\nabla × \vec{E} = -\frac{\partial \vec{B}}{\partial t}$$

The curl is not zero in general, and therefore:

$$\vec{E} ≠ -\nabla V$$

This field is not conservative in general, violating the analysis we obtained from Coulomb's law.

Coulomb's law is a special case of Faraday's law when: $$\frac{\partial \vec{B}}{\partial t} = 0$$, In general Coulomb's law does not hold, and in general the closed line integral of E is not zero.

Congrats. You have observed the basic principle behind how a radio works, or better, how we have encapsulated that principle into theory.

The idea that the closed loop does no work only works in the case of a static electric field. But when the electric field is dynamic, this need no longer be the case, and going in a loop can and does do net work, i.e. your integral need not be zero at all, and in many real-life situations (in theory, at fine enough resolution at least, in all real-life situations).

The reason that's not a problem for energy conservation is because such fields carry energy from one point to another, and doing such work absorbs energy from the field, or else gives it to it, and thus changes it closer to or further from, respectively, a static field. When the antenna in your radio (including a smart phone or a WiFi) is subject to the rapidly-changing magnetic field (so a large $$\frac{\partial \mathbf{B}}{\partial t}$$) of the signal from a remote transmitter (at 2.4 GHz, it reverses direction 4.8 billion times every second), the electrons in the antenna are compelled by the field to move back and forth in closed loops, gaining energy as they do so and transferring it to the rest of the device.

And also, it's more than just radios - in a sense, this is how all electromagnetic transmission of energy happens, including the transmission of thermal energy from the Sun to the Earth across the 149 Gm of intervening near-vacuum. It's just here that the "current" involved is on a much smaller scale due to the much shorter wavelengths, including at the level of single atoms and molecules - a place where quantum mechanics is required to make the most accurate picture.