Why do electromagnetic waves have the magnetic and electric field intensities in the same phase?












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$begingroup$


My question is: in electromagnetic waves, if we consider the electric field as a sine function, the magnetic field will be also a sine function, but I am confused why that is this way.



If I look at Maxwell's equation, the changing magnetic field generates the electric field and the changing electric field generates the magnetic field, so according to my opinion if the accelerating electron generates a sine electric field change, then its magnetic field should be a cosine function because $frac{d(sin x)}{dx}=cos x$.










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    3












    $begingroup$


    My question is: in electromagnetic waves, if we consider the electric field as a sine function, the magnetic field will be also a sine function, but I am confused why that is this way.



    If I look at Maxwell's equation, the changing magnetic field generates the electric field and the changing electric field generates the magnetic field, so according to my opinion if the accelerating electron generates a sine electric field change, then its magnetic field should be a cosine function because $frac{d(sin x)}{dx}=cos x$.










    share|cite|improve this question









    New contributor




    Bálint Tatai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      3












      3








      3





      $begingroup$


      My question is: in electromagnetic waves, if we consider the electric field as a sine function, the magnetic field will be also a sine function, but I am confused why that is this way.



      If I look at Maxwell's equation, the changing magnetic field generates the electric field and the changing electric field generates the magnetic field, so according to my opinion if the accelerating electron generates a sine electric field change, then its magnetic field should be a cosine function because $frac{d(sin x)}{dx}=cos x$.










      share|cite|improve this question









      New contributor




      Bálint Tatai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      My question is: in electromagnetic waves, if we consider the electric field as a sine function, the magnetic field will be also a sine function, but I am confused why that is this way.



      If I look at Maxwell's equation, the changing magnetic field generates the electric field and the changing electric field generates the magnetic field, so according to my opinion if the accelerating electron generates a sine electric field change, then its magnetic field should be a cosine function because $frac{d(sin x)}{dx}=cos x$.







      electromagnetic-radiation






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      Bálint Tatai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question









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      edited 21 mins ago









      David Z

      63.5k23136252




      63.5k23136252






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      asked 6 hours ago









      Bálint TataiBálint Tatai

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          $begingroup$

          The Maxwell equations that relate electric and magnetic fields to each other read (in vacuum, in SI units) as
          begin{align}
          nabla times mathbf E & = -frac{partialmathbf B}{partial t} \
          nabla times mathbf B & = frac{1}{c^2} frac{partialmathbf E}{partial t},
          end{align}

          where the notation $nabla times{cdot}$ is a spatial derivative (the curl). This means that both sides have derivatives, and if you're applying them to a function like $cos(kx-omega t)$, then they will both change the cosine into a sine. This is what locks the phase of both waves to equal values.






          share|cite|improve this answer









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            $begingroup$

            The Maxwell equations that relate electric and magnetic fields to each other read (in vacuum, in SI units) as
            begin{align}
            nabla times mathbf E & = -frac{partialmathbf B}{partial t} \
            nabla times mathbf B & = frac{1}{c^2} frac{partialmathbf E}{partial t},
            end{align}

            where the notation $nabla times{cdot}$ is a spatial derivative (the curl). This means that both sides have derivatives, and if you're applying them to a function like $cos(kx-omega t)$, then they will both change the cosine into a sine. This is what locks the phase of both waves to equal values.






            share|cite|improve this answer









            $endgroup$


















              5












              $begingroup$

              The Maxwell equations that relate electric and magnetic fields to each other read (in vacuum, in SI units) as
              begin{align}
              nabla times mathbf E & = -frac{partialmathbf B}{partial t} \
              nabla times mathbf B & = frac{1}{c^2} frac{partialmathbf E}{partial t},
              end{align}

              where the notation $nabla times{cdot}$ is a spatial derivative (the curl). This means that both sides have derivatives, and if you're applying them to a function like $cos(kx-omega t)$, then they will both change the cosine into a sine. This is what locks the phase of both waves to equal values.






              share|cite|improve this answer









              $endgroup$
















                5












                5








                5





                $begingroup$

                The Maxwell equations that relate electric and magnetic fields to each other read (in vacuum, in SI units) as
                begin{align}
                nabla times mathbf E & = -frac{partialmathbf B}{partial t} \
                nabla times mathbf B & = frac{1}{c^2} frac{partialmathbf E}{partial t},
                end{align}

                where the notation $nabla times{cdot}$ is a spatial derivative (the curl). This means that both sides have derivatives, and if you're applying them to a function like $cos(kx-omega t)$, then they will both change the cosine into a sine. This is what locks the phase of both waves to equal values.






                share|cite|improve this answer









                $endgroup$



                The Maxwell equations that relate electric and magnetic fields to each other read (in vacuum, in SI units) as
                begin{align}
                nabla times mathbf E & = -frac{partialmathbf B}{partial t} \
                nabla times mathbf B & = frac{1}{c^2} frac{partialmathbf E}{partial t},
                end{align}

                where the notation $nabla times{cdot}$ is a spatial derivative (the curl). This means that both sides have derivatives, and if you're applying them to a function like $cos(kx-omega t)$, then they will both change the cosine into a sine. This is what locks the phase of both waves to equal values.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 6 hours ago









                Emilio PisantyEmilio Pisanty

                83.4k22203417




                83.4k22203417






















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