Consider a linear antenna of length d (d << λ)
with a narrow gap in the center for the purposes of excitation. Assume
that the current is sinusoidal and in the same direction in each half of the
antenna, having a value of I_{0} at the gap and falling linearly to zero
at the ends. Find the power radiated in the electric dipole approximation.

Solution:

- Concepts:

Radiation from antennas, dipole radiation - Reasoning:

We can treat the antenna as made up of small sections, each section emitting dipole radiation. - Details of the calculation:

Consider a small section of the antenna of length dz' located at z' . Treat it like a dipole.

Let p(z') = q(z')dz'(**z**/z), q(z') = q_{0}(z')cosωt.

Then I(z') = dq(z')/dt = -ωq_{0}(z')sinωt = I_{0}(z')sin(ωt), I_{0}(z') = -ωq_{0}(z').

p(z') = p_{0}(z')cosωt. p_{0}(z') = q_{0}(z')dz = -(I_{0}(z')/ω)dz.

For dipole at the origin we have**E**_{R}(**r**,t) = -(1/(4πε_{0}c^{2}r))ω^{2}p_{0}cos(ω(t - r/c))sinθ (**θ**/θ).

A small section of the antenna of length dz' located at z' a distance**R**=**r**- (**z**/z)z' from the observation point therefore produces

d**E**_{R}(**r**,t) = (1/(4πε_{0}c^{2}R))ω^{2}(I_{0}(z')/ω)cos(ω(t - R/c))sinθ'dz' (**θ'**/θ').

d**B**_{R}(**r**,t) = (1/(4πε_{0}c^{3}R))ω^{2}(I_{0}(z')/ω)cos(ω(t - R/c))sinθ'dz'(**φ**/φ).

Assume r >> λ, r >> z', d << λ.

Then θ' ≈ θ = constant, for all sections of the antenna.

We can also replace 1/R by 1/r and cos(ωt - kR) by cos(ωt - kr) because the changes in kR are small compared to π/2 over the length of the antenna. (k = ω/c)

I_{0}(z) = I_{0}- (2I_{0}/d)|z|, the current falls linearly to zero at the ends.

E_{θ}(**r**,t) = ∫_{z=-d/2}^{z=d/2}dE_{θ}(**r**,t) = 2 ∫_{0}^{z=d/2}dE_{θ}(**r**,t)

= (I_{0}ω/(2πε_{0}c^{2}r))cos(ω(t - r/c))sinθ ∫_{0}^{z=d/2}(1 - 2z/d)dz'

= (I_{0}d ω^{2}/(8πε_{0}c^{3}))(1/(kr))cos(ω(t - r/c))sinθ.

B_{φ}(**r**,t) = ∫_{z=-d/2}^{z=d/2}dB_{φ}(**r**,t) = 2 ∫_{0}^{z=d/2}dB_{φ}(**r**,t)

= (μ_{0}I_{0}d ω^{2}/(8πc^{2}))(1/(kr))cos(ω(t - r/c))sinθ.

**S**= (1/μ_{0})**E**×**B**= (**r**/r)[(I_{0}^{2}d^{2}ω^{4}/(64π^{2}ε_{0}c^{5}k^{2}r^{2}))cos^{2}(ω(t - r/c))sin^{2}θ.]

P = (I_{0}^{2}d^{2}ω^{2}/(64π^{2}ε_{0}c^{3}r^{2}))cos^{2}(ω(t - r/c)) ∫2πr^{2}sin^{3}θ dθ

= (I_{0}^{2}d^{2}ω^{2}/(32πε_{0}c^{3}))cos^{2}(ω(t - r/c))(4/3).

<P> = ½(I_{0}^{2}d^{2}ω^{2}/(24πε_{0}c^{3})) is the average power radiated in the electric dipole approximation.

Assume an antenna is aligned with the z-axis and extends
from z = -λ/4 to z = +λ/4. Assume the current in the antenna varies as I(z,t) = I_{0}sin(ωt)cos(kz).
Calculate the radiation fields E_{R} and B_{R} and the average
Poynting vector.

Solution:

- Concepts:

Radiation from antennas, dipole radiation - Reasoning:

We can treat the antenna as made up of small sections, each section emitting dipole radiation. - Details of the calculation:

Consider a small section of the antenna. Treat it like a dipole.

Let**p**(z) = q(z) dz (**z**/z), q(z) = q_{0}(z)cosωt.

Then I(z) = dq(z)/dt = -ωq_{0}(z)sinωt = I_{0}(z)sin(ωt), I_{0}(z) = -ωq_{0}(z).

p(z) = p_{0}(z)cosωt. p_{0}(z) = q_{0}(z)dz = -(I_{0}(z)/ω)dz.

Here I_{0}(z) = I_{0}cos(kz)

For dipole at the origin we have**E**_{R}(**r**,t) = -(1/(4πε_{0}c^{2}r))ω^{2}p_{0}(0) cos(ω(t - r/c))sinθ (**θ**/θ).

A small section of the antenna of length dz' located at z' a distance**R**=**r**- (**z**/z)z' from the observation point therefore produces

d**E**_{R}(**r**,t) = (1/(4πε_{0}c^{2}R))ω^{2}(I_{0}/ω)cos(ω(t - R/c))sinθ'cos(kz')dz' (**θ**/θ).

d**B**_{R}(**r**,t) = (1/(4πε_{0}c^{3}R))ω^{2}(I_{0}/ω)cos(ω(t - R/c))sinθ'cos(kz')dz'(**φ**/φ).

dE_{θ}(**r**,t) = (ω^{2}I_{0}/(4πε_{0}c^{3}))[cos(ω(t - R/c))/kR]sinθ'cos(kz')dz',

dB_{φ}(**r**,t) = (ω^{2}I_{0}/(4πε_{0}c^{4}))[cos(ω(t - R/c)/kR]sinθ'cos(kz')dz',

with k = ω/c.

E_{θ}(**r**,t) = (ω^{2}I_{0}/(4πε_{0}c^{3}))∫[cos(ω(t - R/c))/kR]sinθ'cos(kz')dz',

B_{φ}(**r**,t) = (ω^{2}I_{0}/(4πε_{0}c^{4}))∫[cos(ω(t - R/c)/kR]sinθ'cos(kz')dz'.

The integration limits are -λ/4 and λ/4.

Assume r >> λ, r >> z'.

Then θ' ≈ θ = constant for all sections of the antenna.

But cos(ωt - kR) ≠ cos(ωt - kr) because kR changes by π/2 over the length of the antenna. Although kR >> π/2, this changes the phase drastically.

We use kR = kr - kz'cosθ for the argument of the cosine function. We, however, can safely replace 1/R by 1/r.

We then have

E_{θ}(**r**,t) = (ω^{2}I_{0}sinθ/(4πε_{0}c^{3}kr))∫cos(ωt - kr + kz'cosθ)cos(kz')dz',

B_{f}(**r**,t) = (μ_{0}ω^{2}I_{0}sinθ/(4πc^{2}kr))∫cos(ωt - kr + kz'cosθ)cos(kz')dz'.

∫_{-λ/4}^{λ/4}cos(ωt - kr + kz'cosθ)cos(kz')dz'

= (1/k)cos(ωt - kr)∫_{-π/2}^{π/2}cos(kz'cosθ)cos(kz')dkz'

- (1/k)sin(ωt - kr)∫_{-π/2}^{π/2}sin(kz'cosθ)cos(kz')dkz'.

The second integral is zero, since it is an integral over an odd function of z'.

∫_{-π/2}^{π/2}cos(ax) cos(x)dx = sin((a-1)x)/(2(a-1)) + sin((a+1))x/(2(a+1))|_{-π/2}^{π/2}

= [1/(2(a-1))]{sinaxcosx-sinxcosax} + [1/(2(a+1))]{sinaxcosx+sinxcosax}|_{-π/2}^{π/2}

= [1/(2(a-1))]{-2cos(a(π/2))} + [1/(2(a+1))]{2cos(a(π/2))}

= [1/(2(a^{2}-1))]{-4cos(a(π/2))}

With a = cosθ ωe have

∫_{-π/2}^{π/2}cos(cosθx)cos(x)dx = 2cos(cosθ(π/2))/(sin^{2}θ).

E_{θ}(**r**,t) = [I_{0}/(2πε_{0}cr)]cos(ωt - kr)cos[(π/2)cosθ)]/sinθ.

B_{f}(**r**,t) = [μ_{0}I_{0}/(2πr)]cos(ωt - kr)cos[(π/2)cosθ)]/sinθ.

**S**= (1/μ_{0})**E**×**B**= (**r**/r)[I_{0}^{2}/(4π^{2}ε_{0}cr^{2})]cos^{2}(ωt -kr)cos^{2}[(π/2)cosθ)]/sin^{2}θ.

The plot below compares the angular radiation pattern of a dipole (sin^{2}θ) and a half-wave antenna (cos^{2}[(π/2)cosθ)]/sin^{2}θ).

A very thin, straight, conducting wire is centered on the
origin of coordinates and is oriented along the z-axis. The wire carries a current
I = I_{0}cos(ω_{0}t) everywhere along its length
l.

(a) Obtain the vector potential everywhere outside the source region for distances
r >> l. Hint: Use the Lorentz gauge. Make no assumption about the value of the
wavelength, λ_{0 }= 2πc/ω_{0}.

(b) Obtain the dependence of the
intensity of the radiation pattern emitted by the wire on the
angle θ in the regime r >> l >> λ_{0}.

Solution:

- Concepts:

Potentials in the Lorentz gauge - Reasoning:

In the Lorentz gauge the expressions for the potentials are

Φ(**r**, t) = [1/(4πε_{0})]∫_{v'}dV' ρ(**r**', t_{r})/|**r**-**r**'|,**A**(**r**,t) = [μ_{0}/(4π)]∫_{v'}dV'**j**(**r**', t_{r})/|**r**-**r**'|.

t_{r}= t - |**r**-**r**'|/c. - Details of the calculation:

(a) Let r >> l. Then |**r**-**r**'| ≈ r[1 -**r**∙**r**'/r],

1/|**r**-**r**'| ≈ (1/r)[1 +**r**∙**r**'/r].

Let**j**(**r**', t) =**j**(**r**')exp(-iω_{0}t), then**A**(**r**,t) =**A**(**r**)exp(-iω_{0}t), with

**A**(**r**) = [μ_{0}/(4π)]∫_{v'}dV'**j**(**r**')exp(ik|**r**-**r**'|)/|**r**-**r**'| with k = ω_{0}/c.

Let**j**= I_{0}δ(x)δ(y)**k**for -l/2 < z < l/2,**j**= 0 elsewhere.

Then**A**(**r**) =**k**[I_{0}μ_{0}/(4π)](e^{ikr}/r)∫_{-l/2}^{l/2}dz'exp(-ik**r**∙**r**'/r).

(In the denominator keep only the zeroth order term and in the exponent keep terms to first order.)

**A**(**r**) =**k**[I_{0}μ_{0}/(4π)](e^{ikr}/r)∫_{-l/2}^{l/2}dz'exp(-ikz' cosθ)

=**k**[I_{0}μ_{0}/(4π)](e^{ikr}/r)[2/(k cosθ)]sin(k(l/2) cosθ)

=**k**[1/(4πε_{0})][1/(cω_{0}cosθ)](2I_{0}/r)sin(k(l/2) cosθ)e^{ikr}.

(b) For r >> l >> λ_{0}we have only a radiation fields, proportional to 1/r.

Then**B**=**∇**×**A**= (1/r)[∂(**rA**_{θ})/∂r - ∂**A**_{r}/∂θ](**φ**/φ)

≈ (1/r)[∂(**rA**_{θ})/∂r](**φ**/φ) ∝ [sinθ/cosθ]sin(k(l/2) cosθ).

The power per unit solid angle is proportional to |**B**^{2}|.

dP/dΩ ∝ tan^{2}θ sin^{2}(k(l/2) cosθ).

k(l/2) = 2

k(l/2) = 20

As the ratio l/λ_{0}increases, more and more power is emitted perpendicular to the wire.