Electromagnetic waves transport energy through
space. In free space this energy is transported by the wave with speed c. The
**magnitude of the energy flux** S is the amount of energy that crosses a unit area
perpendicular to the direction of propagation of the wave per unit time.
It is given by the energy density times the speed c.

The energy per unit volume in the electric field is u_{E} = ½ε_{0}E^{2},

the energy per unit volume in the magnetic field is u_{B} = B^{2}/(2μ_{0}), and B = E/c.

We also have μ_{0}ε_{0} = 1/c^{2}.
We can therefore write

S = c(u_{E} + u_{B}) = ½(cε_{0}E^{2}
+ E^{2}/(cμ_{0})) = ½(cε_{0}E^{2}
+ cε_{0}E^{2}) = cε_{0}E^{2} =
EB/μ_{0.}The units of the energy flux S are J/(m^{2}s) = W/m^{2}.

Note:

The energy** **transported by an electromagnetic wave
is proportional to the square of the amplitude**,** E_{max}^{2}, of the wave.

For all waves, the energy transported is proportional
to the square of the amplitude.

The energy flux vector is
**S**. It has magnitude S and points in
the direction of propagation, i.e. in the direction of **E** × **B**.

It is called the **Poynting vector**
and can also be written as **S** = (

For a sinusoidal plane wave both E and B vary as a function of time as sin(kx - ωt + φ), and S varies as sin

To find the magnitude of the

I = <S> = ½(cε

How much electromagnetic energy per cubic meter is contained in sunlight if
the intensity of the sunlight at the earth's surface under a fairly clear sky is
1000 W/m^{2}?

Solution:

The intensity <S> is the energy per unit area per unit time or the power per
unit area. Multiplying <S> by the time ∆t yields the energy that falls on an unit area in
the time ∆t. This energy is contained in a volume of unit
area and height h, where h = c∆t. Light travels with speed c, so it
travels a distance of h = c∆t in a time interval ∆t. If we set h = 1 m, then ∆t = (1/3)*10^{-8
}s.

So Energy = <S>*(1/3)*10^{-8 }s *(1 m^{2}) is the energy
that is contained in a cubic meter.

Energy = (1000 W/m^{2})*(1/3)*10^{-8
}s*(1 m^{2}) = 3.33*10^{-6 }J.

If the intensity of the sunlight at the earth's surface is 1000 W/m^{2},
what is the rate P at which the sun radiates light energy in W?

Solution:

The average Earth-Sun distance is R =1.496*10^{8} km. At a
distance r from the Sun, the light energy emitted by the Sun has spread over
an area of 4πr^{2}, the surface area of a sphere with radius r,
centered at the Sun.

So at the earth distance R, the intensity is I = P/(4πR^{2}), where P is the power of the sun, (the rate at which it
emits light energy.)

We have P = I*4πR^{2} = (1000 W/m^{2})*4π(1.496*10^{11} m)^{2} = 2.81*10^{26} W.

(The total energy output (not just visible light) of the Sun is 3.86*10^{26} W.)

Assume the helium-neon lasers commonly used in student physics laboratories
have power outputs of 0.5 mW.

(a) If such a laser beam is projected onto a circular spot 1 mm in
radius, what is its intensity?

(b) Find the peak electric field strength.

(c) Find the peak magnetic field strength.

Solution:

(a) Intensity I = energy/(area*time) = power/area = 0.5*10^{-3}
W/(π*10^{-6} m^{2}) = 1.59*10^{3} W/m^{2}.

(b) I = ½(cε_{0}E_{max}^{2}).

E_{max}^{2} = 2I/(cε_{0}) = (2*1.59*10^{3}
W/m^{2})/(3*10^{8} m/s *8.854*10^{-12} C/(V m) =1 .2*10^{6} N^{2}/C^{2}.

E_{max} = 1095 N/C.

(c) B_{max} = E_{max}/c = (1095 N/C)/(3*10^{8}
m/s) = 3.65*10^{-6} T.

Electromagnetic waves transport energy. ** EM wave
also transport
momentum**. The **momentum flux** is
<**S**>/c.
It is the
amount of momentum that crosses a unit area perpendicular to the direction of
propagation of the wave per unit time. If an electromagnetic wave is absorbed,
momentum conservation requires that the object acquires momentum. The
object is therefore acted on by a force, **F** = ∆**p**
/∆t. The radiation
exerts **radiation pressure** P = F/A on the
object. If the radiation is reflected instead of absorbed, then its momentum
changes direction. The radiation pressure on an object that reflects the
radiation is therefore twice the radiation pressure on an object that absorbs
the radiation.

A plane electromagnetic wave of intensity 6 W/m^{2}
strikes a small pocket mirror of area 40 cm^{2}
held perpendicular to the approaching wave.

(a) What momentum does the wave transfer to the mirror each second?

(b) Find the force that the wave exerts on the mirror.

Solution:

(a) The wave is reflected by the mirror. The momentum transferred to
the mirror per unit area per second is twice the momentum of the light
striking the mirror per unit area per second.

∆p/∆t = 2(1/c)<S>A.

<S>A = (6 W/m^{2})(4*10^{-3 }m^{2}) = 0.024 W.

∆p/∆t = 2*0.024 W/(3*10^{8 }m/s) = 1.6*10^{-10
}kgm/s^{2}.

Each second the wave transfers 1.6*10^{-10
}kgm/s of momentum to the mirror.

(b) F = ∆p/∆t = 1.6*10^{-10
}N.