Getting E-Field from magnetic field intensity, H

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SUMMARY

The discussion focuses on deriving the electric field, E, from the magnetic field intensity, H, given by H=\hat{y}6cos(2z)sin((2x10^7)t - 0.1x). The key equation used is the Maxwell's equation \nabla \times E = \frac{- \partial B}{\partial t}, where B is related to H through the equation B = \muH. The solution involves taking the partial derivative of B with respect to time and applying Stokes's theorem to "uncurl" the electric field E. The conversation emphasizes that such problems are more suited for advanced physics rather than introductory physics courses.

PREREQUISITES
  • Understanding of Maxwell's equations
  • Familiarity with vector calculus, specifically Stokes's theorem
  • Knowledge of electromagnetic theory, particularly the relationship between H and B
  • Basic concepts of time-varying fields in physics
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  • Study the application of Stokes's theorem in electromagnetic fields
  • Learn about the derivation of electric fields from magnetic fields in electromagnetic theory
  • Explore advanced topics in vector calculus relevant to physics
  • Investigate the implications of time-varying magnetic fields on electric fields
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Students and professionals in physics, particularly those studying electromagnetism, as well as educators looking for advanced problem-solving techniques in electromagnetic theory.

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Homework Statement


Given the magnetic field intensity, H, find E.
H=\hat{y}6cos(2z)sin((2x10^7)t - 0.1x)


Homework Equations


\nabla \times E = \frac{- \partial B}{\partial t}


The Attempt at a Solution


Since we have H, we can use the relationship that \muH = B and then take the partial derivative of that with respect to time. That would give us this quantity...

\frac{- \partial B}{\partial t}

Now, how do I get to E-field from that? How do you "uncurl" that?
 
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You could use Stokes's theorem to "uncurl" E. Do you see where it's headed?

Also, just to help you out in the future, anything with Calc 3 doesn't really belong in intro physics, and you may be helped quicker in advanced physics for questions like this.
 

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