Timeline for Why is current a scalar quantity?
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19 events
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| Jan 7 at 18:37 | comment | added | Daniel Davee | It seems most people are confused by this. It's a vector for sure. The addition seems like it's a scalar addition only because the wires are 1 dimensional. Maxwell equations are explicitly vector equations. But we hardly ever think about current in higher dimensions. So consider water flow instead. In a river flow also adds like a scalar, but on a flood plane or in the ocean, it behaves like a vector. Attach a circuit to a sphere, anode on one side cathode on the other. On the surface of the sphere, the current would look like a vector field that obeys Gauss's law and the hairy ball theorem. | |
| Sep 17, 2020 at 19:15 | comment | added | Antonios Sarikas | @JerrySchirmer Your two previous answers covered my doubts. It is the net charge therefore and also the direction is a matter of convention. If we think two boxes and a net charge flowing from one box to the other then at every moment we can calculate their amount of charge. If we have a positive flux (convention from left to right) of 5 C/s then the left box will decrease its amount by 5 C/s whereas the right will increase it by 5. | |
| Sep 17, 2020 at 16:16 | comment | added | Zo the Relativist | @AntoniosSarikas, if you're talking about through a closed surface, yes. typically, when we talk about current, we are not talking about a closed surface, more like the cross-section of a wire. | |
| Sep 10, 2020 at 22:18 | comment | added | Zo the Relativist | Another way of thinking of it: I have a box. The box contains 20 C of charge. The current is how much charge is entering or leaving the box per unit time. I can get to 21 C by either adding 1 C worth protons, or removing 1 C worth of electrons, but either way, the net charge of the box is 21 C, and so the rate of change of that charge won't care whether I'm adding protons, removing electrons, or some combination of the two. It just cares about my net charge. | |
| Sep 10, 2020 at 17:44 | comment | added | Zo the Relativist | @AntoniosSarikas as far as current is concerned, the flow of an electron is just the same thing as the flow of an antiproton. So, if you have 1 C/s of rightward flow of electrons and 1 C/s of leftward flow of protons, you just have a current of 2 C/s leftward | |
| Sep 10, 2020 at 17:37 | comment | added | Antonios Sarikas | @JerrySchirmer So the sign for the derivative $\frac{dq}{dt}$ should be such that to compensate the fact that we have motion in both directions right? Because if we consider just charge (i.e. we use the same sign for flow of electron and protons) then is the same as flow of bananas, balls, rain etc. If we have the same amount of charge moving in left and right through a surface then the above integral would be zero. | |
| Sep 10, 2020 at 15:49 | comment | added | Zo the Relativist | @AntoniosSarikas, no, because net charge is crossing the surface in question. The current is zero if an equal number density of protons and electrons are moving in the <i>same</i> direction at the same speed.. | |
| Sep 10, 2020 at 12:32 | comment | added | Antonios Sarikas | @JerrySchirmer Current density is a vector. When we multiply with the unit vector perpendicular to the differential area then we will get either a positive or a negative flow of rate. So when both electrons and protons are moving in opposite directions is the current zero? | |
| Jan 29, 2015 at 11:29 | history | protected | CommunityBot | ||
| Jan 29, 2015 at 6:08 | answer | added | Zo the Relativist | timeline score: 10 | |
| Dec 5, 2014 at 17:52 | comment | added | Zo the Relativist | THe direction should really be considered to be "sign" more than a proper vector direction. In particular, Current is defined by defining a surface, and then counting the number of particles that cross that surface per unit time. It only depends on the relative orientation of the surface and the charges, it has no absolute notion of distance. More mathematically, the vector nature of these things is dotted out: $I=\int{\vec j} \cdot d{\vec A}$ | |
| Jul 9, 2014 at 16:01 | answer | added | R004 | timeline score: 33 | |
| Jun 15, 2014 at 11:09 | answer | added | Umar | timeline score: 11 | |
| Jun 15, 2014 at 10:53 | answer | added | krrish nagpal | timeline score: -5 | |
| Dec 21, 2013 at 19:39 | history | edited | Brandon Enright | CC BY-SA 3.0 |
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| Dec 21, 2013 at 19:11 | answer | added | Stan Liou | timeline score: 18 | |
| Dec 21, 2013 at 18:42 | review | First posts | |||
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| Dec 21, 2013 at 18:30 | history | edited | David Z |
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| Dec 21, 2013 at 18:26 | history | asked | user36159 | CC BY-SA 3.0 |