TrojanPoem
  • TrojanPoem
If I is the electric current, A the area of the conductor, the number of electrons for the unit of mass is n_o the speed of this electrons is v so the equation I = 0.25 n_oeva is
Physics
jamiebookeater
  • jamiebookeater
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TrojanPoem
  • TrojanPoem
Michele_Laino
  • Michele_Laino
the relòationship between the current density \(j\) and the number of electrons, is: \(j=nev\), where \(e=1.6 \cdot 10^{-19} coulombs\)
Michele_Laino
  • Michele_Laino
relationship*

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Michele_Laino
  • Michele_Laino
\(n \) is the number of electrons over volume
Michele_Laino
  • Michele_Laino
and \(j=I/A\)
Michele_Laino
  • Michele_Laino
I don't understand your coefficient 0.25
TrojanPoem
  • TrojanPoem
It's written like this in the question
TrojanPoem
  • TrojanPoem
so the unit of n_o is 1/kg in (SI) ?
Michele_Laino
  • Michele_Laino
yes!
Michele_Laino
  • Michele_Laino
from my steps, i can write this: \(I=n_oevA\)
TrojanPoem
  • TrojanPoem
M.L^2T^-2* M^-1* L * T^-2 * L^2 L^4 T^-4 ?
TrojanPoem
  • TrojanPoem
I can't be right in anyway.
Michele_Laino
  • Michele_Laino
I'm thinking at your formula, and it can not be correct if \(a\) is the area and \(e\) is the electron charge since we have this: right side= \[\frac{{coulombs}}{{Kg}} \cdot \frac{m}{{\sec }} \cdot {m^2}\]
Michele_Laino
  • Michele_Laino
which is not a current
TrojanPoem
  • TrojanPoem
coulombs ?
Michele_Laino
  • Michele_Laino
yes! I measure \(e\) in coulombs
TrojanPoem
  • TrojanPoem
I meant, why don't you replace it with A * S
TrojanPoem
  • TrojanPoem
A * T
Michele_Laino
  • Michele_Laino
A is the cross sectional area of your conductor
TrojanPoem
  • TrojanPoem
Yeah.
Michele_Laino
  • Michele_Laino
\(v\) is the average speed of conduction electrons, so we have: \(I=0.25n_0\;e\;v\;A=\) \[0.25 \cdot \frac{{coulombs}}{{Kg}} \cdot \frac{m}{{\sec }} \cdot {m^2}\]
Michele_Laino
  • Michele_Laino
is 0.25 a quantity with a specific unit of measure?
Michele_Laino
  • Michele_Laino
for example, can 0.25 be a density?
TrojanPoem
  • TrojanPoem
In dimension problems, It's not necessary to bring to us an equation which we know. Today was the first time to see this. but I think 0.25 is just a constant like last 2 pi
TrojanPoem
  • TrojanPoem
Do you suspect that 0.25 has a unit ?
Michele_Laino
  • Michele_Laino
yes! since the right side of your formula, is not a current, whereas if 0.25 is a density, then its unit of measure is \(Kg/m^3\), so we can write this: \[\begin{gathered} 0.25 \cdot \frac{{coulombs}}{{Kg}} \cdot \frac{m}{{\sec }} \cdot {m^2} = \frac{{Kg}}{{{m^3}}} \cdot \frac{{coulombs}}{{Kg}} \cdot \frac{m}{{\sec }} \cdot {m^2} = \hfill \\ \hfill \\ = \frac{{coulombs}}{{\sec }} = amperes \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
furthermore if \(\rho=m/V\), then we can write this: \[\frac{{dn}}{{dm}} = \frac{{dn}}{{\rho dV}} = \frac{1}{\rho }\frac{{dn}}{{dV}}\]
TrojanPoem
  • TrojanPoem
What does dn/dm mean?
Michele_Laino
  • Michele_Laino
it is the number of electrons over mass of conductor
TrojanPoem
  • TrojanPoem
I looks like a derivative xD
Michele_Laino
  • Michele_Laino
yes! whereas \(dn/dV\) is my density of electrons, namely number of electrons over volume of conductor
TrojanPoem
  • TrojanPoem
You said: "since the right side of your formula, is not a current, whereas if 0.25 is a density, then its unit of measure is Kg/m3" Well that's convincing but how to ensure so ? the question is as written here.
Michele_Laino
  • Michele_Laino
It was my hypothesis, since it is the first time that I see your formula
Michele_Laino
  • Michele_Laino
when I was at university, I have always used this one: \[j = {n_0}ev\]
TrojanPoem
  • TrojanPoem
choices: a) correct,b) correct according to dimensions c) wrong d)wrong " " "
Michele_Laino
  • Michele_Laino
I think it is wrong!
TrojanPoem
  • TrojanPoem
So it's wrong :) thanks again again
Michele_Laino
  • Michele_Laino
:)
Michele_Laino
  • Michele_Laino
I understand the presence of 0.25
Michele_Laino
  • Michele_Laino
if we have this formula: \[I = {n_0}evA\] with \(n_0\) as number of electrons over mass, we can say that it is wrong according to the dimensions. nevertheless, due to the presence of the coefficient 0.25, we have to say that your formula is wrong, since using dimensional analysis we are not able to write numerical coefficients
Michele_Laino
  • Michele_Laino
here is why there is the coefficient 0.25
TrojanPoem
  • TrojanPoem
So 0.25 makes the formula wrong not wrong according to the dimensions ?
Michele_Laino
  • Michele_Laino
yes!
TrojanPoem
  • TrojanPoem
I tried googling this formula and no results
Michele_Laino
  • Michele_Laino
please wait
Michele_Laino
  • Michele_Laino
I'm sorry, as I wrote before, using units of measure, the right side is not a current, so I think, that your formula is wrong according to dimensions
TrojanPoem
  • TrojanPoem
so when to say wrong, when to say wrong according to dimensions
Michele_Laino
  • Michele_Laino
I say wrong according dimensions when, as in our case, the right has the wrong dimensions
Michele_Laino
  • Michele_Laino
right side*
TrojanPoem
  • TrojanPoem
so wrong when the constant is wrong ?
Michele_Laino
  • Michele_Laino
yes I think so!
TrojanPoem
  • TrojanPoem
But there is no way to know that In the last question you said correct as you knew that T= 2 pi sqrt(L/g) you know the formula but I don't.
Michele_Laino
  • Michele_Laino
for example if I write this formula: \[T = 3\sqrt {\frac{l}{g}} \] what can we conclude?
TrojanPoem
  • TrojanPoem
Wrong, but if It's the first time to see it ?
TrojanPoem
  • TrojanPoem
wait wait, in Problem #21 , Prove that I = n_oeva according to dimensions
TrojanPoem
  • TrojanPoem
See the 6th formula in this table: https://quizlet.com/3348670/physics-formulae-and-constants-flash-cards/
Michele_Laino
  • Michele_Laino
it is wrong according to dimensions, since N is the number of free electrons/m^3, whereas A is the cross sectional area so m^2
TrojanPoem
  • TrojanPoem
I made a mistake :o
TrojanPoem
  • TrojanPoem
I have combined two parts of two questions
Michele_Laino
  • Michele_Laino
no worries! :)
TrojanPoem
  • TrojanPoem
n is the number of electrons for the unit of volume , SORRY
Michele_Laino
  • Michele_Laino
and A is the cross sectional area whose unit of measure is m^2
TrojanPoem
  • TrojanPoem
so it's I = neva L^-3 * A * T * L * T^-1 * L^2 = A
TrojanPoem
  • TrojanPoem
so as you said it's wrong
TrojanPoem
  • TrojanPoem
What's the difference between G , g ?

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