Electric Machines, Lenz's Law, and Convention in Standard Textbooks

Question

I am currently reading through PC Sen's Principles of Electric Machines and Power Electronics (3e), and I continue to be confused by the tacit convention that the minus sign in Faraday's Law (and ultimately in the so-called "flux rule" derived therefrom) is omitted. For instance, consider the derivation of hysteresis losses (due to the rotation of magnetic domains in the iron) that he provides, as seen in the first two attached pictures below. Given how the polarity of e and i are defined in the second picture, I would have imagined that e should be negative when i is increasing. The only potential explanation I have is that the e of (1.26) is in fact an e due to a source that is driving the current, and we argue that it must be equal and opposite to the back emf induced by the changing flux (and thus changing current). But why is this so, if indeed I am right? Certainly the author does not mention this crucial detail.

Things are made more confusing when I go to another book. For example, in Analysis of Electric Machinery and Drive Systems by Krause et al., one finds the two pages attached in the last two pictures. Here, once again, there is no minus sign in front of the \$d\lambda/dt\$ term of equation (1.2-3), as would be expected given how polarities are defined for both voltages and fluxes.

What am I missing in both cases? Sorry for the wall of text, but I don't know how else to provide the requisite background for someone who's going to answer.

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Answer 1

May be this simpler example will help. Here is a circuit with Ohm's law written below.

First one is what you usually see. But there is nothing that restricts us from changing positive direction of current. And so we can have Ohm's law with a minus sign.

First thing that you do when analyzing electrical circuit is setup a system of coordinates (reference frame) by selecting positive directions of all values under consideration, that is by drawing arrows.

You can choose any direction of the arrows. You usually select the directions that help your intuition. Hence the minus sign in the Faraday's law. But you actually can't say is it minus or plus before you define arbitrary directions of the current, voltage, flux and also which way you wind the coil.

Answer 2

You are right that Lenz's law states that the induced voltage of an inductor \$e=-N\frac{d\phi}{dt}\$. In this equation, \$e\$ is the total voltage drop across all the inductor windings, which must be driven by an external voltage source. If the external voltage source is at the terminals of the circuit, then its voltage would be equal to \$e\$.

The polarity of the voltage drop is defined so that it already opposes the driving voltage source (as shown above). Since the polarity of \$e\$ is already defined so that it opposes the voltage source, \$e\$ is positive. By defining the polarity in this way, are multiplying \$e=-N\frac{d\phi}{dt}\$ by -1.

Sidenote: all inductors behave in this way and have a voltage drop \$V_L=L\frac{di}{dt}\$, where \$L=\frac{N\phi}{I}\$. In order to model an inductor this way, we assume that a linear B-H curve so that \$I\$ and \$\phi\$ are directly proportional.

Answer 3

Let's assume i is increasing (in clockwise direction in your figure) at time t around its neighborhood .

Based on above assumption , we can find what will be the linked flux through​ coil . And according to Faraday law there will be an induce electric field which causes an induce emf and this induce emf causes a current .

But what should be the direction of current due to induce emf?

From ,Lenz's law -it should be opposite of i because it try to oppose the increasing of i.

So we can say at time t a current Ia is there solely due to induce emf (Ia>0 ,in anticlockwise direction from your figure).

So , $$i= -I_a+I_b$$ , where Ib (>0) is current (solely due to Electrostatic field).

Let's calculate, work done by total electric field (Electrostatic +induce electric field) on moving an unit positive charge along the path of wire from left end of coil to right end of coil = $$iR= (-I_a+I_b)R=-I_aR+I_bR=-\frac{d\phi}{dt}+I_bR$$ Since$$ I_aR=\frac{d\phi}{dt}=L\frac{di}{dt}>0 $$because we assumed already that i is increasing function around neighborhood of t.

Now , let's calculate work done on unit positive charge by external force along the path of wire and starting from -ve terminal of battery moving in clockwise direction to reach again at the same point (360 degree rotation)=W=work done against Electrostatic electric field +work done against induce electric field =V-iR

We know ,work done against Electrostatic electric field =0 because Electrostatic electric field is a conservative field Hence

Work done against induce electric field =Ia R (+ve because path is from left end to right end of coil (since unit charge moving in clockwise direction) =$$\frac{d\phi}{dt}=L\frac{di}{dt} $$

Hence,$$Ia R=V-iR \implies \frac{d\phi}{dt}=V-iR\implies V=iR+\frac{d\phi}{dt}$$

Answer 4

You need to verify this hypothesis :

By convention we always consider the current flowing from positive to the negative, although in reality it is the opposite.

https://www.mi.mun.ca/users/cchaulk/eltk1100/ivse/ivse.htm#:~:text=Conventional%20Current%20assumes%20that%20current,negative%20terminal%20of%20the%20source.&text=Electron%20Flow%20is%20what%20actually,positive%20terminal%20of%20the%20source.

This leads to discrepancies on text books, where electrical engineering textbook will usually consider the flow as positive to negative while more scientific ones will consider the other, resulting in discrepancies in polarity, the minus sign.