Problem 1:

Why are "classical atoms" unstable?

Solution:

• Concepts:
  Accelerating charges produce electromagnetic radiation.
• Reasoning:
  In the Rutherford atomic model, electrons move around the nucleus in elliptical orbits.  Classical electrodynamics requires radiation to be emitted when a charged particle accelerates.  This means that the electrons would lose energy continuously and ultimately be captured by the nucleus.

Problem 2:

Consider the current loop of width a and length b shown in the figure.

Find an expression for the vector potential that is valid anywhere.

Solution:

• Concepts:
  The vector potential A(r)
• Reasoning:
  ∇·B = 0 --> B = ∇×A.
  In magnetostatics we choose ∇·A = 0. 
  Then  ∇²A = -μ₀j,  A(r) = (μ₀/(4π))∫_V'dV' j(r')/|r - r'|.
• Details of the calculation:
  (a)  The vector potential due to the filamentary current is given by
  A(r) = (μ₀/(4π))∫_loopI(r')dl'/|r - r'|.
  Since I(r')dl' has no x-component, A_x = 0.
  A_z(r) = (μ₀/(4π))[∫₀^aIdz'/(x²+y²+(z-z')²)^½ - ∫₀^aIdz'/(x²+(y-b)²+(z-z')²)^½]
  = (μ₀I/(4π))[∫_z-a^zdz''/(x²+y²+z''² )^½ - ∫_z-a^zdz''/(x²+(y-b)²+z''²)^½]
  = (μ₀I/(4π))[ln(z''+(x²+y²+z''²)^½) - ln(z''+(x²+(y-b)²+z''²)^½)]^z_z-a
  = (μ₀I/(4π))[ln{(z+(x²+y²+z²)^½)/(z+(x²+(y-b)²+z²)^½} - ln{(z-a+(x²+y²+(z-a)²)^½)/(z-a+(x²+(y-b)²+(z-a)²)^½)}]
  = (μ₀I/(4π))ln{(z+(x²+y²+z²)^½)(z-a+(x²+(y-b)²+(z-a)²)^½)/((z+(x²+(y-b)²+z²)^½)(z-a+(x²+y²+(z-a)²)^½))}.
  Similarly
  A_y(r) = (μ₀I/(4π))[∫₀^bdy'/(x²+(y-y')²+(z-a)²)^½ - ∫₀^bdy'/(x²+(y-y')²+ z²)^½]
  = (μ₀I/(4π))ln{(y+(x²+y²+(z-a)²)^½)(y-b+(x²+(y-b)²+z²)^½)/((y+(x²+y²+z²)^½)(y-b+(x²+(y-b)²+(z-a)²)^½))}.
  
  Supplement, not part of the requested solution:
  The vector potential of a magnetic dipole m at the origin is A(r) = (μ₀/(4π))m×r/r³, with magnitude μ₀m/(4πr²).
  The loop has a dipole moment m = -Iab i, therefore at large distance r the vector potential should be
  A(r) = -(μ₀Iab(4πr²))i×(r/r).
  On the y-axis at y >> a,b we expect A = A_zk, with A_z = -μ₀Iab(4πy²).
  We can check this.  On the y-axis x = z = 0. 
  A_z(y) = (μ₀I/(4π))ln(y/(y-b)) - ln[(-a + (y²+a²)^½)/(-a +((y-b)²+a²)^½)].
  If y >> a,b we have
  ln(y/(y-b)) = ln(y/(y(1-b/y))) = ln(1/(1-b/y)) ≈ ln(1+b/y)) ≈ b/y,
  since ln(1+x) ≈ for x << 1.
  Similarly,
  ln[(-a+(y²+a²)^½)/(-a+((y-b)²+a²)^½)] ≈ ln[(-a+y)/(-a+y-b))]
  = ln[1/(1 - b/(y-a))]≈ ln[1 + b/(y-a)]≈ b/(y-a) ≈ b/y(1 + a/y) = b/y + ab/y².
  Therefore A_z(y) ≈ -(μ₀I/(4π))(ab/y²), as expected.

Problem 3:

A non-relativistic positron of charge q_e and velocity v₁ (v₁ << c) impinges head-on on a fixed nucleus of charge Zq_e.  The positron which is coming from far away (∞), is decelerated until it comes to rest and then accelerated again in the opposite direction until it reaches a terminal velocity v₂.  Taking radiation loss into account (but assuming it is small), find v₂ as a function of v₁ and Z.

Solution:

• Concepts:
  Motion in a central field, energy conservation, the Larmor formula
• Reasoning:
  The positron moves in a central field.  It has no angular momentum about the position of the nucleus (at the origin).  If we neglect radiation losses, we can find its velocity and acceleration as a function of distance from energy conservation.  Using the Larmor formula, we then can find the energy lost in a small time interval dt or a small distance dr.  Integrating these losses over the path, we find E₂ and v₂.  This works if the energy radiated is much less than the characteristic energy of the system.
• Details of the calculation:
  E = ½mv₁² = ½mv² + Ze²/r.  v(r) = (v₁² - 2Ze²/(mr))^½, with e² = q_e²/(4π²ε₀).
  a = dv/dt = (dv/dr)v = Ze²/(mr²).
  The Larmor formula gives the power radiated by an accelerating charge.
  P = 2e²a²/(3c³) = [2e²/(3c³)](Ze²/(mr²))² = -dE/dt = -(dE/dr)v.
  dt = dr/v.
  ΔE = -∫Pdt = -∫[2e²/(3c³)](Ze²/(mr²))²dt = -2∫_rmin^∞[2e²/(3c³)](Ze²/(mr²))²dr/v
  = -[4e⁶Z²/(3c³m²)]∫_rmin^∞r^-4dr/v(r).
  v(r) = v₁(1 - 2Ze²/(mrv₁²))^½.  v(r_min) = 0.
  r_min = 2Ze²/(mv₁²).
  ΔE = -[4e⁶Z²/(3c³m²v₁)]∫_rmin^∞r^-4dr/(1 - r_min/r)^½.
  Let x = r_min/r,  dx = -r_mindr/r².
  ΔE = -[4e⁶Z²/(3c³m²v₁r_min³)]∫₀¹x²dx/(1 - x)^½.
  From an integral table:
  ∫₀¹x²dx/(1 - x)^½ = [2(8 + 4x + 3x²)/15](1 - x)^½|₀¹ = -16/15.
  ΔE = [4e⁶Z²/(3c³m²v₁r_min³)]16/15 = (8/45)mv₁⁵/(Zc³).
  ΔE/E =  (16/45)Z^-1(v/c)³.
  For a nonrelativistic v₁ we have ΔE << E.
  ΔE is the radiation loss during the whole trip.
  ½mv₂² = ½mv₁² - ΔE,  v₂² = v₁² - (8/45)v₁⁵/(Zc³).
  v₂ ≈ v₁(1 - (8/45)Z^-1(v/c)³).

Problem 4:

Problem 5:

A long thin cylinder carries a fixed uniform magnetization M parallel to its axis.  Find the density, location and direction of the bound currents and the magnetic field at the geometrical center of the cylinder.

Solution:

• Concepts:
  Magnetic materials
• Reasoning:
  The total current density is due to free and to magnetization current densities.  In this problem the free current densities are zero.
• Details of the calculation:
  Let M point into the z-direction, and the axis of the cylinder be the z-axis.
  j_m = ∇×M,  k_m = M×n,  j_m = volume current density, k_m = surface current density.
  j_m = 0, k_m = M φ/φ   (right-hand rule).
  B = B k at the geometrical center of the cylinder.  B = μ₀k_m = μ₀M (infinitely long cylinder approximation).