Energy in the fields of a wave
The LC circuit stored energy in two fields: an electric field between the plates and a magnetic field in the coil. An electromagnetic wave carries both fields, so it carries energy in both. The chapter Capacitance and Dielectrics found the energy per unit volume of an electric field, \(\tfrac12\epsilon_0E^2\). A magnetic field holds \(B^2/2\mu_0\) per unit volume. The energy density of the wave is the sum.
\[ B = \dfrac{E}{c} \]
Why. in the wave the two fields are in phase, and \(E = cB\) at every point.
\[ \dfrac{B^2}{2\mu_0} = \dfrac{E^2}{2\mu_0c^2} \]
Why. put \(B = E/c\) into the magnetic energy density.
\[ \dfrac{E^2}{2\mu_0c^2} = \dfrac12\epsilon_0E^2 \]
Read it. \(c^2 = 1/\mu_0\epsilon_0\), so \(1/\mu_0c^2 = \epsilon_0\).
So the magnetic field, tiny in tesla, carries exactly as much energy as the electric field. At every point \(u = \epsilon_0E^2\). The field goes as \(\sin(kx - \omega t)\), and the average of \(\sin^2\) over a cycle is \(\tfrac12\). So the average energy density is \(\bar u = \tfrac12\epsilon_0E_0^2\).
A plane wave has an electric amplitude of \(300\,\text{V/m}\). Find its average energy density, and show that its magnetic amplitude, \(1\,\mu\text{T}\), holds the same peak energy density as its electric amplitude.
(a) \(\bar u = \tfrac12\epsilon_0E_0^2\) \(= \tfrac12\times8.85\times10^{-12}\times9\times10^4 = 4.0\times10^{-7}\,\text{J/m}^3\).
(b) Electric: \(\tfrac12\epsilon_0E_0^2 = 3.98\times10^{-7}\,\text{J/m}^3\). Magnetic: \(B_0^2/2\mu_0\) \(= 10^{-12}/(8\pi\times10^{-7}) = 3.98\times10^{-7}\,\text{J/m}^3\). The two are equal.
Values used: \(\epsilon_0 = 8.85\times10^{-12}\,\text{C}^2/(\text{N m}^2)\), \(\mu_0 = 4\pi\times10^{-7}\,\text{T m/A}\), \(c = 3\times10^8\,\text{m/s}\).
In a light wave the magnetic field is only \(1/c\) of the electric field, in SI units. How does the energy in the magnetic field compare with the energy in the electric field?
A wave has an electric amplitude of \(600\,\text{V/m}\). What is its average energy density, in units of \(10^{-6}\,\text{J/m}^3\)?
How much power a wave carries
The intensity of a wave is the power it carries across unit area held at right angles to its travel. Picture a box of cross-section \(A\) and length \(c\times1\,\text{s}\), lying along the beam. All the energy in the box crosses its far face in one second. That energy is \(\bar u\times Ac\). So the power per unit area is \(\bar uc\).
For the wave of \(300\,\text{V/m}\), \(I = 4\times10^{-7}\times3\times10^8 = 120\,\text{W/m}^2\). A source that radiates evenly spreads its power \(P\) over a sphere of area \(4\pi r^2\). Sound did the same in the chapter Energy, Power, Intensity and Loudness. So the intensity falls as \(1/r^2\). The field amplitude goes as \(\sqrt I\), so it falls as \(1/r\).
An ideal \(100\,\text{W}\) source radiates evenly in all directions. Find the intensity and the electric amplitude at \(1\,\text{m}\) and at \(2\,\text{m}\).
The power spreads over a sphere of radius \(r\). The field amplitude then follows from \(I = \tfrac12\epsilon_0cE_0^2\).
\(I = P/4\pi r^2\) and \(E_0 = \sqrt{2I/\epsilon_0c}\).
At \(1\,\text{m}\): \(I = 100/4\pi = 7.96\,\text{W/m}^2\) and \(E_0 = \sqrt{2\times7.96/2.66\times10^{-3}} = 77\,\text{V/m}\). At \(2\,\text{m}\): \(I = 1.99\,\text{W/m}^2\) and \(E_0 = 39\,\text{V/m}\).
About \(8\,\text{W/m}^2\) and \(77\,\text{V/m}\) at \(1\,\text{m}\); \(2\,\text{W/m}^2\) and \(39\,\text{V/m}\) at \(2\,\text{m}\).
A mobile tower's antenna sends out a few tens of watts, about \(20\,\text{W}\) from each. Treat it as a point source. At \(100\,\text{m}\) the intensity is \(20/(4\pi\times100^2) = 1.6\times10^{-4}\,\text{W/m}^2\), and the field amplitude is about \(0.35\,\text{V/m}\). Sunlight on the ground, about \(1000\,\text{W/m}^2\), has a field amplitude near \(870\,\text{V/m}\).
Sunlight reaching the ground carries about \(1000\,\text{W/m}^2\). Estimate the amplitude of its electric field, in V/m.
At \(3\,\text{m}\) from a small lamp the electric amplitude is \(E_0\). What is it at \(6\,\text{m}\)?
An ideal \(50\,\text{W}\) source radiates evenly in all directions. What is the intensity \(5\,\text{m}\) away, in W/m²?
Light carries momentum
Maxwell's theory shows that a wave carrying energy \(U\) also carries momentum \(U/c\) along its direction of travel. The electric field of the wave drives charges in a surface to and fro. The magnetic field then pushes those moving charges forward, along the travel. When a surface absorbs the wave, it takes up this momentum. By Newton's second law in the form of the chapter Momentum and Its Conservation, the momentum taken each second is a force.
On an area \(A\) the absorbed energy per second is \(IA\), so the momentum per second is \(IA/c\). The force per unit area, the radiation pressure, is \(I/c\). A mirror reflects the wave, so its momentum is reversed. The change of momentum is then doubled, and so is the pressure, \(2I/c\).
Sunlight of \(1000\,\text{W/m}^2\) falls at right angles on a surface of \(1\,\text{m}^2\). Find (a) the momentum it delivers each second and the pressure if it is absorbed, and (b) the pressure if it is fully reflected.
(a) Each second \(1000\,\text{J}\) arrives, carrying momentum \(1000/(3\times10^8) = 3.3\times10^{-6}\,\text{kg m/s}\). The pressure is \(I/c = 3.3\times10^{-6}\,\text{Pa}\), a force of \(3.3\times10^{-6}\,\text{N}\) on the square metre.
(b) Reflection doubles the change of momentum: \(2I/c = 6.7\times10^{-6}\,\text{Pa}\).
Why does nobody feel sunlight push? The air presses on us with about \(10^5\,\text{Pa}\), some \(3\times10^{10}\) times as much. The push of sunlight on a hand is well under a micronewton. It matters only for very light things with large areas. The dust in a comet's tail is pushed away from the Sun by light. A thin sail of a few hundred square metres in space can be steered by it.
A spacecraft's sail of \(1000\,\text{m}^2\) reflects sunlight of \(1360\,\text{W/m}^2\) falling on it at right angles. What force does the light exert on it, in millinewtons?
Sunlight exerts a pressure on everything it falls on. Why do we never notice it?
Sunlight of \(1000\,\text{W/m}^2\) gives a pressure of only \(I/c\), about \(3\times10^{-6}\,\text{Pa}\), while the air presses with about \(10^5\,\text{Pa}\) on both sides of us. The force of sunlight on a hand is well under a micronewton, far too small to feel. It matters only for very light bodies with large areas, such as dust in a comet's tail or a solar sail.
At Byalalu, near Bengaluru, ISRO's Indian Deep Space Network points a dish \(32\,\text{m}\) across at spacecraft near the Moon and beyond. A spacecraft's transmitter sends only a few watts. Its power spreads out over a sphere whose radius is the distance to the Earth. For a craft near the Moon, \(3.8\times10^8\,\text{m}\) away, that sphere has an area of about \(2\times10^{18}\,\text{m}^2\). So even the large dish collects well under a picowatt, \(10^{-12}\,\text{W}\). The inverse-square law is why deep-space antennas are so large.
- Thinking the magnetic field of light carries little energy.\(B^2/2\mu_0 = \tfrac12\epsilon_0E^2\); the two halves are always equal.
- Forgetting the \(\tfrac12\) of the average.The average of \(\sin^2\) is \(\tfrac12\), so \(\bar u = \tfrac12\epsilon_0E_0^2\) and \(I = \tfrac12\epsilon_0cE_0^2\).
- Letting the field fall as \(1/r^2\).The intensity falls as \(1/r^2\); the field amplitude falls as \(1/r\).
- Using \(I/c\) for a mirror.A reflected wave reverses its momentum, so the pressure is \(2I/c\).