Question

I know that the drift speed can be found by v=J/ne where J is the current density and ne is the carrier charge density.
My question is that is ne constant regardless of the cross-sectional area?

Answer 1

I think no, since "n" is the density of free electrons with respect to the volume of the conductor

Answer 2

Then, would the drift speed be inversely proportional to the cross-sectional area?

Answer 3

no, since J is inversely proportional to the cross sectional area too

Answer 4

the drift speed is proportional to the distance traveled by our free electrons

Answer 5

Wait. Since drift speed is proportional to the current density, and the current density is inversely proportional to the cross-sectional area, shouldn't the drift speed be inversely proportional to the cross-sectional area?

Answer 6

I think that the reasoning is as below:
inside a conductor, we can write:
\[\Large {\mathbf{J}} = \sigma {\mathbf{E}}\]
so what we have to consider as constant is the electric field inside the conductor

Answer 7

\sigma is the conductibility of that conductor

Answer 8

now we can write J as follows:
\[\Large {\mathbf{J}} = Ne{\mathbf{v}}\]

Answer 9

Then, using this, how can I solve this problem?

Figure shows a rectangular solid conductor of edge lengths L, 2L, and 3L. A potential difference V is to be applied uniformly between pairs of opposite faces of the conductor. First, V is applied between the left–right faces, then between the top–bottom faces, and then between the front–back faces. Rank those pairs, greatest first, according to the drift speeds of the electrons.

Answer 10

from which we can see that J is proportional to a length, since, dimensionally speaking we have:
\[v = \frac{J}{{Ne}} = \frac{{Coulomb/{m^2}}}{{\left( {1/{m^3}} \right)Coulomb}} \sim m\]

Answer 11

okay

Answer 12

I try to solve the first case, namely V is applied between left-right faces

Answer 13

You meant to say that v is proportional to the length, right?

Answer 14

I mean the length of the space traveled by free electrons

Answer 15

alright

Answer 16

But, isn't the dimensions for the current density A/m^2?

Answer 17

sorry you are right!
\[v = \frac{J}{{Ne}} = \frac{{Coulomb/\left( {s \times {m^2}} \right)}}{{\left( {1/{m^3}} \right)Coulomb}} \sim \frac{m}{s}\]

Answer 18

as we can see the drift speed is proportional to the space traveled by free electrons inside an unitary interval of time

Answer 19

okay

Answer 20

ok! for your first case, we can write:
\[J = \sigma \frac{V}{{3L}}, \Rightarrow nev = \sigma \frac{V}{{3L}}\]

Answer 21

now we can suppose to have N free electrons, inside your conductor, so we can write:
\[n = \frac{N}{{volume}} = \frac{N}{{L \times 2L \times 3L}}\]

Answer 22

so substituting we have:
\[\Large v = \sigma \frac{V}{{3L}}\frac{1}{{ne}} = \sigma \frac{V}{{3L}}\frac{{6{L^3}}}{{Ne}} = \sigma \frac{{2V}}{3}\frac{{{L^2}}}{{Ne}}\]

Answer 23

okay

Answer 24

or equivalently:
\[\Large v = \sigma \frac{V}{3}\frac{A}{{Ne}},\quad {\text{since}}:A = 2{L^2}\]

Answer 25

namely, the speed drift is proportional to the cross sectional area