The curl of certain vector fields

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SUMMARY

The discussion centers on the mathematical relationship between the electric vector field \(\vec E\) and the magnetic vector field \(\vec B\) as described by Maxwell's equations. Specifically, it highlights the identity curl(\(\vec E\)) = -\(\frac{\partial \vec B}{\partial t}\) and the subsequent derivation involving curl(curl(\(\vec E\))). The user attempted to prove these identities using specific vector field definitions but encountered issues leading to a curl of zero for \(\vec E\). The conclusion drawn is that a static magnetic field results in a zero time derivative, aligning with Maxwell's equations.

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  • Understanding of vector calculus, particularly curl operations.
  • Familiarity with Maxwell's equations in electromagnetism.
  • Knowledge of electric and magnetic field representations.
  • Basic concepts of time derivatives in physics.
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hholzer
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Given the two vector fields:

[tex] \vec E and \vec B[/tex]

Where the first is the electric vector field
and the second is the magnetic vector field, we
have the following identity:

[tex] curl(\vec E) = -\frac{\partial \vec B } { \partial t }[/tex]

and further that:

[tex] curl(curl(\vec E)) = curl(-\frac{\partial \vec B } { \partial t }) <br /> = -\frac{\partial curl(\vec B) } {\partial t }[/tex]

I tried to prove these by defining the vector fields:

[tex] \vec E = C\frac{ \mathbf e_r } {p^2}[/tex]

and

[tex] \vec B = <0, 0, B>[/tex]

where C and B are constants. But I ended up with
zero for curl(E), which cannot be right. So my reasoning
is in error somewhere.

Any insight appreciated.
 
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If p is your radial variable, then the electric field varies only radially and hence has curl zero, so that makes sense. Further, since you have a static magnetic field, its time derivative should also be zero, putting the two equal and satisfying Maxwell's equation.
 
Hm, then in general how was the result reached that they are equal?
 

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