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(Most Probably a duplicate, but I couldn't find it)

I have heard about 'Skin Effect' in AC, but currently Im only and only concerned about DC.

When current flows in simple DC circuit, does the current flow slower at the central region of the wire cross-section? (Like the picture shows)

In other words is the current density higher towards the edges?

OR, Is it equal density throughout?

If it does flow faster towards edges, what causes this to happen? What hindrance is arising at central region?

Also, how much is this difference? Like is the density at edges 10 or 100 times more than centre, or is it like 10% more only.

[My very uneducated guess would be that maybe most energy transferred to wire (according to Poynting Theorem) is procured by the electrons on outer edges and doesn't reach central electrons much]

enter image description here

Rohit Shekhawat
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    have a look at my answer here for links https://physics.stackexchange.com/questions/376452/why-is-current-slowed-down-by-resistance – anna v Mar 25 '23 at 18:24
  • Thanks but I couldnt find answer to my question there. Can you give a short yes/no answer with one line explanation here? – Rohit Shekhawat Mar 26 '23 at 11:48
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    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/miccur.html the link has the formula: It is proportional to the area A , with no functional dependence of the area on the radius. This means it is assumed uniform. – anna v Mar 26 '23 at 12:04

2 Answers2

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Suppose the wire is along the $z$-axis, with uniform cross section and conductance. From Faraday's law, $$ \vec \nabla\times\vec E=\left(\frac{\partial E_z}{\partial y} - \frac{\partial E_y}{\partial z}\right)\hat x + \left(\frac{\partial E_x}{\partial z} - \frac{\partial E_z}{\partial x}\right)\hat y + \left(\frac{\partial E_x}{\partial y} - \frac{\partial E_y}{\partial x}\right)\hat z=0.$$ Since the wire is uniform along $z$, the E-field should not vary along $z$, i.e. $\partial\vec E/\partial z = 0$. This means $$ \vec \nabla\times\vec E=\frac{\partial E_z}{\partial y}\hat x - \frac{\partial E_z}{\partial x}\hat y + \left(\frac{\partial E_x}{\partial y} - \frac{\partial E_y}{\partial x}\right)\hat z=0.$$ From the $x$ and $y$ components, we have $$\frac{\partial E_z}{\partial y}=\frac{\partial E_z}{\partial x}=0.$$ Since the component of the E-field along the wire is uniform across the wire cross section, so is the current density $J_z = \sigma E_z$ (Ohm's law).

Puk
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A metal conductor, usually Cu, is a polycrystalline material. The conductivity is greater inside each individual crystal compared with grain boundaries. The free surface also has a lower conductivity. But it is not relevant for normal electrical wires. However it is important for very thin conductors, necessary for micro-electronics. In this case, the ratio surface to bulk is much higher than for commercial wires. This articles discusses the issue.

Claudio Saspinski
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