Question: 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.

[![enter image description here](https://i.stack.imgur.com/4Q2s0.jpg)](https://i.stack.imgur.com/4Q2s0.jpg)

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 .

Answer: 
> 
> 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 →→→) 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.
**Solution:**

The potential at A and B are not equal. When R=0, most people blindly say that the potential across the wire is always zero, this is completely false. E=0 is an abstraction as the pd across the wire is small incomparison to the resistor, but definitely not zero

This question is more simply rephrased as:

**"is E = 0 in a zero resistance wire?"**

Let's imagine a scenario where there is a battery and a zero resistance wire.

From maxwells equations we have:

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

For this situation Starting at the positive plate, let's compute the closed line integral

$$\oint \vec{E} \cdot \vec{dl} = \epsilon\_{batt} + \epsilon\_{wire}$$

$$\epsilon\_{batt} = V\_{0}$$

$$\epsilon\_{wire} = 0$$

Thus

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

by assuming E=0 and subsequently the emf of the wire is zero,we have shown that this violates maxwells equations and thus is obviously not true.

**Edit:**

in a zero resistance wire zero potential difference is needed to maintain a constant current, yes an initial E field is needed to get that current going, but once it's there, zero energy is required to maintain it. This does still not mean current is constant here, its not.