# How does current actually flow in a wire?

When I was in my school I was taught that the electric field due to the battery is along the wire (from $$A \rightarrow B \rightarrow C \rightarrow D$$) and these are responsible for electrons at each location to move which constitutes the current.

And we also learned that the potential at A and B is the same and that at C and D are same too (ideal circuits) while solving circuit problems.

But when I tried to merge the above two ideas , I got a contradiction.

From electrostats we know that there is a potential drop along the direction of electric field. And if that's true then there can't be any component of electric field along the length of the wire as it would result in a potential drop without any resistor. So this indicates that the electric field must be perpendicular to the wire itself .

But if the above statement is true how can there be current along the wire ?

This is confusing me a lot. Please help me figure out what's happening in there .

• Be careful using the phrase "potential drop without any resistor". This could mean an infinite resistance as if the component were removed from the circuit. or it could mean having no (zero) resistance. Aug 17, 2022 at 3:59
• There is an electric field in all directions always. So there is an electric field perpendicular to the wire always. Electric field gradient along the wire cause current flow. However a forced current flow will also cause a gradient, The small gradient along the wire will come from the current flow caused by the voltage drop across the resistor. Aug 17, 2022 at 4:05

But when I tried to merge the above two ideas , I got a contradiction.

Yes, you are correct. There is a contradiction.

And we also learned that the potential at A and B is the same and that at C and D are same too (ideal circuits) while solving circuit problems.

This statement applies to ideal conductors. In an ideal conductor the potential is the same so there is no E field. Also, since it has no resistance, no potential difference is required for the current in an ideal wire.

the electric field due to the battery is along the wire (from π΄βπ΅βπΆβπ·) and these are responsible for electrons at each location to move which constitutes the current

This statement applies to non-ideal conductors. In a non-ideal conductor there is an E field along the wire so there is a potential difference across the wire. Since a non-ideal wire has some small resistance this potential difference is required for the current.

If you want to be consistent you must decide if you want to model a given wire as an ideal conductor or as a non-ideal conductor. You cannot consistently have it both ways. Practically speaking, for ordinary circuits the differences between the two models is small, so the ideal model is used more often because it is simpler.

there can't be any component of electric field along the length of the wire as it would result in a potential drop without any resistor. So this indicates that the electric field must be perpendicular to the wire itself .

This is incorrect. If there was no resistor, the battery would still cause potential drop along the wire. And if there is resistor, most of potential drop happens in the resistor.

There can and is electric field component along the wire. This field is due to the battery and other surface charges on the wires.

When you short the battery, the circuit is not ideal. The potential varies along the wire. Wire can be considered ideal when there are components in the circuit with much larger resistance than the wire. Then the potential changes mostly near those components.

Try not to think of the battery creating an electric field in the wires. So where is the field gradient created? In the resistor. We say that the voltage across a resistor (applied by the battery) creates an electric field gradient in the resistor.

Because the wires have such a low resistance it is challenging to set up the field. So pretend the resistance in the wires is zero (or insignificant).

So the electrons in the resistor move along the electric field gradient. The nuclei and bound electrons get in the way making the path difficult. It is a resistor after all.

Because the electrons are moving in one direction they leave some protons exposed at one end. These attract electrons from the wire connecting to the battery. And so on down the wire all the way to the battery.

At the other end of the resistor all the electrons are trying to get out. They create a small negative charge that will repel the electrons in the wire, again all the way back to the battery. This is how the very tiny electric field gradient is set up in the wires.

There is more to it but this is a good visualization.

The electrons don't actually move that far. Their drift velocity is small. The energy transfer is actually in the electric field and is is fast.

If the resistor is small enough then we must include the wire resistance in the calculation. The voltage drop along the wire will be in proportion to the total resistance as well as the wire resistance.