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I read some information on a certain blog saying that energy does not flow inside the wires of a circuit, it propagates in the form of an electromagnetic field outside the wires and straight into the electric components.

My intuition told me that electrons moving inside the wires move in the same way a ball falling from the sky towards the ground moves, therefore gaining kinetic energy along the way. Since a lot of recent textbooks treat the battery as a form of a capacitor which drives electrons to one side of the plate using a certain mechanism similar to that of a conveyer belt( which isn't different from raising a ball to give it potential energy). So which idea is wrong or are both ideas correct?

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The energy flows with the electromagnetic wave. As a demonstration, a wire of current $I$ and potential difference $V$, has the common fields

$$ |\mathbf{E}| = \frac{V}{L} $$

and

$$ |\mathbf{B}| = \frac{\mu_0 I}{2\pi r}.$$

You can derive the power emitted in this wire using the Poynting vector $\mathbf{S}$, which describes the intensity of the electromagnetic wave. Then, the power is

$$ P = \oint \mathbf{S} \cdot d\mathbf{a} = \frac{1}{\mu_0}\oint\left(\mathbf{E} \times \mathbf{B}\right)\cdot d\mathbf{a},$$

where $d\mathbf{a}$ is a differential surface area element of the cylindrical wire. You see that

$$ P = \frac{1}{\mu_0}\cdot \frac{V}{L} \cdot \frac{\mu_0 I}{2\pi r} \cdot 2\pi rL = IV. $$

Your intuition for the electron's motion is not completely incorrect. Essentially, charges in wires can modeled like incompressible fluid flow. This is done in the Drude-Lorentz model. Essentially the electrons are modeled as an ideal gas, having elastic collisions with the positive ions of the conductor. As such, there is a total drift speed to the electrons, although each one is always accelerating due to the collisions. The drift speed is directly proportional to the electric field.

The important part is that this does not matter. The information is not transmitted in the arrival of moving electrons, it is transmitted in the electromagnetic field. Electronic information travels from wires at nearly the speed of light, even though the electrons in the conductor move quite slowly.

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  • $\begingroup$ Thank you, now i have got a good rough idea, but the part i don't get is if the electrons move with the same drift velocity throughout why do some textbooks say that after cycling the circuit, the electrons go to the battery to gain energy again. Isn't their energy constant since their velocity is the same everywhere in the circuit? Sorry i do not possess the link, it's a blog i was on about a year ago.. $\endgroup$
    – Energy
    Commented Aug 4, 2018 at 12:16
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I'd like to develop the functionment of the batteries. They are the place of oxydo-reduction chemical reactions (reactions in which electrons are exchanged) :

  • The molecules of one of the reactants, at the (-) electrode (one electric side of the battery cell), oxydate : they release an electron which goes out of the cell in the electrical circuit.

  • The molecules of the other reactant, at the (+) electrode, are reduced : they capture an electron from the electric circuit.

I simplify it, but what matters here, is that electrons are propelled out of the battery and captured back into the battery while exchanging energy with its reactants. The electrons do not circle around endlessly, they only make one trip into the circuit, from one side of the battery to the other, and then stay there, until the battery is charged again. Charging the battery means essentially that you apply an electric field in the other direction to pull the elecrons back into the (-) electrode.

I hope this is clear enough ^^

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