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

1. TrojanPoem

@Michele_Laino

2. Michele_Laino

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

3. Michele_Laino

relationship*

4. Michele_Laino

$$n$$ is the number of electrons over volume

5. Michele_Laino

and $$j=I/A$$

6. Michele_Laino

I don't understand your coefficient 0.25

7. TrojanPoem

It's written like this in the question

8. TrojanPoem

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

9. Michele_Laino

yes!

10. Michele_Laino

from my steps, i can write this: $$I=n_oevA$$

11. TrojanPoem

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

12. TrojanPoem

I can't be right in anyway.

13. 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= $\frac{{coulombs}}{{Kg}} \cdot \frac{m}{{\sec }} \cdot {m^2}$

14. Michele_Laino

which is not a current

15. TrojanPoem

coulombs ?

16. Michele_Laino

yes! I measure $$e$$ in coulombs

17. TrojanPoem

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

18. TrojanPoem

A * T

19. Michele_Laino

A is the cross sectional area of your conductor

20. TrojanPoem

Yeah.

21. Michele_Laino

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

22. Michele_Laino

is 0.25 a quantity with a specific unit of measure?

23. Michele_Laino

for example, can 0.25 be a density?

24. 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

25. TrojanPoem

Do you suspect that 0.25 has a unit ?

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

27. Michele_Laino

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

28. TrojanPoem

What does dn/dm mean?

29. Michele_Laino

it is the number of electrons over mass of conductor

30. TrojanPoem

I looks like a derivative xD

31. Michele_Laino

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

32. 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.

33. Michele_Laino

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

34. Michele_Laino

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

35. TrojanPoem

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

36. Michele_Laino

I think it is wrong!

37. TrojanPoem

So it's wrong :) thanks again again

38. Michele_Laino

:)

39. Michele_Laino

I understand the presence of 0.25

40. 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

41. Michele_Laino

here is why there is the coefficient 0.25

42. TrojanPoem

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

43. Michele_Laino

yes!

44. TrojanPoem

45. Michele_Laino

46. 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

47. TrojanPoem

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

48. Michele_Laino

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

49. Michele_Laino

right side*

50. TrojanPoem

so wrong when the constant is wrong ?

51. Michele_Laino

yes I think so!

52. 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.

53. Michele_Laino

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

54. TrojanPoem

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

55. TrojanPoem

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

56. TrojanPoem

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

57. 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

58. TrojanPoem

59. TrojanPoem

I have combined two parts of two questions

60. Michele_Laino

no worries! :)

61. TrojanPoem

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

62. Michele_Laino

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

63. TrojanPoem

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

64. TrojanPoem

so as you said it's wrong

65. TrojanPoem

What's the difference between G , g ?