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I am reviewing the Maxwell equations hoping to master them quite better than I did when I was on the school bench.

The Faraday's Law of induction leaves me with a question that I can't quite understand.

First Lenz's laws state that the EMF's relationship with the magnetix flux is : $$EMF= -\frac{d\Phi_B}{{dt}}$$

Then, the magentix flux \$\Phi_B\$ would be given by the area integral of the flux density.

$$\Phi_B = \iint_S B\cdot dS $$

Well, now Wikipedia states that the Faraday's law of induction goes as follow.

$$\oint_\Sigma {E \cdot d\ell } = - \int_\Sigma {\frac{\partial B}{{\partial t}}\cdot dA}$$

What confuses me is that the flux derivative is inside the surface integral. A simple substitution of the flux variable in the two first equations would make the flux derivative outside of the area integral.

I know that I can bring the time derivative out of the surface integral if dA is constant, this is purely mathematical. Although, the way I understand the Faraday's equation as it is presented above, a change of the area over time will not affect the EMF.

On the other hand, one of the online demonstrations of professor Walter Lewins explains the case of a sliding copper rod in circuit inducing EMF in a closed loop as the area grows for a constant magnetic field.

If we do this exercise with a constant field B=10and a area that changes, I understand we would get: $$EMF= \oint_\Sigma {E \cdot d\ell } = - \int_\Sigma {\frac{\partial 10}{{\partial t}}\cdot dA(t)} = - \int_\Sigma {0 \cdot dA(t)} = 0$$ Am I wrong ?

What am I missing here ?

Thank you

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What confuses me is that the flux derivative is inside the surface integral.

The partial derivative inside the integral comes from Leibniz Integral Rule (detailed below).

Consider generalized form of Maxwell-Faraday Equation:

$$\oint E \cdot d\ell = -\int_\Sigma \frac{\partial B}{\partial t} \cdot dA$$

This is true for any path \$\partial \Sigma\$, which is any closed-contour bounds the surface \$\Sigma\$.

Now remember generalized form of Leibniz Integral Rule:

$$\frac{d}{dt}\int^{a(x)}_{b(x)} f(x, t) \ dt = f(x, b(x)) \ \frac{d}{dt}b(x) - f(x, a(x)) \ \frac{d}{dt} a(x) + \int^{a(x)}_{b(x)} \frac{\partial}{\partial t}f(x, t) \ dt$$

Did you notice that the bounds of the integral above are not constants?

The right-side term of the generalized Maxwell-Faraday Equation is a surface integral (and, of course, integral around \$\partial \Sigma\$ is a line integral) and the partial derivative inside this integral indicates that any \$\partial \Sigma\$ path is time-dependent. That's why we write Maxwell-Faraday Equation in generalized form because we cannot guarantee that any \$\partial \Sigma\$ is constant.

Now let's look at Leibniz Rule again when the bounds are constants. The first two term of the right-side becomes zero and the integral takes its own special form:

$$\frac{d}{dt}\int^{a}_{b} f(x, t) \ dt = \int^{a}_{b} \frac{\partial}{\partial t}f(x, t) \ dt$$

From this, we can make a conclusion: If the path \$\partial \Sigma\$, which bounds the surface \$\Sigma\$, does not change over time, Maxwell-Faraday Equation turns into:

$$\oint E \cdot d\ell = -\frac{d}{dt}\int_\Sigma B \cdot dA$$

NOTE: I wrote Leibniz Integral Rule for single dimension and made explanations over it just to make things simpler, but the same thing applies for higher dimensions.

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  • \$\begingroup\$ Hmm, what conclusion can I make from this statement ? \$\endgroup\$ – Pier-Yves Lessard Apr 26 '17 at 9:20
  • \$\begingroup\$ @Pier-YvesLessard I thought you were asking where the time dependence of dA comes from. So I tried to show it. Did I misunderstand? \$\endgroup\$ – Rohat Kılıç Apr 26 '17 at 14:57
  • \$\begingroup\$ @ Rohat Kılıç: Maybe I don't get the answer correctly. Having the time derivative outside of the integral like you did makes sense to my intuition and it seems to match with the sliding rod problem. But the equation gotten from Wikipedia shows a time derivative inside the area integral. \$\endgroup\$ – Pier-Yves Lessard Apr 26 '17 at 23:25
  • \$\begingroup\$ @Pier-YvesLessard Ah, ok. Now I got it. Sorry. I'm updating my answer. \$\endgroup\$ – Rohat Kılıç Apr 27 '17 at 6:17
  • \$\begingroup\$ Right, I did not know about Leibniz rule. I still am a little confused. First, the conclusion you propose is something I said I knew in my question (although without the solid math background you showed). On the other hand, the left term is what I would get by substitution of Lenz's Law and definition of Magnetix flux while Faraday's law only include one of the three rightmost terms. I have trouble relating all the equations together. I edited my question to show my mathematical reasoning (which is most likely wrong), maybe that'll make it more clear. Thanks \$\endgroup\$ – Pier-Yves Lessard Apr 28 '17 at 14:24

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