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A typical electrical circuit with resistors only can be shown as in the fig.

enter image description here

This is how every book and every website indicates the direction of current in the circuit diagrams.

But I have a doubt whether we can assign current in the plain wire section of the circuit assuming the wires to be ideal.

We know that points A , B and C in the above circuit are at equal potentials (since the wire is ideal) but this means that the direction of electric field in this section of the wire will be perpendicular to the wire but this again indicates that the electrons in this plain section do not feel any electrical force due to the battery along the length of the wire and thus it does not drift and thus has only random motions which means that there is no current in the plain sections of the wire.

Now my question is that is the picture correct or are the books wrong?

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    $\begingroup$ there is an additional 'oversimplification' problem in your statement. You are assuming the conductor to have dimension 1, i.e. of not having an interior. In reality conduction, and in particularly the establishing of the electric field E = j/sigma inside the wire is due to the distribution of surface charge on the surface of the wire. By denying the existence of a surface you make it impossible for the wire to do what is supposed to do. Even superconductors need that surface charge - and current - to make (keep?) the E field inside zero. $\endgroup$
    – Peltio
    Sep 23 at 16:18
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the electrons in this plain section do not feel any electrical force due to the battery along the length of the wire

Ideally, it is true that they do not feel any electrical force, but ideally the electrical force required to move through an ideal wire is 0.

Of course, that is all an idealization. In reality (except for superconductors) there is a small but non-zero resistance in the wires so there is a small but non-zero electrical force. Since the resistance is small this small force is sufficient. By treating the wire as ideal, we are simply ignoring the small non-zero values. We know that will make our solution inexact, but it is a close enough approximation that we don't care since other sources of error probably dominate over what is essentially a rounding error.

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  • $\begingroup$ @Dake thanks for your answer 😊.. so can an ideal wire placed isolated have current in it ? $\endgroup$
    – Ankit
    Aug 27 at 16:44
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    $\begingroup$ @Ankit Yes, as long as it forms a loop. This is the basis of MRI machines (using superconductors) $\endgroup$
    – Dale
    Aug 27 at 17:23
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We know that points A , B and C in the above circuit are at equal potentials

You know that is true at steady state. Prior to steady-state, there may be differing potentials due to the inductance of the wire. Unlike superconductors which do have zero resistance, there are no zero inductance circuits. When resistance is very low, the inductance becomes significant to describe the transitory behavior.

In going from an open to a closed circuit, there are transitory fields within the wire. These fields create the initial current. Once the current is created, then no field is necessary for it to remain (assuming no resistance).

the electrons in this plain section do not feel any electrical force due to the battery along the length of the wire and thus it does not drift

No, the lack of a field means the electron drift velocity does not change. It doesn't mean the drift velocity is zero (in an ideal, zero-resistance conductor).

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