#### Problem:

An electric dipole p0 vibrates with frequency ω.
(a)  How will the total radiated power change if the frequency is doubled?
(b)  Find the ratio of the differential power in the direction of θ = 45o to the dipole's axis to the differential power in the direction perpendicular to the axis.

Solution:

• Concepts:
• Reasoning:
We are asked to characterize the power radiated by an oscillating dipole.
• Details of the calculation:
(a)  The radiation field of an electric dipole oscillating around the origin at with p(t) = p0cos(ωt) is  ER(r,t) = -(1/(4πε0c2r))(d2p(t-r/c)/dt2).  (SI units).
Here p0 = p0(z/z), so ER(r,t) = -(1/(4πε0c2r))ω2p0 cos(ω(t-r/c))sinθ (θ/θ).
E proportional to sinθ.  P proportional to E2 proportional to sin2θ.  E proportional to ω2.  P proportional to ω4
If the frequency is doubled, the total power radiated will increase by a factor of 16.
(b)  P(45o)/P(90o) = sin2(45o)/sin2(90o) = ½.

#### Problem:

An electric dipole p(t) = p0(z/z)cosωt is placed at the origin.
(a)  Determine the radiation fields on the z-axis.
(b)  Determine the radiation fields for a direction normal to the z-axis.
(c)  How does the intensity of the radiation vary as a function of θ?
(d)  How does the total power radiated vary as a function of ω?

Solution:

• Concepts:
The radiation field of a point charge moving non-relativistically, the Larmor formula
• Reasoning:
A dipole can be viewed as a positive and a negative point charge oscillating 180o out of phase.  Oscillating charges are accelerating charges.
• Details of the calculation:
(a)  The radiation field of an electric dipole oscillating around the origin at with p(t) = p0cos(ωt) is
ER(r,t) = -(1/(4πε0c2r))(d2p(t - r/c)/dt2).  (SI units)
Here p0 = p0(z/z), so ER(r,t) = -(1/(4πε0c2r))ω2p0 cos(ω(t - r/c))sinθ(θ/θ).
BR(r,t) = (r/(rc))×ER(r,t) = -[1/(4πε0c3r)]ω2p0 cos(ω(t - r/c))sinθ(φ/φ).
On the z-axis sinθ = 0, there is no radiation along the z-axis.
(b)  Perpendicular to the z-axis we have sinθ = 1,
ER(r,t) = -(1/(4πε0c2r))ω2p0 cos(ω(t - r/c))(θ/θ),
BR(r,t) = -[1/(4πε0c3r)]ω2p0 cos(ω( t- r/c))(φ/φ).
(c)  S(r,t) = [1/(4πε0)2][1/(c5r2μ0)]ω4p02 cos2(ω(t - r/c))sin2θ (r/r).  The intensity varies as sin2θ.
(d)  <P> = <∫S∙dA> = ω4p02/(12πε0c3), proportional to ω4.

#### Problem:

A thin spherical shell of radius a has a surface potential Φ(a,θ) = Φ0 cosθ, where θ is the usual polar angle relative to the z-axis, and Φ0 is a constant.
(a)  Find the electric potential Φ(r,θ) everywhere outside the shell.  Calculate the equivalent electric dipole moment p that produces this potential Φ.
(b)  Assume that previously calculated dipole moment acquires a cos(ωt) time dependence.  Calculate the magnetic field B and the electric field E in the radiation zone.
(c)  Find the time-averaged power d<P>/dΩ radiated per unit solid angle.  Express your result in terms of ω, θ, a, Φ0, ε0 and c.
(d)  Find the total power <P> radiated.

Solution:

• Concepts:
Boundary value problems, the radiation of an oscillating dipole
• Reasoning:
The electric field outside the shell is a pure dipole field.  A dipole can be viewed as a positive and a negative point charge oscillating 180o out of phase.  Oscillating charges are accelerating charges.
Details of the calculation:
• Details of the calculation:
(a)  Φ(r,θ) = ∑n=0[Anrn + Bn/rn+1]Pn(cosθ) is the most general solution inside and outside of the sphere, since ρ = 0 inside and outside of the sphere.  To find the specific solution we apply boundary conditions.
Assume Φ2(r,θ) = (B/r2)P1(cosθ) = (B/r2)cosθ outside the sphere.  The symmetry of the situation suggests this form of the solution.  If we can satisfy the boundary conditions, the uniqueness theorem guaranties that we have found the only solution.
Boundary conditions:
The potential vanishes as r goes to infinity.
Φ is continuous across the boundary.
Φ0 cosθ = (B/a2)cosθ,  B = Φ0a2,  the potential outside the shell is Φ(r,θ) = Φ0 (a2/r2)cosθ.
The potential of a dipole p = pk is  Φ(r,θ) = k p cosθ/r2, where k = 1/(4πε0).
So p = 4πε0 a2Φ0 is the magnitude of the equivalent electric dipole moment.

(b)  The radiation field of an electric dipole oscillating around the origin at with p(t) = p0cos(ωt) is
ER(r,t) = -(1/(4πε0c2r))(d2p(t - r/c)/dt2).  (SI units)
Here p0 = p0(z/z), so ER(r,t) = -(1/(4πε0c2r))ω2p0 cos(ω(t - r/c))sinθ (θ/θ).
BR(r,t) = (r/(rc))×ER(r,t) = -[1/(4πε0c3r)]ω2p0 cos(ω(t - r/c))sinθ (φ/φ).

(c)  S(r,t) = [1/(4πε0)2][1/(c5r2μ0)]ω4p02 cos2(ω(t - r/c))sin2θ (r/r).  The intensity varies as sin2θ.
d<P>/dΩ = r2<S(r,t) > = 2πε02ω4a4Φ02sin2θ/c3.

(d)  <P> = <∫S∙dA> = ω4p02/(12πε0c3) =  4πε0ω4a4Φ02/(3c3), proportional to ω4.

#### Problem:

An electric dipole p0 makes an angle of θ 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:
• 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 Prad = <(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,  Prad = ω4p02sin2θ/(6πε0c3).

#### Problem:

A small current loop of area A and N turns carries a sinusoidally varying current I(t) = I0sin(ωt).
(a)  Find the magnetic moment of the loop.
(b)  Find the average power radiated by the loop.

Solution:

• Concepts:
• Reasoning:
We are asked to find the average power radiated by a sinusoidally varying magnetic dipole.
• Details of the calculation:
(a)  m(t) = NAI0sin(ωt).  The direction of m is given by the right-hand rule.
(b)  Larmor formula:  The magnetic dipole will radiate energy at a rate Prad = <(d2m/dt2)2>/(6πε0c5).
d2m(t)/dt2 = -ω2NAI0sin(ωt).

#### Problem:

A rotating neutron star slows down due to dipole radiation.   Assume that a neutron star has a surface magnetic field strength of 100 Mega Tesla at the poles, a radius of 10 km, and nuclear density.  The neutron star's magnetic axis is tilted by 45 degrees relative to the axis of rotation of the neutron star.  How long will it take for the neutron star's rotation rate to decrease by a factor of 3 if the initial rotation rate is 105 revolutions per second?

Solution:

• Concepts:
• Reasoning:
Treat the neutron star as a uniform magnetized sphere.  The magnetic field inside the sphere is uniform and the magnetic field outside the sphere is a pure dipole field.
Bout = (μ0/4π)[3(m0∙r)r/r5m0/r3].  Her m0 is the dipole moment of the sphere.
At the poles the sphere Bout = Bin = (μ0/4π)(2m0/R3).
The dipole moment of the sphere therefore has magnitude m0 = (4π/μ0)(BinR3/2).
We therefore have m0 = 5*1026 Am2.
The rotating magnetic dipole will radiate energy at a rate Prad =  <(d2m/dt2)2>/(6πε0c5).
• Details of the calculation:
Let the rotation axis be the z-axis.
m(t) = m0 cosθ k + m0 sinθ (cos(ωt) i + sin(ωt) j),  where θ is tilt angle.
d2m/dt2 = -ω2m0 sinθ [cos(ωt) i + sin(ωt) j].
<(d2m/dt2)2> = ω4m02sin2θ,  Prad = ω4m02sin2θ/(6πε0c5).