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

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The discussion focuses on deriving the relationship between the electric field vector (E) and the magnetic field vector (B) in the context of electromagnetic waves using Maxwell's equations. It explains that E and B satisfy the homogeneous wave equation, leading to the plane-wave solution where E is transverse to the wave vector k. The relationship between E and B is established through the equation B = (1/c) k × E, confirming that E, B, and k form a right-handed orthogonal set. The distinction between Heaviside-Lorentz and SI units is also clarified, particularly regarding the factor of 1/c in the magnetic field representation. Understanding these relationships is crucial for comprehending electromagnetic wave propagation in a vacuum.
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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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