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TrojanPoem
 one year ago
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
TrojanPoem
 one year ago
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

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the relòationship between the current density \(j\) and the number of electrons, is: \(j=nev\), where \(e=1.6 \cdot 10^{19} coulombs\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\(n \) is the number of electrons over volume

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I don't understand your coefficient 0.25

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0It's written like this in the question

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0so the unit of n_o is 1/kg in (SI) ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1from my steps, i can write this: \(I=n_oevA\)

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0M.L^2T^2* M^1* L * T^2 * L^2 L^4 T^4 ?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I can't be right in anyway.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I'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
 one year ago
Best ResponseYou've already chosen the best response.1which is not a current

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! I measure \(e\) in coulombs

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I meant, why don't you replace it with A * S

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1A is the cross sectional area of your conductor

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\(v\) 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
 one year ago
Best ResponseYou've already chosen the best response.1is 0.25 a quantity with a specific unit of measure?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1for example, can 0.25 be a density?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0In 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
 one year ago
Best ResponseYou've already chosen the best response.0Do you suspect that 0.25 has a unit ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! 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
 one year ago
Best ResponseYou've already chosen the best response.1furthermore if \(\rho=m/V\), then we can write this: \[\frac{{dn}}{{dm}} = \frac{{dn}}{{\rho dV}} = \frac{1}{\rho }\frac{{dn}}{{dV}}\]

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0What does dn/dm mean?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1it is the number of electrons over mass of conductor

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I looks like a derivative xD

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! whereas \(dn/dV\) is my density of electrons, namely number of electrons over volume of conductor

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0You 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
 one year ago
Best ResponseYou've already chosen the best response.1It was my hypothesis, since it is the first time that I see your formula

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1when I was at university, I have always used this one: \[j = {n_0}ev\]

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0choices: a) correct,b) correct according to dimensions c) wrong d)wrong " " "

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think it is wrong!

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0So it's wrong :) thanks again again

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I understand the presence of 0.25

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1if 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
 one year ago
Best ResponseYou've already chosen the best response.1here is why there is the coefficient 0.25

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0So 0.25 makes the formula wrong not wrong according to the dimensions ?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I tried googling this formula and no results

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I'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
 one year ago
Best ResponseYou've already chosen the best response.0so when to say wrong, when to say wrong according to dimensions

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I say wrong according dimensions when, as in our case, the right has the wrong dimensions

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0so wrong when the constant is wrong ?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0But 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
 one year ago
Best ResponseYou've already chosen the best response.1for example if I write this formula: \[T = 3\sqrt {\frac{l}{g}} \] what can we conclude?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Wrong, but if It's the first time to see it ?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0wait wait, in Problem #21 , Prove that I = n_oeva according to dimensions

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0See the 6th formula in this table: https://quizlet.com/3348670/physicsformulaeandconstantsflashcards/

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1it 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
 one year ago
Best ResponseYou've already chosen the best response.0I made a mistake :o

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I have combined two parts of two questions

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0n is the number of electrons for the unit of volume , SORRY

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1and A is the cross sectional area whose unit of measure is m^2

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0so it's I = neva L^3 * A * T * L * T^1 * L^2 = A

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0so as you said it's wrong

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0What's the difference between G , g ?
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