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I read about an experiment from a very reliable source (my textbook). The basic idea is a circuit with a battery, ammeter, switch and wire. A segment of the wire is sort of a test wire. We replace the original wire with a wire of twice the length. The ammeter reading falls by half. Try it out with different length and then we empirically declare that current is inversely proportional to length and consequently that resistance is directly proportional to length (By Ohm's law).

I am confused with the explanation of why current reduces when length is doubled. Current is defined simply as the amount of charge through a given cross section in a given period of time. So why does this change when length is altered. Why charge per given time decreases when length is increased? The only possible explanation i can think of is that charge in a circuit is fixed and when it is made to flow through a longer wire, lesser charge flows through a given cross section or it takes longer for the original charge to move through a given C.S. Am I right?

Edit: As rightly pointed out in the comments, the voltage in this experiment is constant.

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    $\begingroup$ I think the confusion here is in the fact that you should put $\Delta V$ constant when you have you cable with lenght $l$ and that with lenght $2l$; in this other case the electric field intensity gets cut in half, so there is less drift and less current. $\endgroup$
    – Rob Tan
    Sep 23, 2020 at 11:08
  • $\begingroup$ @YasirSadiq the statement you said is what I wanted to prove through the experiment. I want an explanation in terms of electrons. $\endgroup$
    – user248823
    Sep 23, 2020 at 11:14

2 Answers 2

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  1. The voltage drop across something with resistance $R$ is given by $V_R=IR$.

  2. If the battery and the wire with non-negligible resistance are the only parts of the circuit with potential drops, then by Kirchoff's loop law it must be that $V_B=V_R$.

  3. Doubling the length of the wire will double the amount of collisions electrons have while moving through the wire, hence we will get twice the resistance as the original wire.

Putting this all together, we see that for a constant battery potential that the current must decrease by a factor of $2$ when the wire lengthens by a factor of $2$:

$$I=\frac VR=\frac{V_B}{R}\to\frac{V_B}{2R}=\frac I2$$

As for a more physical explanation, the same potential drop over a longer distance means that the electric field in the wire is smaller. If we are assuming an Ohmic wire, a field that is cut in half will produce a current density that is also half.

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The fact is that the Electric field is inversely proportional to length of wire ie. $$E=\frac{V}{L}$$ So if the length of wire is increased by factor of two, the electric field will reduce to half.It's clear if the electric field is reduced ,the electron will be move slow than before and thus the resistance will be increased.

Mathematically, for constant $V$ $$E \propto \frac{1}{L}$$

For most of substance we have linear relationship $$J\propto E$$ or $$I \propto J$$ and finally we conclude that $$I \propto E \propto \frac{1}{L}.$$

that proves. These are vector quantities but we are taking everything lined up.

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