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
Best Answer
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.