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How does electrical current have energy? (when electrons move very slowly through circiut )

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In an electrical circuit, an electrical current $I$ in a conductor with resistance $R$ produces an energy per time (as heat) $$W=I V=R I^2$$ This energy corresponds to the energy gained from the electric field by each charge carrier (electron) when traversing the potential difference (voltage) over the resistance $R$ $$I=Q/t=ne/t$$ Thus $$W=neV/t $$ On the other hand, a stationary electric current $I$ produces a magnetic field $B$ around it which has the magnetic field energy density $$u=\frac {B^2}{2\mu_0}$$ The work corresponding to this total magnetic field energy has to be done in order to set up the stationary current. This is related to the phenomenon of self-induction.

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You're right that the electron drift velocity is very low, and given the small mass of the electrons to begin with, it must carry very little kinetic energy. What's more, the time averaged current has to be constant in any steady state circuit, so the kinetic energy content of the electrons flowing is undiminished throughout the circuit.

So, how is energy being delivered from source (eg battery) to sink (eg resistor)? The answer is that the energy is moving through the electromagnetic fields in and around the wires. I've not seen or done a detailed analysis on this point, but my understanding is that if you do you'll find that the Poynting vector field accounts for all of the energy flow in the circuit.

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