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Does a thicker wire mean more charge carriers in I = n A q V?

If you have two wires A & B made of same material with A being twice as thick as B, when using the formula I = n A q V will the value of n and Q remain the same for both wires or will they change?

I think because the A wire is more thick it therefore has more valence electrons that can carry charge and therefore the number of charge carriers should increase compared to wire B. But i think i heard someone say that n & q remains constant and only A & V value changes.
the free electrons ∝ volume of metal... so if twice as thick then twice as many free electrons.
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Kyx
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Original post by Jyashi
If you have two wires A & B made of same material with A being twice as thick as B, when using the formula I = n A q V will the value of n and Q remain the same for both wires or will they change?

I think because the A wire is more thick it therefore has more valence electrons that can carry charge and therefore the number of charge carriers should increase compared to wire B. But i think i heard someone say that n & q remains constant and only A & V value changes.


The answer:

Original post by the bear
the free electrons ∝ volume of metal... so if twice as thick then twice as many free electrons.


:h:
Original post by the bear
the free electrons ∝ volume of metal... so if twice as thick then twice as many free electrons.
Not quite because surface area is a product function and volume is cubic. Hence doubling the diameter of a cylindrical cable will increase the surface area in the ratio of π(d2)2\pi (\frac{d}{2})^2 and the volume by Lπ(d2)2L\pi(\frac{d}{2})^2
(edited 8 years ago)
Original post by uberteknik
Not quite because surface area is a product function and volume is cubic. Hence doubling the diameter of a cylindrical cable will increase the surface area in the ratio of π(d2)2\pi (\frac{d}{2})^2 and the volume by Lπ(d2)2L\pi(\frac{d}{2})^2


but the charge carriers are inside the wire ?

so this would be like the number of currants in spotted dick... twice the volume, twice the currants.
Original post by the bear
but the charge carriers are inside the wire ?

so this would be like the number of currants in spotted dick... twice the volume, twice the currants.
Current is the rate of flow of charge.

The analogy with fluid being litres per second.

Resistance is a function of the cross sectional area of the conductor.

i.e. increase the c.s.a. and the volume of water needed to give the same flow rate (current) decreases accordingly with the fluid pressure (e.m.f.) remaining constant.

In your analogy that would be the number of currants from the spotted dick passing into a mouth per second - open the mouth wider and more currants get in per second without needing to push (pressure) them in faster.
(edited 8 years ago)
Original post by uberteknik
Current is the rate of flow of charge.

The analogy with fluid being litres per second.

Resistance is a function of the cross sectional area of the conductor.

i.e. increase the c.s.a. and the volume of water needed to give the same flow rate (current) decreases accordingly with the fluid pressure (e.m.f.) remaining constant.

In your analogy that would be the number of currants from the spotted dick passing into a mouth per second - open the mouth wider and more currants get in per second without needing to push (pressure) them in faster.


thank you for that; there will be a limiting radius beyond which no more dick can be inserted whatever the pressure.
:yep:
Original post by uberteknik

In your analogy that would be the number of currants from the spotted dick passing into a mouth per second - open the mouth wider and more currants get in per second without needing to push (pressure) them in faster.


Da*** did I just read
Original post by Student403
Da*** did I just read
It's all in the best possible taste.
Original post by uberteknik
It's all in the best possible taste.


:colonhash: I hope that pun was not intended
Original post by uberteknik
It's all in the best possible taste.


You sick sick man!
I like you
Original post by the bear
the free electrons ∝ volume of metal... so if twice as thick then twice as many free electrons.


free electrons / unit length ∝ cross-sectional area of the wire

double the thickness, you quadruple the cross-sectional area - so there's 4x as many free electrons.
Original post by Student403
:colonhash: I hope that pun was not intended
:colone:

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