In chapter on waves, my sir taught us that intensity of wave is directly proportional to the square of amplitude of wave. So when we were asked what is contribution of electric field and magnetic field to intensity of light, I used the relation that amplitude of electric field is $c$ times amplitude of magnetic field. When I used this I got the contributions of magnetic field to electric field to be in ratio of $1:c^2$. But this was wrong and the answer was $1:1$. When I asked my sir the reason for this, he said both cases are different, but got angry when I asked him to elaborate. Could anyone please explain this to me?
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3 Answers
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The energy per unit volume, $U_\text{B}$, in a magnetic field is given by $$U_\text{B}=\frac{1}{2\mu_0} B^2.$$ The energy per unit volume, $U_\text{E}$ in an electric field is $$U_\text{E}=\frac12\epsilon_0 E^2.$$ As you say, in an e-m wave, $$E=cB.$$ Therefore $$U_\text{E}=\frac12 \epsilon_0 c^2B^2=\frac{1}{2 \mu_0} B^2=U_B.$$ Here we have used the relationship $\mu_0 \epsilon_0=\frac{1}{c^2}$. The mean rate of flow of energy in a plane wave per unit area of wavefront is then $$\bar I=c\left[\frac{1}{2\mu_0}\overline{B^2} +\frac12\epsilon_0 \overline{E^2} \right].$$ The two terms in the square brackets are equal, that is the electric field energy and magnetic field energy contribute equally to the wave energy per unit area, that is to the wave intensity.
answered Jan 19, 2021 at 13:29
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$\begingroup$ Could you please be a bit direct and state how intensity contributions are equal due to the two fields? Because I'm confused after reading your answer.. but thanks for answering! $\endgroup$ Commented Jan 19, 2021 at 13:48
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$\begingroup$ You were quite right to ask for more clarity. I've added a bit to my answer. $\endgroup$ Commented Jan 19, 2021 at 14:48
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$\begingroup$ I understood your answer(thanks!) but I also had another question, why can't we use the relationship between intensity and amplitude of the wave (intensity directly proportional to square of amplitude of wave) in this case? $\endgroup$ Commented Jan 20, 2021 at 6:02
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$\begingroup$ Is the reason because that relationship is valid only for mechanical waves? $\endgroup$ Commented Jan 20, 2021 at 6:07
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1$\begingroup$ You can use the relationship between amplitude and intensity, with the (peak) electric field as amplitude. That's because equal energies are contained in the electric and magnetic fields, so the contents of my square bracket can be written as a single term, $\epsilon_0 \overline{E^2}$. $\endgroup$ Commented Jan 20, 2021 at 9:16
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In addition to the other answers, here is actually a simple argument (partially hinted at in Philip's answer):
The EM wave is fundamentally an oscillation between electric and magnetic fields. In other words, the energy goes back-and-forth between the electric and the magnetic component. Therefore, the electric and magnetic energies necessarily have to be equal !
How this translates to electric field and magnetic field values depends on the unit system used.
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$\begingroup$ Good point (note that your example propagates in the z direction). In fact maxwell's law tells you that E increases most in time where curl of B (so dBy/dz) is maximum, so the zero of E matches the zero of B. $\endgroup$– NicolasCommented Jan 21, 2021 at 18:03
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$\begingroup$ [This comment has been corrected. It originally preceded Nicholas's reply.] "the energy goes back-and-forth between the electric and the magnetic component." How does this work? For a plane wave polarised wave propagating in the 𝑥 direction we have, do we not? $$𝐸_y=𝐸_0 \sin(𝜔𝑡−𝑘𝑥)\ \ \ \ \text{and}\ \ \ \ \ B_z=𝐵_0 \sin(𝜔𝑡−𝑘𝑥).$$ In other words the two fields vary in phase. [You can include a phase constant if you wish, but it's the same constant for each field.] $\endgroup$ Commented Jan 21, 2021 at 19:25
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$\begingroup$ Thank you. But I still don't get this "energy back and forth" claim. For oscillations in an electric circuit consisting of an inductor, a capacitor (and, if you like, a little resistance) I agree that energy does swap backwards and forwards between inductor and capacitor (that is between B field and E field), but I don't see how this applies to an e-m wave. $\endgroup$ Commented Jan 21, 2021 at 19:42
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$\begingroup$ Every wave/oscillation that I can think of is a periodic exchange between two forms of energy. Pendulum: kinetic and potential. LC (as you said) : magnetic and electric. Sound (air): thermal and kinetic.Sound (solid): elastic potential and kinetic. Tsunami: gravitation potential and kinetic… In all those cases, both forms store half the energy. Where the maximum is located (the phase) doesnt matter so much, and is a matter for dynamic analysis (my previous comment), as long as the total energy is conserved in time. It is true that for EM waves in vacuum, both energy components are in phase. $\endgroup$– NicolasCommented Jan 22, 2021 at 14:12
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1$\begingroup$ I'm not really sure what to make of that peculiarity, apart from the fact that E and B maxima have to be at the same place so they can be swapped in case of a change of frame. $\endgroup$– NicolasCommented Jan 22, 2021 at 14:13
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The ratio of the energy densities can't be $1/c^2$, because their ratio has to be unitless.
The easiest way to see that the energy densities must be equal is by considering the case where two wave packets collide head-on, requiring that energy be conserved regardless of the polarization. This argument goes through regardless of what system of units you use.
