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We know for a metallic conductor

Current $\frac{I}{enA} = v$ where $v$ is drift velocity , $e$ is the charge of an electron, $n$ is no of electrons per unit volume and $A$ is area of cross section.

Ok my questions that I am struggling with:

  1. Does it depend on length of wire? (let's say if the current is the same and area is also the same but one wire is longer and one wire is shorter then is the drift velocity for both wires the same?)

  2. Does it depend on the cross sectional area of the wire? (Yes, according to the relation, if the area is greater, drift velocity is lower, correct?)

I have digged university physics still I couldn't get my answer :( .

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5 Answers 5

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The drift velocity does not depend on the length or the cross sectional area of the wire, when dealing with a macroscopic (ordinary, everyday life) wire. However, if the wire is, say, too short, e.g. comparable to the average distance a charge carrier travels before undergoing a collision, then it might begin to depend on the wire length, but for all practical intents and purposes a wire won't be that short.

The reason v does not depend on the wire cross sectional area is that the ratio I/A is constant (assuming the applied electric field within the wire is not changing), also called the current density, denoted by J=I/A. So, for example, if A doubles, I will also double (wire capacity doubles), keeping J constant.

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  • $\begingroup$ What about the skin effect? I thought almost all conduction happened near the surface of the wire. $\endgroup$ Commented Nov 10, 2014 at 16:50
  • $\begingroup$ That's an effect in AC (and depends on the frequency). For DC, it's not a consideration. $\endgroup$
    – BowlOfRed
    Commented Nov 10, 2014 at 16:53
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    $\begingroup$ @Pooya Jannaty the equation says directly that for a fixed current and a varying cross-sectional area, the velocity changes... can you elaborate on the independence of A please? $\endgroup$
    – kηives
    Commented Jan 24, 2015 at 19:16
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No.It is not dependent on Cross sectional area or the length of the wire.

Let's study an analogy,

$$F=m\times a$$

$$\frac{F}{a}=m$$

Does that mean mass of an object is dependent upon the force applied on it?

No.

Drift velocity is dependent upon electron mobility and electric field applied. $$v_s=\mu\times E$$

None of which are dependent upon cross sectional area or the length of the wire.

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  • $\begingroup$ The linked answer is for a resistor with a fixed voltage between the ends. A wire in a circuit with a constant current doesn't have a fixed voltage (or electric field). $\endgroup$
    – BowlOfRed
    Commented Oct 14, 2016 at 17:23
  • $\begingroup$ @bowl of red do you know about the equation: current density= specific resistance × electric field for a conductor? There is a fixed electric field in this case. $\endgroup$ Commented Jan 25, 2017 at 6:36
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    $\begingroup$ @Mockingbird This question is not about constant voltage, but constant current. Given constant current and changing resistance, the voltage drop in the wire (and electric field) must change $\endgroup$
    – BowlOfRed
    Commented Jan 25, 2017 at 7:28
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Yes, it does depend upon the length of the wire. Drift velocity (${ v }_{ d }$) can be found using this formula, ${ v }_{ d }=\frac { eE\tau }{ m }$ (where $e$ is the charge on an electron, $E$ is the electric field inside the wire, $\tau$ is the average time between the collision of electrons, and $m$ is the mass of an electron). Since $E=\frac { V }{ l } $, the first equation rearranges to ${ v }_{ d }=\frac { eV\tau }{L m }$. From this, it is clearly visible that the drift velocity is inversely proportional to the length of a wire, so if the length of the wire doubles, the drift velocity is halved.

And yes, it also depends upon the area of the cross-section of the wire, which is clear from the equation you mentioned itself.

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    $\begingroup$ That formula is suitable for a resistor. We tend to model wires as having nearly no resistance (or voltage drop within the circuit). To consider the voltage drop across a conductor as constant when the length changes but current is held constant is incorrect. $\endgroup$
    – BowlOfRed
    Commented Oct 14, 2016 at 16:31
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It will not depend on area bcz as area doubled then current will also doubled but as length doubled the current will be half ...so drift velocity inversely proportional to the length. Use equations- 1...... I=neav 2. ..... R=pa/L Where p =resistivity

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  • $\begingroup$ Why do you say the current is doubled when the original question presumes "if the current is the same"? $\endgroup$
    – BowlOfRed
    Commented Jan 25, 2017 at 7:48
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I think drift velocity depends of length but not on area of cross section This is because v=eE/m where e is electronic charge, E is electric field and m is mass of electron As V/L=E and replacing the value of E in the above equation, we get v inversely proportional to v

Also, it is independent on Area of cross section as it is not included in the above equation Coming to your formula of I=neAv, even I (that is current) doesn't depend upon any of the parameters used to calculate it in the above equation. I only depends on V and R

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  • $\begingroup$ "we get v inversely proportional to v" doesn't make sense to me (may be it is a typo). Also, please consider using MathJax. $\endgroup$ Commented Sep 8, 2019 at 16:04
  • $\begingroup$ I meant v(that is drift velocity) inversely proportional to L(length) $\endgroup$ Commented Sep 19, 2019 at 12:15

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