(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 = 10^{12}W/cm^{2}, 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 4πr^{2}with distance r.

(b) The magnitude of the Poynting vector**S**= (1/μ_{0})**E**×**B**relates the intensity of an electromagnetic wave to the field strength. For a sinusoidal electromagnetic wave we have <S> = (|E_{0}^{2}|/2)(ε_{0}/μ_{0})^{½}.

The electron in hydrogen atom experiences the electric field of a point charge q = 1.6*10^{-19}C a distance a_{0}= 5.29*10^{-11}m from the point charge, E_{q}= q/(4πε_{0}a_{0}^{2}) . - Details of the calculation:

(a) For a light bulb we have: Intensity = energy/(area * time) proportional to 1/r^{2}.

Intensity at 6 m = (1/9) * intensity at 2 m.

A laser beam with a Gaussian profile has a waist d_{0}. 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/e^{2}of its value on the beam axis.

For a Gaussian beam of a particular wavelength, the product d_{0}θ is constant. A highly collimated beam with small θ has a large beam waist d_{0}. Over short distances, the change in d is negligible to d_{0}. The intensity is therefore practically the same at 2 m and at 6 m.(b) 10

^{12}W/(0.01 m)^{2}= (|E_{0}^{2}|/2)(ε_{0}/μ_{0})^{½}, E_{0}= 2.74*10^{9}N/C is the electric field of the light wave.

E_{q}= q/(4πε_{0}a_{0}^{2}) = 5.15*10^{11}N/C is the electric field experienced by the electron in a Hydrogen atom.

E_{q}>> E_{0}.

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 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/m^{2})(4*10^{-3 }m^{2}) = 0.024 W.

dp/dt = 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 in the incident wave direction.

(b) F = dp/dt = 1.6*10^{-10}N.

A beam of light with an intensity of 10^{8} W/m^{2} 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 dp_{perp}/dt = (1/c)SA_{perp}, where A_{perp}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 = (10^{8}W/m^{2})/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*(1/3) N/m^{2}+ 0.6*(2/3) N/m^{2}= 8/15 N/m^{2}

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 10^{6} watts uniformly in an area of 5 m^{2}
and
the beam is reflected, how long would it take for a 1000 kg spaceship to reach a
final velocity of 10^{6} 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*10^{5}W/m^{2}= energy flux.

<S>/c = momentum flux.

dp/dt = 2*<S>A/c = 2*10^{6 }W/c = (2/3)*10^{-2}N.

dv/dt = (2/3)*10^{-5}m/s^{2}.

t = 10^{6 }s/((2/3)*10^{-5}) = 1.5*10^{11 }s.

d = ½*(2/3)*10^{-5}(1.5*10^{11})^{2 }m = 7.5*10^{16}m = 7.9 ly.

By considering the Poynting vector
**S** = (1/μ_{0})(**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**= σ**E**. Let**j**= j**k**, I = jπR^{2}= σEπR^{2}. E = I/(σπR^{2}).

On the surface of the wire**E**= I/(σπR^{2})**k**.

Outside of the wire**B**= (**φ**/φ) μ_{0}I/(2πr). On the surface**B**= (**φ**/φ) μ_{0}I/(2πR).

**S**= (1/μ_{0})(**E**×**B**).**S**= -(**ρ**/ρ) I^{2}/(2σπ^{2}R^{3}).

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 = I^{2}/(σπR^{2}).

The thermal energy dissipated per unit length is P = I^{2}R_{res per unit length}.

R_{res per unit length}= 1/(σπR^{2}), since j = σE, πR^{2}j = πR^{2}σE, or,

I = πR^{2}σV/l = V/R_{res}, R_{res}= l/(πR^{2}σ).

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/(4πε_{0})] (q/r^{2}) (**r**/r),**B**= [μ_{0}/(4π)] (m/r^{3})(2cosθ (**r**/r) + sinθ (**θ**/θ)).

**S**= (1/μ_{0})(**E**×**B**),**S**= 1/(16π^{2}ε_{0}) (mq/r^{5}) sinθ (**φ**/φ).

Field energy circulates about the z-axis in the (**φ**/φ) direction.

An "air core" toroid of mean radius R and cross-sectional
area πr^{2} 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) = μ_{0}nI(t) = μ_{0}NI(t)/(2πR) = μ_{0}NKt/(2πR)

The energy stored in the magnetic field inside the torus is

U_{mag}= (1/(2μ_{0}))B^{2}*πr^{2}*2πR = μ_{0}N^{2}K^{2}t^{2}r^{2}/(4R)

(b) The flux of B through a loop encircling the inner surface of the torus is B(R)πr^{2}. 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 = πr^{2}∂B/∂t = μ_{0}Nkπr^{2}/(2πR). E = μ_{0}Nkr/(4πR)

**S**= (1/μ_{0})(**E**×**B**),**S**= μ_{0}N^{2}k^{2}tr/(8π^{2}R^{2}) pointing everywhere towards the center circle (r = 0) of the torus.

(c) P = ∫_{A}(**S∙n**)dA = S*2πR*2πr = μ_{0}N^{2}K^{2}tr^{2}/(2R) = dU_{mag}/dt.

A long coaxial cable consists of two concentric cylindrical conducting sheets
of radii R_{1} and R_{2} respectively (R_{2 }> R_{1}).
The two conductors are connected to a battery which maintains a voltage V_{0}
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)
= C

(c) Calculate the Poynting vector. Show that the power carried by the cable is P = V

Solution:

- Concepts:

Gauss' law, Ampere's law, the Poynting vector**S**= (1/μ_{0})(**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 = μ_{0}I_{through Γ}.**B**= B (**φ**/φ).

r > R_{2}: B = 0.

r < R_{1}: B = 0.

R_{1}< r < R_{2}:**B**= (**φ**/φ) μ_{0}I/(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

2πrE_{r}L = λ_{inside}L/ε_{0}.**E**_{r}_{inside}/(2πε_{0}r)(**r**/r)

For a current to flow in the conductor in a static situation we need a voltage.

∫_{R1}^{R2}**E**∙d**r**= V_{0}. ∫_{R1}^{R2}**E**∙d**r**= λ/(2πε_{0})[lnR_{2}- lnR_{1}), λ = 2πε_{0}V_{0}/ln(R_{2}/R_{1}).

**E**_{r}_{0}/(ln(R_{2}/R_{1}))(**r**/r^{2}).

If the conductors have no resistance, the electric field only has a radial component.

(c) Assume**E**= V_{0}/(ln(R_{2}/R_{1}))(**r**/r^{2}), i.e. the conductors have no resistance.

Then**S**= (1/μ_{0})(**E**×**B**),**S**=**k**IV_{0}/[2πr^{2}ln(R_{2}/R_{1})].

P = ∫**S**∙**n**dA = ∫_{R1}^{R2}2πrdr IV_{0}/[2πr^{2}ln(R_{2}/R_{1})] = [IV_{0}/ln(R_{2}/R_{1})]∫_{R1}^{R2}(1/r)dr = IV_{0}.

P is the power flowing between the conductors into the**k**direction.

- (a) For a circular loop Γ of radius r,
concentric with the z-axis and lying in the x-y plane we have

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/μ_{0})(**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**= [μ_{0}/(4π)](m/r^{3})(2cosθ (**r**/r) + sinθ (**θ**/θ)).

The electric field is nonzero only between the shells.= [1/(4πε

E_{0})](q/r^{2}) (**r**/r) between the shells.

**S**= (1/μ_{0})(**E**×**B**),**S**= [1/(16π^{2}ε_{0})](mq/r^{5}) sinθ(**φ**/φ).

Field energy circulates about the z-axis in the (**φ**/φ) direction.

The momentum density in the field is**g**=**S**/c^{2},

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**- r^{2}**z**.

L = ∫_{V }**r**×**g**dV = -[mq/(16π^{2}ε_{0}c^{2})]∫_{V }dV (cosθ(x**i**+ y**j**+ z**k**)/r^{5}- (1/r^{4})**k**)

= [mq/(16π^{2}ε_{0}c^{2})]**k**[∫_{a}^{b}_{ }4πr^{2}dr/r^{4}- ∫_{a}^{b}_{ }∫_{0}^{π}_{ }2πr^{2}dr sinθdθ rcos^{2}θ/r^{5}]

The x and y components of**L**vanish from symmetry.

L = [mq/(4πε_{0}c^{2})]**k**[(1/a - 1/b) - (1/a - 1/b)∫_{0}^{π}cos^{2}θ sinθdθ]

= ⅔mq/(4πε_{0}c^{2})][(1/a - 1/b)**k**.

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 E_{0}(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 = t_{0}we have**E**(0,t_{0}) =**E**_{0}(1 - at_{0}) cos(ωt_{0}).

The field at z = L and t = t_{0}is the same as the field at z = 0 and t = t_{0}- L/c.(L,t

E_{0}) =**E**_{0}(1 - a(t_{0}- L/c) cos(ω(t_{0}- L/c)).= (1/μ

S_{0})(**E**×**B**) =**k**(1/μ_{0})E^{2}=**k**ε_{0}c E^{2}.points in the z-direction, so the integral of

S**S∙n**over the surface of the cylinder reduces to an integral over the ends of the cylinder.

∮**S**∙**n**dA = ε_{0}cπr^{2}E_{0}^{2}[(1 - at_{0}+ aL/c^{2})cos^{2}(ωt_{0}- L/c) - (1 - at_{0})^{2}cos^{2}(ωt_{0})].

Since a << ω, averaging over one period of the oscillation yields

∮**S**>∙**n**dA = ½ε_{0}cπr^{2}E_{0}^{2}[(1 - at_{0}+ aL/c^{2}) - (1 - at_{0})^{2}]

= ½ε_{0}cπr^{2}E_{0}^{2}[(1 - at_{0})(2aL/c) + (aL/c)^{2}].(b) The energy density is u = (1/(2μ

_{0}))B^{2}+ (ε_{0}/2)E^{2}= ε_{0}E^{2}.

The energy density averaged over one period is <u> = (ε_{0}/2)E^{2}.

The average energy inside the cylinder at t = t_{0}therefore is

U = (ε_{0}/2)πr^{2}E_{0}^{2}∫_{0}^{L}dx (1 - at + ax/c)^{2}= ½ε_{0}cπr^{2}E_{0}^{2}(ax/c - at + 1)^{3}/(3a)|_{0}^{L}.

U = ½ε_{0}cπr^{2}E_{0}^{2}[(aL/c - at + 1)^{3}/(3a) - (1 - at)^{3}/(3a)].

U decreases at a rate

dU/dt = ½ε_{0}cπr^{2}E_{0}^{2}[-(aL/c - at + 1)^{2}+ (1 - at)^{2}]

= -ε_{0}cπr^{2}E_{0}^{2}[(1 - at)(2aL/c) + (aL/c)^{2}].

At t = t_{0}, the rate at which W decreases equals the average outward energy flux from the cylinder.