Why do electromagnetic waves have it's magnetic and electric field intensities in the same phase?
$begingroup$
My question is exactly about that in the elctromagnetic 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 is that this way, if i look at the maxwells 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 cosine function because d(sinx)/dx=cosx
electromagnetic-radiation
asked 5 hours ago
Bálint TataiBálint Tatai
241
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$begingroup$
My question is exactly about that in the elctromagnetic 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 is that this way, if i look at the maxwells 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 cosine function because d(sinx)/dx=cosx
electromagnetic-radiation
asked 5 hours ago
Bálint TataiBálint Tatai
241
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.
$endgroup$
add a comment |
$begingroup$
My question is exactly about that in the elctromagnetic 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 is that this way, if i look at the maxwells 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 cosine function because d(sinx)/dx=cosx
electromagnetic-radiation
asked 5 hours ago
Bálint TataiBálint Tatai
241
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 exactly about that in the elctromagnetic 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 is that this way, if i look at the maxwells 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 cosine function because d(sinx)/dx=cosx
electromagnetic-radiation
electromagnetic-radiation
asked 5 hours ago
Bálint TataiBálint Tatai
241
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.
asked 5 hours ago
Bálint TataiBálint Tatai
241
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.
asked 5 hours ago
Bálint TataiBálint Tatai
241
New contributor
Bálint Tatai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 5 hours ago
Bálint TataiBálint Tatai
241
asked 5 hours ago
Bálint TataiBálint Tatai
241
241
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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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New contributor
Bálint Tatai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1 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.
answered 4 hours ago
Emilio PisantyEmilio Pisanty
83.4k22203417
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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.
answered 4 hours ago
Emilio PisantyEmilio Pisanty
83.4k22203417
$endgroup$
add a comment |
$begingroup$
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.
answered 4 hours ago
Emilio PisantyEmilio Pisanty
83.4k22203417
$endgroup$
add a comment |
$begingroup$
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.
answered 4 hours ago
Emilio PisantyEmilio Pisanty
83.4k22203417
$endgroup$
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.
answered 4 hours ago
Emilio PisantyEmilio Pisanty
83.4k22203417
answered 4 hours ago
Emilio PisantyEmilio Pisanty
83.4k22203417
answered 4 hours ago
Emilio PisantyEmilio Pisanty
83.4k22203417
answered 4 hours ago
Emilio PisantyEmilio Pisanty
83.4k22203417
83.4k22203417
add a comment |
add a comment |
Bálint Tatai is a new contributor. Be nice, and check out our Code of Conduct.
Bálint Tatai is a new contributor. Be nice, and check out our Code of Conduct.
Bálint Tatai is a new contributor. Be nice, and check out our Code of Conduct.
Bálint Tatai is a new contributor. Be nice, and check out our Code of Conduct.
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