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I wish to generalise the equations governing electromagnetic induction. Consider a loop whose size is changing, and is subject to a changing magnetic field. Then,

Definitions: $$\phi \equiv\int_{S(t)}\mathbf{B(t)}\cdot d\mathbf{a} \tag{1.1}$$ $$\mathrm{emf} \equiv\oint \mathbf{f}\cdot d\mathbf{l} = \oint \mathbf{(E+v\times B)}\cdot d\mathbf{l} \tag{1.2}$$ Laws: $$e=-\dfrac{d\phi}{dt}\tag{2.1}$$ $$\nabla \times \mathbf{E}=-\dfrac{\partial \mathbf{B}}{\partial t}\tag{2.2}$$

If the loop is fixed ($v=0$), and the magnetic field changes with time, then it's the $\oint \mathbf{E}\cdot d\mathbf{l}$ that produces the emf (transformer emf).

If the loop boundary changes (for e.g. a metal rod sliding on metallic rails, with speed $v$) and the magnetic field doesn't, then $\nabla \times \mathbf{E}=0$, so $\oint \mathbf{E}\cdot d\mathbf{l}=0$, and only $\oint (\mathbf{v \times B})\cdot d\mathbf{l}$ is responsible for the emf (motional emf).

For these two cases individually, it seems so far so good.

Question (part a): Is this generalisation indeed correct?

What's particularly bothering me is the equation $\nabla \times \mathbf{E}=-\dfrac{\partial \mathbf{B}}{\partial t}$. If we were to put it in the integral form, (using Stokes' theorem), there seems to be no reference to $v$ or $B$, which seems to be incorrect as per definition $(1.2)$.

A possible explanation for this is that Stokes' theorem does not hold when the surfaces/boundaries in question also change.

The lack of literature on the most general case ($B$ changing and the loop changing), makes it difficult to verify this generalisation completely, and brings me to:

Question (part b): Using the "definitions", can one of the "laws" be derived? By assuming the other one as an axiom? (or shockingly, derive both the "laws" from simply the definitions: I am almost certain this is blasphemous)

The basis of this question is more or less mathematical in nature.

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  • $\begingroup$ "there seems to be no reference to V or B, which seems to be incorrect as per definition (1.2)." could you elaborate on this? how do you envision V and B to come $\endgroup$ Commented Mar 2, 2021 at 14:58
  • $\begingroup$ $\mathrm{emf} \equiv\oint \mathbf{f}\cdot d\mathbf{l} = \oint \mathbf{(E+v\times B)}\cdot d\mathbf{l} $ is my equation 1.2. $\endgroup$
    – satan 29
    Commented Mar 3, 2021 at 5:03
  • $\begingroup$ In definition 1.2, there is no V or B, I got the equation you wrote but not about how you want to see V and and B in the equation. Or did you mean 'v' when you wrote 'V' $\endgroup$ Commented Mar 3, 2021 at 5:42
  • $\begingroup$ I meant v, as in the velocity, and not V as in the potential.Edited. $\endgroup$
    – satan 29
    Commented Mar 3, 2021 at 6:02
  • $\begingroup$ Did my answer solve your query or did I misunderstand something in the question? @satan 29 $\endgroup$ Commented Mar 4, 2021 at 10:03

2 Answers 2

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What's particularly bothering me is the equation $\nabla \times \mathbf{E}=-\dfrac{\partial \mathbf{B}}{\partial t}$. If we were to put it in the integral form, (using Stokes' theorem), there seems to be no reference to $V$ or $B$, which seems to be incorrect as per definition $(1.2)$.

$$ \oint \vec{E} \cdot \vec{dl} = \int_{S} -(\frac{\partial }{\partial t} \vec{B})\vec{dS}\tag{1}$$

This is the version of the equation you write the maxwell equation in integral form, the miraculous part of the equation is that it is a nature is independent of any 'physical loop' , it is a nature of space and the fields. Particularly speaking, original EMF equation had a loop term associated due to the $\vec{v} \times \vec{B}$ term, the velocity is a property of the physical object.

Question (part b): Using the "definitions", can one of the "laws" be derived? By assuming the other one as an axiom? (or shockingly, derive both the "laws" from simply the definitions: I am almost certain this is blasphemous)

from equation 1.2 and 2.1, we can write:

$$ -\dfrac{d \int_S \vec{B} \cdot \vec{dS}}{dt}= \oint \mathbf{(E+v\times B)}\cdot \vec{d\mathbf{l}} \tag{2}$$

So, here's the deal for the LHS term, you are differentiating the whole integral, this means you have to take contributions from the loop changing (boundary of surface integral) and also the magnetic field changing, after some work you will find that:

$$ \frac{d }{d t}\int_S \vec{B} \cdot \vec{dS} = -\oint \vec{v} \times \vec{B} \cdot \vec{dl} + \int_S \frac{\partial \vec{B} }{\partial t} \vec{dS} \tag{3}$$

Now plugging in the expression (3) into (2) you will get back equation (1).


Discussed from 56:23 in this lecture on Faraday's law by professor Shankar.

Also there is a proof on wiki but it's sort of hidden, you have to click a button written 'show'

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  • $\begingroup$ It may be noted that the differentiation of integral is actually general leibniz rule $\endgroup$ Commented Mar 3, 2021 at 6:09
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Electromotive force (EMF) already has established meaning that includes non-EM sources, like thermal, chemical gradients (batteries). So I don't recommend redefining EMF in terms of your equation 1.2. That equation is a physical law of limited validity: when the EMF is purely due to external EM field, and there are no thermal, chemical gradient or other sources of EMF.

Maxwell's equation

$$ \nabla \times \mathbf E = - \frac{\partial \mathbf B}{\partial t} $$ being valid everywhere is equivalent to the integral equation $$ \oint_{\gamma} \mathbf E \cdot d\mathbf l = - \frac{d}{dt}\int_S\mathbf B\cdot d\mathbf S $$

being valid for every closed curve $\gamma$ and surface $S$ that has the curve $\gamma$ as boundary.

These equations are universally valid, whether the left-hand side is total EMF or not. It is when loop is stationary and EMF is purely due to induced electric field; it is not if loop moves in presence of magnetic field, or if there is a thermal/chemical gradient in some part of the loop (battery).

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  • $\begingroup$ I think the E should be a B in the integral equation @Jan Lalinsky $\endgroup$ Commented Mar 4, 2021 at 10:58
  • $\begingroup$ Right, fixed. Thanks. $\endgroup$ Commented Mar 4, 2021 at 13:53
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    $\begingroup$ The integral form is equivalent to the differential form, provided the loop $\gamma$ is stationary. See Buraian's answer. $\endgroup$ Commented Jun 26, 2021 at 14:15

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