EM II Problem 39Multipole Radiation for a Small Circular Antenna

Chris MuellerDept. of Physics, University of Florida

5 April, 2010

39. Jackson 9.14: An antenna consists of a circular loop of wire of radius a located in the xyplane with its center at the origin. The current in the wire is

I = I0 cosωt = <I0e−iωt

(a) Find the expressions for E, H in the radiation zone without approximations as to themagnitude of ka. Determine the power radiated per unit solid angle.

(b) What is the lowest nonvanishing multipole moment (Qlm or Mlm)? Evaluate thismoment in the limit ka 1.

Part aThe current density is given by

J = I0e−iωt 1

rδ(r − a)δ(cos θ)φ.

We will seek to describe the fields in the radiation zone in terms of the multipole expansion

H =eikr−iωt

kr

∑l,m

(−i)l+1[aE(l,m)Xlm + aM (l,m)n×Xlm]

E = Z0H× n,

where the multipole coefficients aE(l,m) and aM (l,m) are expressed in terms of the current density.Since the charge density and the intrinsic magnetization are zero and r · J = 0, the electric multipolecoefficients vanish. The magnetic coefficients are given by

aM (l,m) =k2

i√l(l + 1)

∫Y ∗lm∇ · (r× J)jl(kr) d

3x.

Lets calculate

∇ · (r× J) = −∇ · (I0δ(r − a)δ(cos θ)θ)

= − 1

r sin θ

∂

∂θ(I0δ(r − a) sin θδ(cos θ))

= −I0rδ(r − a)

0cos θ

sin θδ(cos θ) +

∂

∂θδ(cos θ)

=I0rδ(r − a) sin θδ′(cos θ)

We can now plug this in to the expression for the magnetic multipole coefficient. Not that since

1

the problem has azimuthal symmetry, we know that the only possible value of m is zero.

aM (l, 0) =I0k

22πa

i√l(l + 1)

jl(ka)

∫dθ Y ∗l0 sin2 θδ′(cos θ)

= −iI0k2√πa

√2l + 1

l(l + 1)jl(ka)

∫dθ Pl(cos θ) sin2 θδ′(cos θ)

= iI0k2√πa

√2l + 1

l(l + 1)jl(ka)

∫dxPl(x)

√1− x2δ′(x)

= iI0k2√πa

√2l + 1

l(l + 1)jl(ka)

[d

dx

(√1− x2Pl(x)

)]x=0

= iI0k2√πa

√2l + 1

l(l + 1)jl(ka)

[−xPl(x)√

1− x2+√

1− x2(−lxPl(x) + lPl−1(x)

1− x2

)]x=0

= iI0k2√πa

√l(2l + 1)

l(l + 1)jl(ka)Pl−1(0)

Since the Legendre polynomials are either odd or even in x with their oddness or evenness deter-mined by l, only the odd l terms will contribute. The fields are therefore given by

H =ieikr−iωtI0k

√πa

r

∑l=odd

((−1)

l+12

√l(2l + 1)

l + 1jl(ka)Pl−1(0)n×Xlm

)

E = −Z0ieikr−iωtI0k

√πa

r

∑l=odd

((−1)

l+12

√l(2l + 1)

l + 1jl(ka)Pl−1(0)n× n×Xlm

)Finally, the power radiated per unit solid angle is

dP

dΩ=

Z0

2k2

∣∣∣∣∣ ∑l=odd

(−1)l+12 iI0k

2√πa

√l(2l + 1)

(l + 1)jl(ka)Pl−1(0)Xlm

∣∣∣∣∣2

=1

2Z0I

20k

2πa2

∣∣∣∣∣ ∑l=odd

(−1)l+12

√l(2l + 1)

(l + 1)jl(ka)Pl−1(0)Xlm

∣∣∣∣∣2

Part bFrom our work above, we can see that the lowest multipole coefficient is aM (1, 0). Since the

induced magnetic multipole moment depends on the magnetization which is zero, we can expressthis coefficient as

M10 =3√2ik3

aM (1, 0)

Hence,

M10 =3√2ik3

(iI0k

2√πa)√3

2j1(ka)P0(0)

=3√

3πI0a

2k

(sin(ka)

k2a2− cos(ka)

ka

)

In the limit ka 1 this reduces to

M10 →3√

3πI0a

2k

(1

ka− ka

6− 1

ka+ka

2

)=

√3πI0a

2

2

2