How to write the relationship of B and E in vector form

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SUMMARY

This discussion focuses on deriving the relationship between the electric field vector, vec{E}, and the magnetic field vector, vec{B}, in the context of electromagnetic waves using Maxwell's equations. The plane-wave solutions are expressed in the form vec{E} = vec{E}_0 exp(-i ω t + i vec{k} · vec{x}), where ω(vec{k}) = c|vec{k}|. The relationship vec{B} = (c/ω) (vec{k} × vec{E}_0) exp(...) is established, confirming that vec{E}_0, vec{B}_0, and vec{k} form a right-handed orthogonal set of vectors, with magnitudes |vec{E}_0| = |vec{B}_0|.

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  • Understanding of Maxwell's equations in Heaviside-Lorentz units
  • Knowledge of plane-wave solutions in electromagnetic theory
  • Familiarity with vector calculus and cross products
  • Basic concepts of wave propagation and phase velocity
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garylau
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http://www.physicspages.com/2014/08/22/electromagnetic-waves-in-vacuum/
Sorry
i have no idea how to get the last step in the vector form

how to convert and mix it?thank you
 

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These are obviously the plane-wave modes of the electromagnetic field. You start, of course, from the Maxwell equations with vanishing charges and currents (in Heaviside-Lorentz units),
$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E}=0.$$
You can use some manipulations to see that ##\vec{E}## and ##\vec{B}## obey the homogeneous wave equation for waves with phase velocity ##c##. From this to get the plane-wave modes you make the ansatz
$$\vec{E}=\vec{E}_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}), \quad \omega(\vec{k})=c|\vec{k}|.$$
From the last Maxwell equation you get
$$\vec{\nabla} \cdot \vec{E} =\mathrm{i} \vec{k} \cdot \vec{E}_0 \exp(\cdots)=0, \; \Rightarrow \; \vec{k} \cdot \vec{E}_0=0,$$
i.e., the electric field is a transverse wave.

The first equation gives
$$\partial_t \vec{B}=-\mathrm{i} c \vec{k} \times \vec{E}_0 \exp(\cdots).$$
Integrating gives
$$\vec{B}=\frac{c \vec{k}}{\omega} \times \vec{E}_0 \exp(\cdots)=\hat{k} \times \vec{E}_0 \exp(\cdots)=\vec{B}_0 \exp(\cdots).$$
It's easy to check that this also solves the 2nd and the 3rd Maxwell equation. This shows that ##\vec{E}_0##, ##\vec{B}_0##, and ##\vec{k}## form a righthanded orthogonal set of vectors and ##|\vec{E}_0|=|\vec{B}_0|##.

The additional factor of ##1/c## in your figure comes most likely from the use of SI units. Indeed we have
$$|\vec{B}_{\text{SI}}|=\sqrt{\mu_0} |\vec{B}_{\text{HL}}|=\sqrt{\mu_0} |\vec{E}_{\text{HL}}|=\sqrt{\mu_0 \epsilon_0} |\vec{E}_{\text{SI}}|=\frac{1}{c} |\vec{E}_{\text{SI}}|.$$
 
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