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Suppose I have a battery of $V$ volts and a circuit is made using it. This battery produces a constant electric field which moves the electrons thus establishing current. Now what will happen to the electric field if I increase the length of wire in the circuit (i.e. make it a longer circuit). Will electric field change? If yes then why, since electric field is produced by the battery having some potential difference, it is only related to the battery i.e. $\Delta V = -\int \vec E.d\vec r $ or $E = -\frac{\Delta V}{\Delta r} $ , here $\Delta r$ is the distance between higher and lower potential point of the battety.

If no then why so ??

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Will electric field change ?

Yes

If yes then why

Because you kept the potential the difference the same but increase the distance over which it is dropped.

electric field is produced by the battery having some potential difference, it is only related to the battery i.e. $\Delta V = -\int \vec E.d\vec r $ or $E = -\frac{\Delta V}{\Delta r} $

$\Delta r$ is the distance along the path of integration, which in this case is along the path of the wire (because it's along the path of the wire that you know you have a uniform electric field). If you increase the length of the wire then you must also change the path of integration to follow the new wire, to know that you are following a path with uniform field and with the field vector aligned to the path elements.

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  • $\begingroup$ also how this field is constant throughout the wire ? $\endgroup$
    – Spencer
    Commented Nov 25, 2021 at 17:57
  • $\begingroup$ The field is constant through the wire because you are using a wire with uniform cross-sectional area and material composition. The current through the wire must be uniform along its length because otherwise charge would be building up in the wire, therefore the field must be uniform along the wire. $\endgroup$
    – The Photon
    Commented Nov 25, 2021 at 17:59
  • $\begingroup$ If you are interested in the field in the space outside the wire then the presence of the wire does affect the field, but very close to the battery it won't be much affected by the length of the wire. It's not easy to give a general solution to the problem---you'd normally solve for it numerically, and you'd have to specify the shape of the wire loop. $\endgroup$
    – The Photon
    Commented Nov 25, 2021 at 18:01
  • $\begingroup$ My answer assumed you were asking about the field inside the wire because your question says "This battery produces a constant electric field which moves the electrons thus establishing current." It's the field inside the wire that establishes the current, so as written it seems that this is the field you are asking about. $\endgroup$
    – The Photon
    Commented Nov 25, 2021 at 18:03
  • $\begingroup$ Okay then electric field inside the battery (due to its voltage) must be different from the constant electric field along the wire (due to same battery). Am I right ? $\endgroup$
    – Spencer
    Commented Nov 25, 2021 at 18:04
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"...here Δr is the distance between higher and lower potential point of the battery." Not so sure about that? Since you integrated the battery in the wire circuit, the leads of the battery extend now to the ends of the two wires. So, dr must include also the length of the wires.

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