Assignment 4

Problem 2:

A thin wire of radius b is used to form a circular wire loop of radius a (a >> b) and total resistance R.
The loop is rotating about the z-axis with constant angular velocity ωk in a region with constant magnetic field B = B₀i. 
At t = 0 the loop lies in the y-z plane and the point A at the center of the wire crosses the y-axis.

 

Let the (θ/θ) direction be tangential to the loop and be equal to the positive z direction at point A.
Let the (φ/φ) direction be tangential to the wire and be equal to the direction indicated in the figure.
(a)  Find the current flowing in the loop.  Neglect the self-inductance of the loop.  What is current density J as a function of time?
(b)  Find the thermal energy generated per unit time, averaged over one revolution.
(c)  Write down an expression for the the Poynting vector S on the surface of the wire.
(d)  Use S to find the field energy per unit time flowing into the wire, averaged over one revolution.

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)  Flux Φ = B∙A = B₀πa² cos(ωt).  (Define the normal to the area by the right-hand rule.)
  emf = -dΦ/dt = ωB₀πa² sin(ωt),  I = ωB₀πa² sin(ωt)/R.  At t = 0 + dt the current at point A flows in the z-direction.
  The current density in the wire is J = (θ/θ) (a²/b²)ωB₀ sin(ωt)/R.
  
  (b)  P = I²R = ω²B₀²π²a⁴ sin²(ωt)/R.  
  <P> = ½ω²B₀²π²a⁴/R is the thermal energy generated per unit time, averaged over one revolution.
  
  (c)  E_wire = (θ/θ)ωB₀πa² sin(ωt)/(2πa) = ½ωB₀a sin(ωt)(θ/θ). 
  B_wire = μ₀I/(2πb)(φ/φ).  B = B₀ + B_wire.
  S = (1/μ₀)(E×B) = (1/μ₀)(E_wire×B_wire) + (1/μ₀)(E_wire×B₀).
  (E_wire×B_wire) always points towards the center of the wire.
  Let us call this direction the -(ρ/ρ) direction.
  Field energy flowing into the wire per unit time per unit length:
  -∫₀^2πS∙(ρ/ρ) bdφ = (1/μ₀)|E_wire×B_wire|2πb = ω²B₀²π²a⁴ sin²(ωt)/(2πaR).
  [(1/μ₀)(E_wire×B₀) has the same direction all along a circumference of the wire and therefore there is no net flux into the wire due to this term.]
  Total field energy flowing into the wire per unit time
  2πa(1/μ₀)|E_wire×B_wire|2πb = ω²B₀²π²a⁴ sin²(ωt)/R.
  Averaged over one revolution:  ½ω²B₀²π²a⁴/R = <P>.

Problem 3:

(a)  From Maxwell's equations, derive the conservation of energy equation,
(∂u/∂t) + ∇∙S = -E∙j,
where u is the energy density and S is the Poynting vector.  Rewrite this equation in integral form and explain why it is a statement of energy conservation.
(b)  A long, cylindrical conductor of radius a and conductivity σ carries a constant current I.  Find S at the surface of the cylinder and interpret your result in terms of conservation of energy.

Solution:

• Concepts:
  Maxwell's equations, energy density and energy flux in the electromagnetic field
• Reasoning:
  We are explicitly instructed to use Maxwell's equations to derive the conservation of energy equation.
• Details of the calculation:
  (a)  Maxwell's equations in SI units are
  ∇∙E = ρ/ε₀,  ∇×E = -∂B/∂t,  ∇∙B = 0,  ∇×B = μ₀j + (1/c²)∂E/∂t.
  E∙j = E∙(1/μ₀)(∇×B) - ε₀(∂E/∂t)∙E
  = -(1/μ₀)∇∙(E×B) + B∙(1/μ₀)(∇×E) - ε₀(∂E/∂t)∙E
  = -(1/μ₀)∇∙(E×B) - (1/μ₀)B∙(∂B/∂t) - ε₀(∂E/∂t)∙E
  = -(1/μ₀)∇∙(E×B) - (∂/∂t)(B²/2μ₀ + ε₀E²/2).
  (∂/∂t)(B²/2μ₀ + ε₀E²/2) + (1/μ₀)∇∙(E×B) = -E∙j.
  Interpretation:
  Let u = B²/2μ₀ + ε₀E²/2 be the energy density in the electromagnetic field.
  Let S = (1/μ₀)(E×B) be the energy flux.
  Then (∂u/∂t) + ∇∙S = -E∙j.
  ∫_V∇∙E  dV = ∮S∙n dA,  -∫_V ∂u/∂t dV = ∮S∙n dA +  ∫_V E∙j  dV.
  This is a statement of energy conservation.  The rate at which the field energy in a volume decreases equals the rate at which it leaves across the boundaries plus the rate at which it gets converted into other forms.
  (b)  In the wire j = σE.  Let j = j k, I = jπa² = σEπa².  E = I/(σπa²).
  On the surface of the wire E = I/(σπa²) k.
  Outside of the wire B = (φ/φ) μ₀I/(2πr).  On the surface B = (φ/φ) μ₀I/(2πa).
  S = (1/μ₀)(E×B).  
  S =  -(ρ/ρ) I²/(2σπ²a³).
  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 = S2πa = I²/(σπa²).
  The thermal energy dissipated per unit length is P = I²R_{unit length}.
  R_{unit length} = 1/(σπa²).
  Therefore thermal energy dissipated per unit length = field energy flowing into the wire per unit length.]

Problem 4:

Show that a radially oscillating spherically symmetric charge distribution does not radiate.
Assume a harmonic time variation of frequency ω and use the complex formalism,
namely ρ(r,t) = Re[ρ(r)exp(-iωt)], D(r,t) = Re[D(r)exp(-iωt)], and so forth.

(a)  First write down Maxwell's equations and the continuity equation using the complex formalism to eliminate time derivatives.
(b)  Note that one can write j(r) = rf(r).  Compare ∇∙D and ∇∙j to relate D and j.
(c)  Show that ∫ρ(r) d³r = 0.
(d)  What are D and B outside the source?  (Hint: Consider ∇∙B and ∇×B.)
(e)  What do you conclude concerning radial oscillations of a spherically symmetric source with an arbitrary time variation?

Solution:

• Concepts:
  Maxwell's equations, the continuity equation
• Reasoning:
  We use the continuity equation to relate ρ and j.  For a given ρ and j we now find E and B outside the source region.  We check if E×B = 0.
• Details of the calculation:
  (a)  ∇·E = ρ/ε₀,  ∇×E = -∂B/∂t, ∇·B = 0,  ∇×B = μ₀j + (1/c²)∂E/∂t, D = ε₀E.
  (In a lih medium change ρ to ρ_f, j to j_f, ε₀ to ε, μ₀ to μ, and c² to 1/(εμ).)
  Let ρ(r,t) = ρ(r)exp(-iωt),  E(r,t) = E(r)exp(-iωt), etc.
  Then  ∇·E(r) = ρ(r) /ε₀,  ∇×E(r) = iωB(r), ∇·B(r) = 0,  ∇×B(r) = μ₀j(r) - (iω/c²)E(r).
  The continuity equation ∇∙j = -∂ρ/∂t becomes ∇∙j(r)  = iωρ(r).
  
  (b)  Let j(r) = rf(r), independent of θ and φ.
  Then ∇∙j = r∙∇f + f(r)∇∙r and ρ(r) are also independent of θ and φ.
  From symmetry and Gauss' law we now have that E and D are radial, independent of θ and φ.
  Therefore ∇×E = 0 --> B = 0 everywhere --> j(r) = (iω/(μ₀c²))E(r) = iωD(r).
  
  (c)  ∫ρ(r) d³r ∝ ∫∇∙j d³r = ∫_surfacej∙n dA = 0 for a finite charge distribution.
  
  (d)  B = E = 0 outside the charge distribution --> no radiation.
  (e)  Requiring all quantities to vary sinusoidally yields E = 0 outside.  The source must contain equal amounts of positive and negative charge.  Arbitrary time variations do not require E = 0 outside.  But if ρ is independent of θ and φ, the E is radial and ∇×E = 0 --> B = 0 for a finite current distribution.