Electric Fields inside circuits and conductors / Faraday's law - Page 1
Forum section: electronics / projects

author: ElectronicsGeek
author_rank: Contributor
timestamp: July 12, 2017, 06:07:35 pm
content: Alright,so the electric field is the negative gradient of the voltage: E = -grad(V)So, if the electric field at a point is a vector pointing along the direction R, then the voltage along that direction is always decreasing, that is, the electric field points along the direction of decreasing voltage, so that at infinity the voltage is zero.Of course the gradient operator always gives you the direction of maximum increase, so by assigning it a negative sign, we point to the other direction of maximum decrease, and this is because we use the infinity reference as zero.So far so good.If we now look at an electronic circuit, where we have a voltage source and a resistor, and the polarity of the resistor voltage is opposite that of the source, then applying kirchiff's law along this loop has two outcomes:First, if you go from negative to positive on the voltage source, then you get a positive sign, since the voltage is higher at the + side, then crossing the resistor gives you a negative sign.This is the way I prefer doing it, since the electric field goes from + to -, and so going from - to + on the voltage source means you are going to a place of higher potential as you go against the electric field an do work against it, hence increasing your potential energy. Then you go across the resistor and you go along the electric field, hence losing potential energy.However in faraday's law, you have the integral of the electric field along a closed loop being equal to the negative time derivative of the magnetic flux penetrating an open surface. And if you apply faraday's law to a circuit with an inductor, then it only works if you assume that going against the electric field gives you a voltage drop, and going in the same direction as the field gives you a positive drop, or an increase in voltage, and then the total is -dPhi/dt.But clearly the electric field points along decreasing voltage, so faraday's law has a minus sign on the dPhi/dt, if you go with this convention. Why is it this way?If you use the opposite convention the minus sign disappears.Faradays law in a resistor inductor circuit gives -V + Vr = -Ldi/dt, and this is opposite the convention of decreasing if you go in the direction of the electric field. That equation is equivalent to V - Vr = L di/t, which is in accordance with E = -grad(V). So why is faradays law the opposite ?Can anyone explain where I'm going wrong?Thank you

author: ElectronicsGeek
author_rank: Contributor
timestamp: July 12, 2017, 07:42:58 pm
content: It seems nobody knows it here

author: DaJMasta
author_rank: Super Contributor
timestamp: July 12, 2017, 08:18:15 pm
content: The wording and explanation is both extremely dense and sort of confusing to try and sift through, so I imagine many people just haven't tried to stick it out and give you a reply.I would caution you in the way you're applying your theorems to an inductor with an unspecified voltage source. Since we're talking mathematics, I assume you're thinking of ideal parts, in which case an inductor with a DC voltage source has no impedance and the inductance makes no matter because you are analyzing in a window between the power on and power off states when it is assumed to be stabilized. Thus meaning you just have a circuit with a lossless short on the voltage source.You're also talking a lot about an electric field.... and while you can translate to those terms, I think conceptually, this is a much more difficult way to consider a basic circuit than in terms of voltage, current, power, etc. I think some of the confusing-ness of the original post for me is that when you're talking about electric field and inductor... you are not talking about just regular circuit potential and current, you are talking about the field lines coming out from the wires of the circuit. This is the way I'm used to hearing "electric field" be described, but this is a very roundabout way of trying to evaluate the flow of charge in a closed loop, DC circuit and I think that's tripping up your understanding somewhere.

author: orin
author_rank: Frequent Contributor
timestamp: July 12, 2017, 08:50:31 pm
content: There isnoE field in the inductor, other than that due to its resistance. The tinkering with the sign of Ldi/dt is done to make Faraday's Law look like Kirchoff's 2nd Law. If you can't remember which way the sign goes, remember it is always such as to oppose a change in current.This should help:http://web.mit.edu/8.02/www/Spring02/lectures/lecsup4-1.pdfIt immediately dismisses E = -grad(V) for circuits with time varying magnetic fields and goes on to use Faraday's Law to analyze the LR circuit.

author: ElectronicsGeek
author_rank: Contributor
timestamp: July 12, 2017, 09:23:12 pm
content: Quote from: orin on July 12, 2017, 08:50:31 pmThanks buddy. That solves my problem. If E != -gradV then who cares what direction the electric field gives a positive voltage drop isn't it mate?Good on you mate, you got it. Do you have a boyfriend by the way ?