More Problems

Problem 1:

An infinitely long cylinder with radius a and permeability μ is placed into an initially uniform magnetic field B₀ with its axis perpendicular to B₀.  Find the resultant field inside and outside of the cylinder.

Solution:

• Concepts:
  The principle of superposition, boundary conditions, B = μH.
• Reasoning:
  The external magnetic field will magnetize the cylinder.   The total magnetic field will be the superposition of the magnetic field of a cylinder with uniform magnetization M = mk and  the external field B₀.
• Details of the calculation:
  Assume that the magnetic field will be of the form
  B_in = B₂(cos(φ)[ρ/ρ] - sin(φ)[φ/φ]) = B₂k.
  B_out = B₀ + (B₁/ρ²)(cos(φ)[ρ/ρ] + sin(φ)[φ/φ]).
  If this form of the field can satisfy the boundary conditions, then the uniqueness theorem guaranties that it is the correct and only solution.
  The boundary conditions at ρ = a are B∙n = continuous, H∙t = continuous.
  B∙n = continuous -->  B₀cos(φ) + (B₁/a²)cos(φ) = B₂cos(φ), 
  B₀ + (B₁/a²) = B₂.
  H∙t = continuous -->  -(B₀/μ₀)sin(φ) + (B₁/(μ₀a²))sin(φ) = -(B₂/μ)sin(φ), 
  -(B₀/μ₀) + (B₁/(μ₀a²)) = -(B₂/μ).
  Combining:  -(B₀/μ₀) + (B₁/(μ₀a²)) = -(B₀/μ) - (B₁/(μa²)),  (B₁/a²)(μ + μ₀) = B₀(μ - μ₀).
  B₁ = a²B₀(μ - μ₀)/(μ + μ₀),  B₂ = B₀2μ/(μ + μ₀).
  B_in = [B₀2μ/(μ + μ₀)]k.
  B_out = B₀ + [B₀(μ - μ₀)/(μ + μ₀)](a²/ρ²)(cos(φ)[ρ/ρ] + sin(φ)[φ/φ]).

Problem 2:

In SI units, the Larmor formula for the average power radiated by an electric dipole p is P_rad = <(d²p/dt²)²>/(6πε₀c³).
What is the Larmor formula for the average power radiated by a magnetic dipole m?

Solution:

• Concepts:
  Magnetic dipole radiation
• Reasoning:
  The magnetic dipole will radiate energy at a rate P_rad = <(d²m/dt²)²>/(6πε₀c⁵).
  This is something you should remember.
  
  Memory aid:
  The static electric field E produced by an electric dipole p has the same dependence on the spatial coordinates as the magnetic field B produce by a magnetic dipole m pointing in the same direction as p.
  E(r) = (1/(4πε₀))[3(p₁∙r)r/r⁵ - p/r³]. 
  B(r) = (μ₀/4π)[3(m∙r)r/r⁵ - m/r³] = (1/(4πε₀c²))[3(m∙r)r/r⁵ - m/r³].
  From the symmetry of Maxwell's equations in free space, we may expect the radiation field to also have the same spatial dependence.
  For radiation fields B = E/c or E = Bc.  In SI units, we therefore expect the electric radiation field produced by the magnetic dipole to be  a factor of 1/c smaller that the radiation electric radiation field produced by the electric dipole of the same magnitude.  The power radiate is proportional to E². 
  We could expect that P_rad = <(d²m/dt²)²>/(6πε₀c⁵).

Problem 3:

A non-conducting sphere of radius R = 0.1 m, mass M = 10 kg and uniform mass density carries a surface charge density σ = σ₀cosθ, with σ₀ = 10 microCoulomb.
(a)  Find the dipole moment p₀ of the sphere.
(b)  Assume that at t = 0 the sphere receives an angular impulse and starts rotating about the x-axis with angular velocity ω₀ = 1000/s.  Calculate the power radiated by the sphere when it rotates with angular frequency ω.
(c)  Estimate the time it takes for the rotation rate of the sphere to decrease by a factor of 2.

Solution:

• Concepts:
  Dipole moment, the Larmor formula
• Reasoning:
  The y and z components of the dipole moment are oscillating with angular frequency ω.  The rotating electric dipole will radiate energy at a rate P_rad = <(d²p/dt²)²>/(6πε₀c³).
• Details of the calculation:
  (a)  Dipole moment: p = ∫σ(r)r dA.  Symmetry dictates that p = p k.
  p = k ∫₀^πσ₀cosθ Rcosθ 2πR²sinθdθ = k σ₀(4π/3)R³ = k p₀.
  p₀ = 4.19*10^-8 Cm.
  
  (b)  p(t) = p₀cos(ωt) k + p₀sin(ωt) j.
  d²p/dt² = -ω²p,  <(d²p/dt²)²> = ω⁴p₀².
  P_rad = <(d²p/dt²)²>/(6πε₀c³) = ω⁴p₀²/(6πε₀c³).
  <P> = ω⁴*3.9*10^-31 W s⁴.
  For ω = ω₀ = 1000/s we have <P> = 3.9*10^-19 W.
  
  (c)  <P> = -dE/dt.  E = ½Iω² = (1/5)MR²ω².
  -dE/dt = -(2/5)MR²ω dω/dt  = ω⁴p₀²/(6πε₀c³).
  -5p₀²dt/(12 MR²πε₀c³) = dω/ω³.
  [-5p₀²/(12 MR²πε₀c³)] ∫₀^T dt = ∫_ω0 ^ω0/2 dω/ω³.
  [5p₀²/(12 MR²πε₀c³)] T = ½((ω₀/2)^-2 - ω₀^-2).
  T ≈ 10²³ s.