EM wave field strength and energy

AI Thread Summary
The discussion centers on calculating the peak value of the electric or magnetic field of an electromagnetic (EM) wave given its total energy. The energy density of an EM wave is expressed as εE² + μH², where ε is permittivity, μ is permeability, E is the electric field amplitude, and H is the magnetic field amplitude. In a vacuum, the energy density is equally distributed between the electric and magnetic components, with the electric energy density given by u_E = (ε₀E²)/2 and the magnetic energy density by u_B = (B²)/(2μ₀). It is noted that in many cases, the magnetic component can be neglected since B is proportional to E/c. The relationship B = E/c and the speed of light c = 1/√(ε₀μ₀) are also highlighted, confirming that the energy densities are equal in a vacuum.
Usaf Moji
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Another noobish question: Let's say there is an electromagnetic wave of constant wavelength and constant peak amplitude that travels a known distance. Let's say that we also know the total energy of this EM wave. Is there some formula that can tell us what the peak value of the electric field is (or the peak value of the magnetic field, it doesn't matter which)?

All responses appreciated.
 
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The energy density of an EM wave is given by \epsilon E^{2}+ \mu H^{2} where \epsilon is the permittivity, \mu the permeability, E the electric field amplitude, and H the magnetic field amplitude. For the majority of cases, the magnetic component is much less than the electric component, B ~ E/c, and can be neglected.
 
Actually, in an EM wave, in vacuum at least, the energy density is equally distributed between the electric and magnetic portions of the wave. The electric and magnetic energy densities in vacuum are

u_E = \frac{\epsilon_0 E^2}{2}

u_B = \frac{B^2}{2 \mu_0}

In an EM wave, B = E/c and

c = \frac{1}{\sqrt{\epsilon_0 \mu_0}}

which allow you to show that u_E = u_B.
 
Thank you Andy and jtbell, that was very helpful.
 
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