Question

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

Answer 1

@Michele_Laino

Answer 2

the relòationship between the current density \(j\) and the number of electrons, is:
\(j=nev\), where \(e=1.6 \cdot 10^{-19} coulombs\)

Answer 3

relationship*

Answer 4

\(n \) is the number of electrons over volume

Answer 5

and \(j=I/A\)

Answer 6

I don't understand your coefficient 0.25

Answer 7

It's written like this in the question

Answer 8

so the unit of n_o is 1/kg in (SI) ?

Answer 9

yes!

Answer 10

from my steps, i can write this:
\(I=n_oevA\)

Answer 11

M.L^2T^-2* M^-1* L * T^-2 * L^2
L^4 T^-4 ?

Answer 12

I can't be right in anyway.

Answer 13

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=
\[\frac{{coulombs}}{{Kg}} \cdot \frac{m}{{\sec }} \cdot {m^2}\]

Answer 14

which is not a current

Answer 15

coulombs ?

Answer 16

yes! I measure \(e\) in coulombs

Answer 17

I meant, why don't you replace it with A * S

Answer 18

A * T

Answer 19

A is the cross sectional area of your conductor

Answer 20

Yeah.

Answer 21

\(v\) is the average speed of conduction electrons, so we have:
\(I=0.25n_0\;e\;v\;A=\)
\[0.25 \cdot \frac{{coulombs}}{{Kg}} \cdot \frac{m}{{\sec }} \cdot {m^2}\]

Answer 22

is 0.25 a quantity with a specific unit of measure?

Answer 23

for example, can 0.25 be a density?

Answer 24

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

Answer 25

Do you suspect that 0.25 has a unit ?

Answer 26

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:
\[\begin{gathered}
0.25 \cdot \frac{{coulombs}}{{Kg}} \cdot \frac{m}{{\sec }} \cdot {m^2} = \frac{{Kg}}{{{m^3}}} \cdot \frac{{coulombs}}{{Kg}} \cdot \frac{m}{{\sec }} \cdot {m^2} = \hfill \\
\hfill \\
= \frac{{coulombs}}{{\sec }} = amperes \hfill \\
\end{gathered} \]

Answer 27

furthermore if \(\rho=m/V\), then we can write this:
\[\frac{{dn}}{{dm}} = \frac{{dn}}{{\rho dV}} = \frac{1}{\rho }\frac{{dn}}{{dV}}\]

Answer 28

What does dn/dm mean?

Answer 29

it is the number of electrons over mass of conductor

Answer 30

I looks like a derivative xD

Answer 31

yes! whereas \(dn/dV\) is my density of electrons, namely number of electrons over volume of conductor

Answer 32

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.

Answer 33

It was my hypothesis, since it is the first time that I see your formula

Answer 34

when I was at university, I have always used this one:
\[j = {n_0}ev\]

Answer 35

choices:
a) correct,b) correct according to dimensions c) wrong d)wrong " " "

Answer 36

I think it is wrong!

Answer 37

So it's wrong :) thanks again again

Answer 38

:)

Answer 39

I understand the presence of 0.25

Answer 40

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

Answer 41

here is why there is the coefficient 0.25

Answer 42

So 0.25 makes the formula wrong not wrong according to the dimensions ?

Answer 43

yes!

Answer 44

I tried googling this formula and no results

Answer 45

please wait

Answer 46

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

Answer 47

so when to say wrong, when to say wrong according to dimensions

Answer 48

I say wrong according dimensions when, as in our case, the right has the wrong dimensions

Answer 49

right side*

Answer 50

so wrong when the constant is wrong ?

Answer 51

yes I think so!

Answer 52

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.

Answer 53

for example if I write this formula:
\[T = 3\sqrt {\frac{l}{g}} \]
what can we conclude?

Answer 54

Wrong, but if It's the first time to see it ?

Answer 55

wait wait, in Problem #21 , Prove that I = n_oeva according to dimensions

Answer 56

See the 6th formula in this table: https://quizlet.com/3348670/physics-formulae-and-constants-flash-cards/

Answer 57

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

Answer 58

I made a mistake :o

Answer 59

I have combined two parts of two questions

Answer 60

no worries! :)

Answer 61

n is the number of electrons for the unit of volume , SORRY

Answer 62

and A is the cross sectional area whose unit of measure is m^2

Answer 63

so it's
I = neva
L^-3 * A * T * L * T^-1 * L^2 = A

Answer 64

so as you said it's wrong

Answer 65

What's the difference between G , g ?