Why does Faraday's law involve a partial vs an ordinary derivative?

  • Thread starter Nick R
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Main Question or Discussion Point

Faraday's law is often stated in SI units as

[tex]\nabla x E(x,y,z,t) = \frac{\partial B(x,y,z,t)}{\partial t}[/tex]

But x, y, z (or some other set of coordinates) are variables not functions and thus have no "t dependence".

So it would seem that the "total derivative" of B is the same as the partial:

[tex] \frac{dB}{dt}=\frac{\partial B}{\partial t}+\frac{\partial B}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial B}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial B}{\partial z}\frac{\partial z}{\partial t} = \frac{\partial B}{\partial t} + 0 + 0 + 0[/tex]

It seems that there isn't any difference between the partial of B wrt to t and the "full derivative" of B wrt t.

The reason I'm concerned is the issue where you CANNOT do the following (or maybe sometimes... like in this case you can?):

[tex]\int^a_b\frac{\partial f(x,y)}{\partial x}dx = f(b,y) - f(a,y)[/tex]
 
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Answers and Replies

  • #2
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It's the change in flux that matters so B can be constant and area change.
 

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