More Problems

Problem 1:

A sphere of radius R is uniformly magnetized with magnetization M and has a magnetic dipole moment m = 4πR³M/3.
The magnetic field due to this sphere is a pure dipole field outside the sphere and it is uniform inside the sphere.
(a)  Find B_inside and H_inside in terms of M.
(b)  If the sphere is made of lih material with magnetic susceptibility Χ_m > 0 and it is magnetized because it is located in a uniform external field B₀, find M in terms B₀ or H₀

Solution:

• Concepts:
  Magnetostatics, magnetic materials
• Reasoning:
  Orient your coordinate system so that M points in the z-direction and place the center of the sphere at the origin
  B is continuous on the z-axis at at z = R.
• Details of the calculation:
  (a)  B(r) = (μ₀/4π)(3(m·r)r/r⁵ - m/r³) = (μ₀m/(4πr³))[2 cosθ (r/r) + sinθ (θ/θ)] outside the sphere.
  On the z-axis at z = R we have B = 2μ₀m/(4πR³) = 2μ₀M/3.
  B_in = 2μ₀M/3, H_inside = B_in/μ₀ - M = -M/3.
  
  (b)  The total field inside the sphere is the superposition of the field of a uniform magnetized sphere and the constant external field B₀.  We need to find M.
  B_in = B₀ + 2Mμ₀/3,  μH_in = B_in,  M = χ_mH_in = (χ_m/μ)B_in.
  (1 - 2μ₀χ_m/(3μ))B_in = B₀. 
  μ = μ₀(1 + Χ_m).  Therefore
  B_in = B₀(3 + 3χ_m)/(3 + χ_m).
  μ₀M = B₀([(3 + 3χ_m)/(3 + χ_m)][ χ_m/(1 + Χ_m)] = 3χ_mB₀/(3 + χ_m). 
  M = 3χ_mH₀/(3 + χ_m).

Problem 2:

An electric dipole p₀ makes an angle of 30 degrees with respect to the z-axis as it rotates about the z-axis with angular speed ω.  Find its initial rate of energy loss.

Solution:

• Concepts:
  Electric dipole radiation
• Reasoning:
  The x and y 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:
  p(t) = p₀ cosθ k + p₀ sinθ (cos(ωt) i + sin(ωt) j), where θ is the tilt angle.
  d²p/dt² = -ω²p₀ sinθ [cos(ωt) i + sin(ωt) j].
  <(d²p/dt²)²> = ω⁴p₀²sin²θ = ω⁴p₀²/4,   P_rad = ω⁴p₀²/(24πε₀c³).

Problem 3:

The Larmor formula gives the power radiated by a particle of charge q with an acceleration a.  Use this formula and the Bohr model of the hydrogen atom to estimate the time for an electron in the lowest Bohr orbit to spiral into the nucleus.  It is a valid approximation to assume that the orbits remain almost circular as the electron spirals.  Show your work!

Solution:

• Concepts:
  The radiation field of a point charge moving non-relativistically, the Larmor formula
• Reasoning:
  The electron orbits the "infinitely heavy" proton.  Its acceleration is a = v²/r, and it therefore radiates away energy,
  P = -dE/dt = [q²/(6πε₀c³)]a².  (SI units)
• Details of the calculation:
  To solve for the total time it takes the electron to spiral into the nucleus, we can express the acceleration a in terms of E.  We then have
  -dE/dt = f(E), ∫dt = Δt = ∫_E0^-∞dE/f(E).
  Alternatively, we can express E as a function of the radius r and derive an expression -dr/dt = f(r).  We then have
  ∫dt = ΔT = ∫_r0⁰dr/f(r).
  To find the relationship between E and a we can use
  F = q²/(4πε₀r²) = mv²/r.  mv² = q²/(4πε₀r),
  E = ½mv² - q²/(4πε₀r) = - q²/(8πε₀r) = -½mv².
  Now we can express a in terms of E.
  a = v²/r = q²/(4πε₀r²m) = 2E²8πε₀/(q²m).
  -dE/dt = E⁴16²πε₀/(6c³q²m²).  dt = -(dE/E⁴) (6c³q²m²)/(16²πε₀).
  Δt = (6c³q²m²)/(16²πε₀)∫_E0^-∞(dE/E⁴)  = -(6c³q²m²)/(16²πε₀)(E₀^-3/3)
  where E₀ = -13.6 eV,  Δt = 1.5*10^-11s.
  Or we can also express both E and a in terms of r.
  a = F/m = q²/(4πε₀r²m),  E = - q²/(8πε₀r),
  dE = [q²/(8πε₀r²)]dr = -Pdt.
  dt = -[q²/(8πε₀r²)]dr/P = -[q²/(8πε₀r²)]dr/[q²a²/(6πε₀c³)].
  dt = -dr[3c³/(4r²a²)] = -r²dr[12c³π²ε₀²m²/q⁴].
  Δt = -[12c³π²ε₀²m²/q⁴]∫_r0⁰r²dr = [12c³π²ε₀²m²/q⁴](r₀³/3)
  where r₀ = Bohr radius = 5.29*10^-11m,  Δt = 1.5*10^-11s.