Find Magnetic Field from Electric Field Using Maxwell's Equations

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SUMMARY

The discussion focuses on deriving the magnetic field \(\mathbf{B}\) from the given electric field \(\mathbf{E} = E_0 \cos(kz - \omega t) \hat{x}\) using Maxwell's equations. The initial attempt resulted in an incorrect expression for \(\mathbf{B}\) due to a miscalculation during integration. The correct magnetic field is \(\mathbf{B} = \frac{kE_0}{\omega} \cos(kz - \omega t) \hat{y}\), which aligns with the expected solution. The error was identified in the integration process of the time derivative of the magnetic field.

PREREQUISITES
  • Understanding of Maxwell's equations, specifically \(\mathbf{\nabla \times E} = \dot{\mathbf{B}}\)
  • Proficiency in vector calculus, particularly cross products and integrals
  • Familiarity with electromagnetic wave equations
  • Knowledge of trigonometric identities and integration techniques
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  • Review the derivation of magnetic fields from electric fields using Maxwell's equations
  • Study integration techniques for trigonometric functions in the context of physics
  • Explore the implications of electromagnetic wave propagation in different media
  • Learn about the physical significance of the terms in Maxwell's equations
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Students studying electromagnetism, physics educators, and anyone interested in the mathematical foundations of electromagnetic theory.

ADCooper
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Homework Statement


An electromagnetic wave has an electric field \mathbf{E} = E_0 \cos(kz-ωt) \hat{x}. Using Maxwell's equations, find the magnetic field.

Homework Equations


\mathbf{∇\times E} = \mathbf{\dot{B}}

The Attempt at a Solution



So this problem appears extremely simple, but other students have told me my answer is incorrect, and I can't figure out what is wrong with my math. I find the cross product, which results in the following equation for the time derivative of the magnetic field:

\mathbf{\dot{B}} = \hat{y}kE_0\sin(kz-ωt)

I now integrate both sides with respect to time. This is where my answer diverges from others, so I'll fully write out my steps:

\mathbf{B} = kE_0∫_0^t \sin(kz-ωt') dt'\hat{y}

I set u = kz-ωt', which means du = -ωdt'

Plugging this in, the new integral is:

\mathbf{B} = -\frac{kE_0}{ω}∫_{kz}^{kz-ωt}\sin(u)du

The result is then

\mathbf{B} = \frac{kE_0}{ω}[\cos(kz-ωt)-\cos(kz)]\hat{y}

However, every student I've talked to has told me that the correct answer should be

\mathbf{B} = \frac{kE_0}{ω} \cos(kz-ωt)\hat{y}

Is there something simple I'm missing? There's nothing else in the problem description I didn't write. The other answer looks more correct but I can't find any reason that mine is incorrect.
 
Last edited:
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ADCooper said:
\mathbf{\dot{B}} = \hat{y}kE_0\sin(kz-ωt)

I now integrate both sides with respect to time. This is where my answer diverges from others, so I'll fully write out my steps:

\mathbf{B} = kE_0∫_0^t \sin(kz-ωt') dt'\hat{y}

The left side of the second equation above is incorrect. Think again about what you get when you integrate the left side of the first equation from t' = 0 to t' = t.
 
Last edited:
TSny said:
The left side of the second equation above is incorrect. Think again about what you get when you integrate the left side of the first equation from t' = 0 to t' = t.

Woops thanks for the help, figured it out!
 

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