An electric dipole p_{0} 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 = 45^{o}
to the dipole's axis to the differential power in the direction
perpendicular to the axis.

Solution:

- Concepts:

Radiation of an oscillating dipole - 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) =**p**_{0}cos(ωt) is**E**_{R}(**r**,t) = -(1/(4πε_{0}c^{2}r))(d^{2}**p**_{⊥}(t-r/c)/dt^{2}). (SI units).

Here**p**_{0}= p_{0}(**z**/z), so**E**_{R}(**r**,t) = -(1/(4πε_{0}c^{2}r))ω^{2}p_{0}cos(ω(t - r/c))sinθ (**θ**/θ).

E proportional to sinθ. P proportional to E^{2}proportional to sin^{2}θ. E proportional to ω^{2}. P proportional to ω^{4}.

If the frequency is doubled, the total power radiated will increase by a factor of 16.

(b) P(45^{o})/P(90^{o}) = sin^{2}(45^{o})/sin^{2}(90^{o}) = ½.

An electric dipole **p**(t) = p_{0}(**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 180^{o}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) =**p**_{0}cos(ωt) is

**E**_{R}(**r**,t) = -(1/(4πε_{0}c^{2}r))(d^{2}**p**_{⊥}(t - r/c)/dt^{2}). (SI units)

Here**p**_{0}= p_{0}(**z**/z), so**E**_{R}(**r**,t) = -(1/(4πε_{0}c^{2}r))ω^{2}p_{0}cos(ω(t - r/c))sinθ(**θ**/θ).

**B**_{R}(**r**,t) = (**r**/(rc))×**E**_{R}(**r**,t) = -[1/(4πε_{0}c^{3}r)]ω^{2}p_{0}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,**E**_{R}(**r**,t) = -(1/(4πε_{0}c^{2}r))ω^{2}p_{0}cos(ω(t - r/c))(**θ**/θ),

**B**_{R}(**r**,t) = -[1/(4πε_{0}c^{3}r)]ω^{2}p_{0}cos(ω( t- r/c))(**φ**/φ).

(c)**S**(**r**,t) = [1/(4πε_{0})^{2}][1/(c^{5}r^{2}μ_{0})]ω^{4}p_{0}^{2}cos^{2}(ω(t - r/c))sin^{2}θ (**r**/r). The intensity varies as sin^{2}θ.

(d) <P> = <∫**S∙**d**A**> = ω^{4}p_{0}^{2}/(12πε_{0}c^{3}), proportional to ω^{4}.

A thin spherical shell of radius a has a surface potential Φ(a,θ) = Φ_{0}
cosθ, where θ is the usual polar angle relative to the z-axis, and Φ_{0}
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, Φ_{0}, ε_{0} 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 180^{o}out of phase. Oscillating charges are accelerating charges.

Details of the calculation: - Details of the calculation:

(a) Φ(r,θ) = ∑_{n=0}^{∞}[A_{n}r^{n}+ B_{n}/r^{n+1}]P_{n}(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 Φ_{2}(r,θ) = (B/r^{2})P_{1}(cosθ) = (B/r^{2})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.

Φ_{0}cosθ = (B/a^{2})cosθ, B = Φ_{0}a^{2}, the potential outside the shell is Φ(r,θ) = Φ_{0}(a^{2}/r^{2})cosθ.

The potential of a dipole p = p**k**is Φ(r,θ) = k p cosθ/r^{2}, where k = 1/(4πε_{0}).

So p = 4πε_{0}a^{2}Φ_{0}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) =**p**_{0}cos(ωt) is

**E**_{R}(**r**,t) = -(1/(4πε_{0}c^{2}r))(d^{2}**p**_{⊥}(t - r/c)/dt^{2}). (SI units)

Here**p**_{0}= p_{0}(**z**/z), so**E**_{R}(**r**,t) = -(1/(4πε_{0}c^{2}r))ω^{2}p_{0}cos(ω(t - r/c))sinθ (**θ**/θ).

**B**_{R}(**r**,t) = (**r**/(rc))×**E**_{R}(**r**,t) = -[1/(4πε_{0}c^{3}r)]ω^{2}p_{0}cos(ω(t - r/c))sinθ (**φ**/φ).

(c)**S**(**r**,t) = [1/(4πε_{0})^{2}][1/(c^{5}r^{2}μ_{0})]ω^{4}p_{0}^{2}cos^{2}(ω(t - r/c))sin^{2}θ (**r**/r). The intensity varies as sin^{2}θ.

d<P>/dΩ = r^{2}<S(**r**,t)> = 2πε_{0}^{2}ω^{4}a^{4}Φ_{0}^{2}sin^{2}θ/c^{3}.(d) <P> = <∫

**S·**d**A**> = ω^{4}p_{0}^{2}/(12πε_{0}c^{3}) = 4πε_{0}ω^{4}a^{4}Φ_{0}^{2}/(3c^{3}), proportional to ω^{4}.

An electric dipole **p**_{0} 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:

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}= <(d^{2}**p**/dt^{2})^{2}>/(6πε_{0}c^{3}). - Details of the calculation:

**p**(t) = p_{0}cosθ**k**+ p_{0}sinθ (cos(ωt)**i**+ sin(ωt)**j**), where θ is the tilt angle.

d^{2}**p**/dt^{2}= -ω^{2}p_{0}sinθ [cos(ωt)**i**+ sin(ωt)**j**].

<(d^{2}**p**/dt^{2})^{2}> = ω^{4}p_{0}^{2}sin^{2}θ, P_{rad}= ω^{4}p_{0}^{2}sin^{2}θ/(6πε_{0}c^{3}).

A small current loop of area A and N turns carries a sinusoidally varying
current I(t) = I_{0}sin(ωt).

(a) Find the magnetic moment of the loop.

(b) Find the average power radiated by the loop.

Solution:

- Concepts:

Magnetic dipole radiation - Reasoning:

We are asked to find the average power radiated by a sinusoidally varying magnetic dipole. - Details of the calculation:

(a) m(t) = NAI_{0}sin(ωt). The direction of**m**is given by the right-hand rule.

(b) Larmor formula: The magnetic dipole will radiate energy at a rate P_{rad}= <(d^{2}**m**/dt^{2})^{2}>/(6πε_{0}c^{5}).

d^{2}m(t)/dt^{2}= -ω^{2}NAI_{0}sin(ωt).

<P_{rad}> = ω^{4}N^{2}A^{2}I_{0}^{2}/(12πε_{0}c^{5}).

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 10^{5}
revolutions per second?

Solution:

- Concepts:

Magnetic dipole radiation - 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.

**B**_{out}= (μ_{0}/4π)[3(**m**)_{0}∙r**r**/r^{5}-**m**/r_{0}^{3}]. Her**m**is the dipole moment of the sphere._{0}

At the poles the sphere B_{out}= B_{in}= (μ_{0}/4π)(2m_{0}/R^{3}).

The dipole moment of the sphere therefore has magnitude m_{0}= (4π/μ_{0})(B_{in}R^{3}/2).

We therefore have m_{0}= 5*10^{26}Am^{2}.

The rotating magnetic dipole will radiate energy at a rate P_{rad}= <(d^{2}**m**/dt^{2})^{2}>/(6πε_{0}c^{5}). - Details of the calculation:

Let the rotation axis be the z-axis.

**m**(t) = m_{0}cosθ**k**+ m_{0}sinθ (cos(ωt)**i**+ sin(ωt)**j**), where θ is tilt angle.

d^{2}**m**/dt^{2}= -ω^{2}m_{0}sinθ [cos(ωt)**i**+ sin(ωt)**j**].

<(d^{2}**m**/dt^{2})^{2}> = ω^{4}m_{0}^{2}sin^{2}θ, P_{rad}= ω^{4}m_{0}^{2}sin^{2}θ/(6πε_{0}c^{5}).

Initially P_{rad}(0) = 2.41*10^{43}J/s.

P_{rad}(t) = -dE/dt = -(d/dt)(½(2/5)MR^{2}ω^{2}) = -(2/5)MR^{2}ω dω/dt

ω^{4}m_{0}^{2}sin^{2}θ/(6πε_{0}c^{5}) = -(2/5)MR^{2}ω dω/dt.

-5 m_{0}^{2}sin^{2}θ dt/(12 MR^{2}πε_{0}c^{5}) = dω/ω^{3}.^{ }[-5 m_{0}^{2}sin^{2}θ/(12 MR^{2}πε_{0}c^{5})] ∫_{0}^{T}dt = ∫_{ω0}^{ω0/3}dω/ω^{3}.^{ }[5 m_{0}^{2}sin^{2}θ/(12 MR^{2}πε_{0}c^{5})] T = ½((ω_{0}/3)^{-2}- ω_{0}^{-2}).

T = ((ω_{0}/3)^{-2}- ω_{0}^{-2})(16 ρ R^{5}π^{2}ε_{0}c^{5})/(10 m_{0}^{2}sin^{2}θ ).

The average radius of a nucleus with A nucleons is R = R_{0}A^{(1/3)}, where R_{0}= 1.2*10^{-15}m.

The mass of a nucleus is A times the mass of a nucleon, m_{nucleon}~ 1.6*10^{-27}kg.

The nuclear density is therefore ρ = m_{nucleon}/((4/3)πR_{0}^{3}) ~ 2*10^{17}kg/m^{3}.

Plugging in numbers: T ~ 1.1*10^{6}s ~ 13 days.