### Energy and momentum flux (examples)

#### Problem:

(a)  Compare the intensity of a light bulb at a distance of 6 m from it to the intensity at 2 m from it.  Repeat this comparison for laser light.  Explain fully but briefly.
(b)  Compute the electric field corresponding to a focused light intensity of I = 1012W/cm2, and compare the result with the electric field experienced by the electron in a hydrogen atom.

Solution:

• Concepts:
The inverse square law, the Poynting vector, the electric field of a point charge
• Reasoning:
(a)  The energy emitted by the bulb is emitted into a solid angle of 4π.  It is spread over an area that increases as 4πr2 with distance r.
(b)  The magnitude of the Poynting vector S = (1/μ0)E×B relates the intensity of an electromagnetic wave to the field strength.  For a sinusoidal electromagnetic wave we have <S> = (|E02|/2)(ε00)½.
The electron in hydrogen atom experiences the electric field of a point charge q = 1.6*10-19 C a distance a0 = 5.29*10-11 m from the point charge, Eq = q/(4πε0a02) .
• Details of the calculation:
(a)  For a light bulb we have:  Intensity = energy/(area * time) proportional to 1/r2.
Intensity at 6 m = (1/9) * intensity at 2 m.
A laser beam with a Gaussian profile has a waist d0.  It either diverges from or converges to this beam waist.  This divergence or convergence is measured by the angle θ  which is subtended by the points on either side of the beam axis where the irradiance has dropped if to 1/e2 of its value on the beam axis.

For a Gaussian beam of a particular wavelength, the product d0θ is constant.  A highly collimated beam with small θ has a large beam waist d0.  Over short distances, the change in d is negligible to d0.  The intensity is therefore practically the same at 2 m and at 6 m.

(b)  1012W/(0.01 m)2 = (|E02|/2)(ε00)½,  E0 = 2.74*109 N/C is the electric field of the light wave.
Eq = q/(4πε0a02) = 5.15*1011 N/C is the electric field experienced by the electron in a Hydrogen atom.
Eq >> E0.

#### Problem:

A plane electromagnetic wave of intensity 6 W/m2 strikes a small pocket mirror of area 40 cm2 held perpendicular to the approaching wave.
(a)  What momentum does the wave transfer to the mirror each second?
(b)  Find the magnitude of the force that the wave exerts on the mirror.

Solution:

• Concepts:
Energy and momentum flux of an electromagnetic wave, radiation pressure
• Reasoning:
The energy flux of an electromagnetic wave is given by the Poynting vector S.  Its magnitude is the intensity.  The momentum flux is S/c.  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.
• Details of the calculation:
(a)  dp/dt = 2(1/c)S∙A.
S∙A = (6 W/m2)(4*10-3 m2) = 0.024 W.
dp/dt = 2*0.024 W/(3*108 m/s) = 1.6*10-10 kgm/s2.
Each second the wave transfers 1.6*10-10 kgm/s of momentum to the mirror in the incident wave direction.
(b)  F = dp/dt = 1.6*10-10N.

#### Problem:

A beam of light with an intensity of 108 W/m2 traveling in free space hits normally on a lossy mirror.  The mirror reflects 60% of the light and absorbs 40%.  What is the resulting radiation pressure on the surface of the mirror?

Solution:

• Concepts:
Momentum transported by an EM wave
• Reasoning:
The momentum of an object absorbing the radiation changes.  The rate of change is dpperp/dt = (1/c)SAperp, where Aperp is the cross-sectional area of the object perpendicular to the direction of propagation of the electromagnetic wave. (We assume a massive object absorbing a negligible fraction of the energy.)
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.
• Details of the calculation:
Momentum flux of incident wave: S/c = (108 W/m2)/c
Force exerted on an area A of the mirror:  0.4*(S/c)A + 0.6*(2S/c)A
Radiation pressure:  P = 0.4*⅓ N/m2 + 0.6*⅔ N/m2 = 8/15 N/m2

#### Problem:

It has been proposed to drive a spacecraft remotely by directing an intense electromagnetic beam to the craft.
(a)  Is it more efficient to absorb the beam or reflect it from the craft?
(b)  If the beam carries 106 watts uniformly in an area of 5 m2 and the beam is reflected, how long would it take for a 1000 kg spaceship to reach a final velocity of 106 m/s, and how far would the craft travel in this time?

Solution:

• Concepts:
Energy and momentum flux for an electromagnetic wave, momentum conservation
• Reasoning:
Electromagnetic waves carry energy and momentum. When an EM wave is absorbed or reflected, momentum must be conserved.
• Details of the calculation:
(a)  Approximately twice as much momentum is transferred by reflecting the beam than by absorbing the beam.  Nonrelativistic kinematics can be used.
(b)  <S> = 2*105 W/m2 = energy flux.
<S>/c = momentum flux.
dp/dt = 2*<S>A/c = 2*106 W/c = ⅔*10-2 N.
dv/dt = ⅔*10-5 m/s2.
t = 106 s/(⅔*10-5) = 1.5*1011 s.
d = ½*⅔*10-5(1.5*1011)2 m = 7.5*1016 m = 7.9 ly.

#### Problem:

By considering the Poynting vector S = (1/μ0)(E×B) comment as quantitatively as possible on the energy flow in the following situations:
(a)  a long cylindrical wire of conductivity σ carrying a current I.
(b)  a stationary electric charge q sitting on top of a magnetic dipole with magnetic moment m.

Solution:

• Concepts:
The Poynting vector, energy conservation
• Reasoning:
The Poynting vector represents the energy flux in the electromagnetic field.  The energy can circulate or flow into an object.  If it flows into an object and is absorbed, energy conservation requires that the field energy is converted into another form of energy.
• Details of the calculation:
(a)  In the wire j = σE.  Let j = j k, I = jπR2 = σEπR2.  E = I/(σπR2).
On the surface of the wire E = I/(σπR2) k.
Outside of the wire B = (φ/φ) μ0I/(2πr).  On the surface B = (φ/φ) μ0I/(2πR).
S = (1/μ0)(E×B).  S = -(ρ/ρ) I2/(2σπ2R3).
Energy is flowing into the wire and is converted into heat.
[The field energy that is flowing into the wire per unit length is P = S2πR = I2/(σπR2).
The thermal energy dissipated per unit length is P = I2Rres per unit length.
Rres per unit length = 1/(σπR2), since j = σE,  πR2j = πR2σE, or,
I = πR2σV/l = V/Rres,  Rres = l/(πR2σ).
Therefore thermal energy dissipated per unit length = field energy flowing into the wire per unit length.]

(b)  Assume the point charge and the magnetic dipole are located at the origin with m = m k.
E = [1/(4πε0)] (q/r2) (r/r),  B = [μ0/(4π)] (m/r3)(2cosθ (r/r) + sinθ (θ/θ)).
S = (1/μ0)(E×B), S = 1/(16π2ε0) (mq/r5) sinθ (φ/φ).
Field energy circulates about the z-axis in the (φ/φ) direction.

#### Problem:

An "air core" toroid of mean radius R and cross-sectional area πr2 is wound with N turns of wire.  (r << R).  A current with a time dependence I(t) = Kt is turned on at t = 0.
(a)  Find the energy stored in the magnetic field at time t.
(b)  Find the direction and magnitude of the Poynting vector at a point just inside the toroid at time t.
(c)  Using the Poynting vector, find the rate of change of the field energy inside the toroid at time t.  Check your answer with that of part (a).

Solution:

• Concepts:
The Poynting vector
• Reasoning:
The Poynting vector represents the energy flux in the electromagnetic field.
• Details of the calculation:

(a)  B inside the torus is tangential the to a circle with radius R (right-hand rule) with approximately constant magnitude
B(t) = μ0nI(t) = μ0NI(t)/(2πR) = μ0NKt/(2πR)
The energy stored in the magnetic field inside the torus is
Umag = (1/(2μ0))B2*πr2*2πR = μ0N2K2t2r2/(4R)
(b)   The flux of B through a loop encircling the inner surface of the torus is B(R)πr2.  If this flux changes an electric field is induced.
The  electric field just inside the torus is tangential to a circle with radius r (Lenz's rule) with magnitude given by
2πr*E = πr2∂B/∂t =  μ0Nkπr2/(2πR).  E = μ0Nkr/(4πR)
S = (1/μ0)(E×B), S = μ0N2k2tr/(8π2R2) pointing everywhere towards the center circle (r = 0) of the torus.
(c)  P = ∫A(S∙n)dA  = S*2πR*2πr = μ0N2K2tr2/(2R) = dUmag/dt.

#### Problem:

A long coaxial cable consists of two concentric cylindrical conducting sheets of radii R1 and R2 respectively (R2 > R1).  The two conductors are connected to a battery which maintains a voltage V0 between them.  They carry equal and opposite currents I.
(a)  Use Ampere's law to calculate the magnetic field everywhere.
Plot the magnetic field as a function of r (use cylindrical coordinates).
(b)  Use Gauss' law to show that the electric field between the two conductors is
E
(r) = Cr/r2.  Express C in terms of R1, R2, and V0.  (Here r is a cylindrical coordinate.)
(c)  Calculate the Poynting vector.  Show that the power carried by the cable is P = V0I.

Solution:

• Concepts:
Gauss' law, Ampere's law, the Poynting vector S = (1/μ0)(E×B)
• Reasoning:
We are instructed to use Gauss' law and Ampere's law and to calculate the Poynting vector.
• Details of the calculation:

• (a)  For a circular loop Γ of radius r, concentric with the z-axis and lying in the x-y plane  we have
2πrB = μ0Ithrough ΓB = B (φ/φ).
r > R2:  B = 0.
r < R1:  B = 0.
R1 < r < R2B = (φ/φ) μ0I/(2πr).
(b)  For a Gaussian surface in the form of a cylinder with radius r and length L and its axis on the z-axis we have
2πrErL = λinsideL/ε0Er = λinside/(2πε0r)(r/r)
For a current to flow in the conductor in a static situation we need a voltage.
R1R2 E∙dr = V0.  ∫R1R2 E∙dr = λ/(2πε0)[lnR2 - lnR1),  λ = 2πε0V0/ln(R2/R1).
Er = V0/(ln(R2/R1))(r/r2).
If the conductors have no resistance, the electric field only has a radial component.
(c)  Assume E = V0/(ln(R2/R1))(r/r2), i.e. the conductors have no resistance.
Then S = (1/μ0)(E×B), S = k IV0/[2πr2 ln(R2/R1)].
P = ∫Sn dA = ∫R1R2 2πrdr IV0/[2πr2 ln(R2/R1)] = [IV0/ln(R2/R1)]∫R1R2 (1/r)dr = IV0.
P is the power flowing between the conductors into the k direction.

#### Problem:

Consider two concentric spherical metal shells of radius a and b, a < b.  There is a charge +q on the inner and a charge -q on the outer sphere.  A magnetic dipole with dipole moment m is in the center of the inner sphere.  Find the angular momentum associated with the electromagnetic field of the system.

Solution:

• Concepts:
The Poynting vector, energy momentum and angular momentum density in the electromagnetic field.
• Reasoning:
S = (1/μ0)(E×B) is the energy flux.  The momentum flux is S/c.  (Photon energy: hf, photon momentum: hf/c.)
The momentum flux equals the moment density times c.
When S is circulating the angular momentum density will not be zero.
• Details of the calculation:
Assume the magnetic dipole is located at the origin with m = m k.
Then B = [μ0/(4π)](m/r3)(2cosθ (r/r) + sinθ (θ/θ)).
The electric field is nonzero only between the shells.
E
= [1/(4πε0)](q/r2) (r/r) between the shells.
S = (1/μ0)(E×B), S = [1/(16π2ε0)](mq/r5) sinθ(φ/φ).
Field energy circulates about the z-axis in the (φ/φ) direction.
The  momentum density in the field is g = S/c2,
and the angular momentum density is r×g.
sinθ (φ/φ)= (r×z)/(rz)
r×(r×z) = (r∙z)r - (r∙r)z = rz cosθ r - r2 z.
L = ∫V r×g dV = -[mq/(16π2ε0c2)]∫V dV (cosθ(xi + yj + zk)/r5 - (1/r4)k)
= [mq/(16π2ε0c2)]k[∫ab 4πr2dr/r4 - ∫ab 0π 2πr2dr sinθdθ  rcos2θ/r5]
The x and y components of L vanish from symmetry.
L = [mq/(4πε0c2)]k[(1/a - 1/b) - (1/a - 1/b)∫0π cos2θ sinθdθ]
= ⅔mq/(4πε0c2)][(1/a - 1/b)k.

#### Problem:

A (locally) plane electromagnetic wave in vacuum is propagating in the positive z-direction.  The angular frequency of the wave is ω.  The wave generator slowly decreases the amplitude of the wave.  At the position z = 0 the amplitude of the wave is E0(1 – at) for times between t = 0 and t = 1/a, with a << ω.
Consider a cylindrical surface as shown.

(a)  Find the average outward energy flux from the cylinder.
(b)  Find the field energy enclosed by the surface, and show that it decreases at a rate equal to the flux found in part (a).

Solution:

• Concepts:
The Poynting vector, energy density
• Reasoning:
The Poynting vector represents the energy flux.
• Details of the calculation:
(a)  At z = 0 and t = t0 we have E(0,t0) = E0(1 – at0) cos(ωt0).
The field at z = L and t = t0 is the same as the field at z = 0 and t = t0 - L/c.
E
(L,t0) = E0(1 – a(t0 – L/c) cos(ω(t0 – L/c)).
S
= (1/μ0)(E×B) = k (1/μ0)E2 = k ε0c E2.
S
points in the z-direction, so the integral of S∙n over the surface of the cylinder reduces to an integral over the ends of the cylinder.
Sn dA = ε0cπr2E02[(1 - at0 + aL/c2)cos2(ωt0 - L/c) - (1 - at0)2cos2(ωt0)].
Since a <<  ω, averaging over one period of the oscillation yields
<S>∙n dA = ½ε0cπr2E02[(1 - at0 + aL/c2) - (1 - at0)2]
= ½ε0cπr2E02[(1 - at0)(2aL/c) + (aL/c)2].

(b)  The energy density is u = (1/(2μ0))B2 + (ε0/2)E2 = ε0E2.
The energy density averaged over one period is <u> =  (ε0/2)E2.
The average energy inside the cylinder at t = t0  therefore is
U = (ε0/2)πr2E020Ldx (1 – at + ax/c)2 = ½ε0cπr2E02(ax/c - at + 1)3/(3a)|0L.
U = ½ε0cπr2E02[(aL/c - at + 1)3/(3a) - (1 - at)3/(3a)].
U decreases at a rate
dU/dt = ½ε0cπr2E02[-(aL/c - at + 1)2 + (1 - at)2]
= -ε0cπr2E02[(1 - at)(2aL/c) + (aL/c)2].
At t = t0, the rate at which W decreases equals the average outward energy flux from the cylinder.