Problem 1:

A point electric dipole located at the origin has a time-dependent dipole moment given by
p(t) = p₀ cos(ωt) k.

In the far field region this gives
E(r, t) =μ ₀p₀ω²/(4πr) sinθ cos(ωt − kr) e_θ,
B(r, t) = μ₀p₀ω²/(4πrc) sinθ cos(ωt − kr) e_φ,
or electric and magnetic field.
(a)  Calculate the time-averaged power radiated per unit solid angle dP/dΩ .
(b)  Integrate over all angles to find the total radiated power.
(c)  Discuss the angular distribution of the radiation and where the radiation is strongest.

Solution:

• Concepts:
  Radiation of oscillating dipoles, the Larmor formula
• Reasoning:
  S(r,t) = (1/μ₀)(E × B) is the energy flux.
• Details of the calculation:
  (a)  S(r,t) = (1/μ₀)(E × B) = [1/(4πε₀)²][1/(c⁵r²μ₀)]ω⁴p₀² cos²(ωt − kr) sin²θ (r/r),
  or
  S(r,t) = (1/μ₀)(E × B) = [1/(4πε₀)][1/(4πc³r²)]ω⁴p₀² cos²(ωt − kr) sin²θ (r/r). 
  S = power per unit area = dP/dA.
  r²S = power per unit solid = dP/dΩ.
  <dP/dΩ> = [1/(4πε₀)] [ω⁴/(8πc³)]p₀² sin²θ.
  <dP/dΩ> is the time-average power radiated per unit solid angle.
  
  (b)  <P> = 2π∫₀^π<dP/dΩ> sinθ dθ = [1/(2ε₀)] [ω⁴/(8πc³)]p₀² ∫₀^πsin³θ dθ
  is the time-average power radiated to the whole space.
  <P> =  ω⁴p₀²/(12πε₀c³).
  This is the Larmor formula applied to an oscillating dipole
  <dP> ∝ ω⁴.
  
  (c)  <dP/dΩ> ∝ sin²θ.   The radiation is strongest at θ = 90^o.

Problem 2:

A proton is accelerated through a voltage of 1000 V and enters a magnetic field perpendicular to its direction of travel.  The magnetic field strength is 1.5 Tesla.  What electric field (direction and magnitude) is needed in this region to compensate for the deflection due to the magnetic field?

Solution:

• Concepts:
  Velocity filter
• Reasoning:
  Crossed electric and magnetic fields can act as a velocity filter.
• Details of the calculation:
  Let the proton travel in the z-direction and the magnetic field point in the y-direction.  Then the  magnetic force F_m = q_ev × B = -F_m i.  The electric force F_e = q_eE = q_eE i must have the same magnitude.
  We need E = vB.  U = q_e1000 V = ½m_pv².  v =  (2000 eV * 9*10¹⁶ m²/s²/938.26*10⁶ eV) ^½ = 4.37*10⁵ m/s.
  E = 6.57*10⁵ N/C i.

Problem 3:

A thin spherical shell of radius R carries a uniform surface charge density σ.  The shell rotates with constant angular velocity ω = ωk.

(a)  Find the surface current density K(θ, φ) on the sphere.
(b)  Compute the magnetic dipole moment m of the rotating shell.
(c)  Use the magnetic dipole approximation to find the magnetic field B at a point far
from the sphere (r >> R).
(d) Compare the magnitude of the field at the equator and the pole.

Solution:

• Concepts:
  The field of a magnetic dipole
• Reasoning:
  We can consider the rotating sphere to be a stack of current loops.
• Details of the calculation:
  (a)  The rotating surface charge density produces a surface current density, k_m = σω x R, 
  k_m = σωRsinθ e_φ.
  
  (b)  To find the dipole moment m, we consider the rotating sphere to be a stack of current loops.
  For a current loop at θ we have dm = IA = Rσω sinθ Rdθ πR²sin²θ = R⁴σωπ sin³θdθ.
  Integrating dm from θ = 0 to θ = π we find m = (4/3)R⁴σωπ, m = mk.
  [∫sin³xdx = -(1/3)cosx (sin²x + 2)]
  
  (c)  For a dipole m at the origin B(r) =  (μ₀/(4π))[3(m∙r)r/r⁵ - m/r³].
  If m = mk then B(r) = (μ₀m/(4πr³))[2 cosθ e_r + sinθ e_θ)].
  
  (d)  At the pole, on the z-axis,  θ = 0, B(z) = (μ₀m/(2πz³) e_z.
  At the equator θ = π/2, B(r) = -μ₀m/(4πr³) e_z.

Problem 4:

At t = 0 a particle of mass m and charge q moves in the xy-plane in a circular orbit of radius R in a uniform magnetic field B = Bk.  The particle will loose kinetic energy through radiation.  Assume that the energy loss per revolution is small compared to the total energy of the particle. 
(a)  Neglecting radiation, what is the magnitude of the particles momentum at t = 0?  Does your answer depend on the speed of the particle (relativistic or non-relativistic)?
(b)  Now assume the particle is moving non-relativistically.  Derive an expression for the time it takes the particle to loose half of its kinetic energy.

Solution:

• Concepts:
  Motion of a charged particle in a magnetic field, the radiation field of a point charge moving non-relativistically, the Larmor formula
• Reasoning:
  In a magnetic field a moving charged particle is acted on by a force perpendicular to the direction of its velocity.  Therefore this force changes the direction of the velocity, but not the energy of the particle.  This acceleration results in power loss via radiation.  The power radiated is given by the Larmor formula.
• Details of the calculation:
  (a)  F = qv×B,  dp/dt = γmdv/dt = γma.
  For circular motion a = v²/R pointing towards the center of the circle.
  γmv²/R = qvB,  R = γmv/(qB) = p/(qB).  p = RqB.
  (b)  For a non-relativistic particle mv = rqB, r = mv/qB.
  We assume that the orbital radius r changes very little while the particle completes one revolution, so that the acceleration is always given by v²/r.
  From the Larmor formula we have
  P = (2/3)[q²/(4πε₀c³)]a² = -dE/dt.
  a = v²/r = 2E/(mr).  Since r = mv/qB,we have
  a² = 2Eq²B²/m³.
  dE/dt = -[q⁴B²/(3π m³ε₀c³)]E.
  E = E₀exp(-t/τ), where τ = (3πε₀c³m³)/(q⁴B²)
  ½ = exp(-t/τ), t = τ ln(2) is the time it takes the particle to loose half of its kinetic energy.