Good day, I am a student of Physics at the university of Padova, I must solve this problem for my exam of electromagnetic fields, but I have got different problems. The text is the follower:
The electric field of an electromagnetic wave is:
$$E = E_0\sin \omega t (\sin \omega z, \cos \omega z, 0 ). $$
I have to find the magnetic field $B$, then I have to verify the Maxwell equation for $E$ and $B$, and finally I have to find the 4-potential $A^\mu$ in the Lorenz gauge.
First of all, I have considered $n$ the direction of propagation of the wave:
$n=(0,0,1)$
So that I thought B is the cross product of $n$ and E, I have obtained:
$B= n \times E = E_0\sin \omega t (-cos \omega z, sin \omega z, 0 )$
And this result seemed me reasonable, because the scalar product between E and B is null.
In empty space, I expect that the divergence of E and B is null, and in this case is verified. The other two Maxwell's equation establish:
$\nabla \times E + \frac{\partial B}{\partial t} =0 $
$E_0 \omega \sin \omega t (sin \omega z, cos \omega z, 0 ) + E_0 \omega \cos \omega t (-cos \omega z, sin \omega z, 0 ) = E_0 \omega (-cos(\omega t + \omega z), sin(\omega t + \omega z),0) $
But in this way the sine and the cosine can not vanish simultaneously, they have got the same argument !
$\nabla \times B – \frac{\partial E}{\partial t} =0 $
$E_0 \omega \sin \omega t (-cos \omega z, sin \omega z, 0 ) – E_0 \omega \cos \omega t (sin \omega z, cos \omega z, 0 ) = E_0 \omega (-sin(\omega t + \omega z), -cos(\omega t + \omega z), 0)$
There is the same problem as before.
Then I have tried to obtain $A^\mu$.
$A = \frac{B}{\omega}$
$\nabla \times A = \nabla \times \frac{B}{\omega} = \frac{1}{\omega} \nabla \times B = B$
Because I noticed
$\nabla \times B = \omega B$
Then I wanted to obtain $A^0$
$E = \frac{\partial A}{\partial t} – \nabla A^0$
But I stopped there because in my opinion there are too many errors in my reasoning. The Maxwell equation are not verified.
If someone has the solution, I will be to him infinitely grateful
Best Answer
I would rather use faraday law to get the magnetic field: $$\nabla \times E = -\frac{\partial B}{\partial t}= E_0 \omega\sin\omega t (\sin\omega z, \cos\omega z, 0) $$ $$B=E_0 \cos\omega t (\sin\omega z, \cos\omega z, 0) $$