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I used the formula Current=charge density× e × area × drift velocity i.e. i=neAV So this yield that drift velocity inversely proportional to area of cross section But the answer to this question is that drift velocity don't depend on cross section Please explain why can't I use this formula

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  • $\begingroup$ The language of question is a constant voltage is applied across a wire of constant length how does the Drift velocity of electrons depend on the area of cross section of the wire $\endgroup$
    – Nandini Yadav
    Sep 12, 2020 at 2:47
  • $\begingroup$ One could only conclude that the drift velocity is inversely proportional to the area if $ne$ and the current would be constant. But of course they aren't. They could take on any value. The drift velocity of the electrons depends on the microscopical properties of the material they flow through and this material could have any size, i.e. any cross section. $\endgroup$
    – Frederic Thomas
    Sep 16, 2020 at 14:02

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Please explain why can't I use this formula

You still have an unknown, the current $i$, in your formula.

You need to use some other knowledge to eliminate $i$ and get the velocity in terms of the independent quantities in your problem (the voltage and the wire geometry) before you can determine whether the drift velocity depends on any particular quantity.

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You can use the formula but for a fixed voltage and conductor length, $I$ and $A$ are not independent variables.

For a given length of conductor, the resistance of the conductor is inversely proportional to the cross sectional area. Then, per Ohm’s law, for a fixed voltage the current is proportional to the cross sectional area, per your equation.

For a given amount of charge per unit volume, the drift velocity $V$ is proportional to the current density, or $I/A$. Since for a fixed voltage if $A$ increases $I$ increases and if $A$ decreases $I$ decreases, the drift velocity remains constant.

Hope this helps.

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  • $\begingroup$ @NandiniYadav So you find the answer acceptable? As pointed out by The Photon rather than "thanks" comments, either an upvote and/or accepting whichever answer you like best is usually done. $\endgroup$
    – Bob D
    Sep 15, 2020 at 16:50

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