10. Jackson problem 5.10

10.1. (a) Loop parallel to interface

The force on a loop parallel to the interface is conveniently given by

F_{z} = −\frac{2πaI}{c} B_{ρ}(a, 2d)  (10.1)

where the z axis is perpendicular to the interface and B_{ρ} is a component of the field of the image loop. Using Jackson’s (5.36), the vector potential of the image loop is

A_{ϕ}(ρ, z) = \frac{I′a}{c} ∫_{0}^{2π} \frac{cos α′ dα′}{\sqrt{a^{2} + ρ^{2} + z^{2} − 2aρ cos α′}}

where

I′ = \frac{μ − 1}{μ + 1} I

and

B_{ρ}(ρ, z) = −\frac{∂}{∂z} A_{ϕ}(ρ, z) = \frac{I′a}{c} ∫_{0}^{2π} \frac{z cos α′ dα′}{(a^{2} + ρ^{2} + z^{2} − 2aρ cos α′)^{3/2}}

Thus F_{z} is obtained directly as a single well-behaved integral, which can be expressed as a hypergeometric function or as a sum of elliptic integrals (see below). One can also obtain F_{z} as a single integral over the product of two Bessel functions, but such integrals are not easy to evaluate numerically or to expand at large separations. Working with spherical harmonics, it is possible to obtain for F_{z} a power series in a/\sqrt{a^{2} + 4d^{2}} that is very useful for large separations. This is probably the answer that Jackson wants (see below).

Using elliptic integrals

B_{ρ}(ρ, z) is given on page 11 of Lecture 12 in terms of elliptic integrals:

B_{ρ}(ρ, z) = \frac{2Iz}{cρ\sqrt{(a + ρ)^{2} + z^{2}}} [\frac{a^{2} + ρ^{2} + z^{2}}{(a − ρ)^{2} + z^{2}} E(k^{2}) − K(k^{2})]

with

k^{2} = \frac{4aρ}{(a + ρ)^{2} + z^{2}}

EM1 spring 96, Assignment 6, Problem 3 2 Hence

F_{z} = − \frac{μ − 1}{μ + 1} \frac{4πdI^{2}}{c^{2}\sqrt{a^{2} + d^{2}}} [\frac{a^{2} + 2d^{2}}{2d^{2}} E (\frac{a^{2}}{a^{2} + d^{2}} ) − K (\frac{a^{2}}{a^{2} + d^{2}} )]  (10.2)

For a << d, we can use the expansions (17.3.11 and 17.3.12 in Abramowitz and Stegun)

K(α) = \frac{π}{2} (1 + \frac{1}{4} α + \frac{9}{64} α^{2} + …)
E(α) = \frac{π}{2} (1 − \frac{1}{4} α − \frac{3}{64} α^{2} + …)

and the square parenthesis in eq. (10.2) reduces to

\frac{π}{2} [(1 + \frac{a^{2}}{2d^{2}} ) (1 − \frac{1}{4} \frac{a^{2}}{a^{2} + d^{2}} − \frac{3}{64} \frac{a^{4}}{d^{4}} ) − 1 − \frac{1}{4} \frac{a^{2}}{a^{2} + d^{2}} − \frac{9}{64} \frac{a^{4}}{d^{4}} ]

which to leading order gives

\frac{π}{2} \frac{3}{16} \frac{a^{4}}{d^{4}}

so that, in agreement with part (c) here below,

F_{z} ≃ −\frac{μ − 1}{μ + 1} \frac{3π^{2}I^{2}}{8c^{2}} \frac{a^{4}}{d^{4}}  (10.3)

Using spherical harmonics B_{r} and B_{θ} are given by Jackson in (5.48) and (5.49). Then

B_{ρ} = B_{r} sin θ + B_{θ} cos θ

where in our case

sin θ = \frac{a}{\sqrt{a^{2} + 4d^{2}}} cos θ = \frac{2d}{\sqrt{a^{2} + 4d^{2}}}

and one obtains

F_{z} = −\frac{μ − 1}{μ + 1} \frac{2π^{2}I^{2}}{c^{2}} ∑_{n=0}^{∞} \frac{(−1)^{n}(2n + 1)!!}{2^{n}n!} (sin θ)^{2n+3}  (10.4)

× [2 sin θ P_{2n+1}(cos θ) − \frac{cos θ}{n + 1} P_{2n+1}^{1}(cos θ)]

For a << d, which is the same as θ << 1, the important term is n = 0. For this term

P_{1} = cos θ P_{1}^{1} = − sin θ

and the square bracket in eq. (10.4) reduces to 3 sin θ. Hence we find again the result (10.3).

10.2. (b) Loop perpendicular to interface

The force on the loop perpendicular to the interface can be computed directly, but leads to a double integral, one over the loop and one over its image. One of the integrals can be expressed as a sum of elliptic integrals, but the other remains. Working with Bessel functions one can obtain F as a single integral over the product of three Bessel functions, not easy to evaluate numerically or to expand at large separations. Working with spherical harmonics, it is possible to obtain a straight power series in a/d that is very useful for large separations. This is probably the answer that Jackson wants.

Using spherical harmonics

Put the loop at \frac{1}{2}R + a, where R = (2d, 0, 0) and a = a(cos α, sin α, 0), and put the image loop at −\frac{1}{2}R + a′, where a′ = a(cos α′, sin α′, 0). The force on the loop is

F = −II′ ∫ dl · dl′ \begin{matrix}R + a − a & ′ \\ |R + a − a′| & 3 \end{matrix} = II′∇_{R}M(R)

where in our case

I′ = −I \frac{μ − 1}{μ + 1}

and M(R) is the mutual inductance

M(R) = ∫ \frac{dl · dl′}{|R + a − a′|}

Since dl is perpendicular to a, and dl′ to a′, we have dl · dl′ = a^{2} cos(α − α′)dα dα′. It is convenient to put a′−a = ρ(cos ϕ, sin ϕ, 0) and to expand

\frac{1}{|R + a − a′|} = ∑_{l=0}^{∞} ∑_{m=0}^{l} \frac{ρ^{l}}{R^{l+1}} \frac{(l − m)!}{(l + m)!} [P_{l}^{m}(0)]^{2} e^{imϕ}

We can evaluate ρ^{l}e^{imϕ} with the help of

ρ cos ϕ = a(cos α′ − cos α) ρ sin ϕ = a(sin α′ − sin α)

These equations give

ρe^{iϕ} = a (e^{iα′} − e^{iα}) = −2iae^{i(α′+α)/2} sin β
ρ = 2a sin β

with β = \frac{1}{2} (α − α′) , when 0 ≤ β ≤ π. It follows that

ρ^{l−m} (ρe^{iϕ})^{m} = (−i)^{m} e^{im(α+α′)/2} (2a sin β)^{l}

and

M(R) = ∑_{l=0}^{∞} ∑_{m=0}^{l} \frac{a^{l+2}}{R^{l+1}} \frac{(l − m)!}{(l + m)!} [P_{l}^{m}(0)]^{2} C_{lm}

where

C_{lm} = (−i)^{m} ∫ e^{imα′}e^{imβ}(2 sin β)^{l} cos 2β dα dα′

Because the integrand is periodic, we can replace the integral over α with an integral over β from 0 to π. The result is that C_{lm} = δ_{m0}C_{l0} with

C_{l0} = 4π ∫_{0}^{π} (2 sin β)^{l} (1 − 2 sin^{2} β) dβ

At this point it is convenient to recall that P_{l}^{m}(0) vanishes for odd l, so that for even l we can write

C_{l0} = 2π ∫_{0}^{2π} [(2 sin β)^{l} − \frac{1}{2} (2 sin β)^{l+2}] dβ

and we can use

∫_{0}^{2π} (2 sin β)^{2n} dβ = 2π\binom{2n}{n} = 2π \frac{(2n)!}{(n!)^{2}}

to obtain C_{00} = 0 and

C_{2n,0} = −4π^{2} \frac{(2n)!}{(n!)^{2}} \frac{n}{(n + 1)} = −4π^{2} \frac{(2n)!}{(n − 1)!(n + 1)!}

for n > 0. We have arrived at the desired result

M(R) = −4π^{2} ∑_{n=1}^{∞} \frac{a^{2n+2}}{R^{2n+1}} [P_{2n}(0)]^{2} \frac{(2n)!}{(n − 1)!(n + 1)!}

where we recall that

|P_{2n}(0)| = \frac{(2n − 1)!!}{2^{n}n!} = \frac{(2n − 1)!}{2^{n}n! 2^{n−1}(n − 1)!} = \frac{(2n)!}{2^{2n} (n!)^{2}} = (\frac{1}{2})^{2n} \binom{2n}{n}

and in partcular P_{2n}(0) = \frac{1}{2} . Then F = II′∂M/∂R gives

F = −4π^{2}I^{2} \frac{μ − 1}{μ + 1} ∑_{n=1}^{∞} \frac{a^{2n+2}}{(2d)^{2n+2}} [P_{2n}(0)]^{2} \frac{(2n + 1)!}{(n − 1)!(n + 1)!}

and for d >> a, in agreement with part (c) here below,

F ≃ −\frac{μ − 1}{μ + 1} \frac{3π^{2}I^{2}}{16c^{2}} \frac{a^{4}}{d^{4}}

10.3. (c) The limit of large d/a

For large d/a, the loop and its image can be regarded as dipoles. The force between two dipoles is F = −∂U/∂r, where

U = \frac{m · m′ − 3(m · n)(m′·n)}{r^{3}}

and n is the unit vector along the line joining them. We must differentiate with respect to r first, then put r = 2d. In our problem n is along the z axis and

m = \frac{πIa^{2}}{c} m′ = \frac{πIa^{2}}{c} \frac{μ − 1}{μ + 1}

In case (a) the dipoles are parallel to each other and to n, hence U = −2mm′/r^{3}.

F = − \frac{6mm′}{r^{4}} = −\frac{3π^{2}I^{2}}{8c^{2}} \frac{a^{4}}{d^{4}} \frac{μ − 1}{μ + 1}

In case (b) the dipoles are opposite to each other and perpendicular to n, hence U = −mm′/r^{3} and the force is half as large as in case (a), F = −3mm′/(2d)^{4}.

EM1 spring 96, Assignment 6, Problem 3 6 Using Bessel functions

\frac{1}{\sqrt{ρ^{2} + a^{2} + z^{2} − 2aρ cos(ϕ − α′)}} = ∑_{m=−∞}^{∞} e^{im(ϕ−α′)} ∫_{0}^{∞} dk e^{−k|z|}J_{m}(ka)J_{m}(kρ) (10.5)

which is a particular case of the Green function formula given by Jackson in Problem 3.14(b). Using

∫_{0}^{2π} cos(α − α′)e^{im(ϕ−α′)} dα′ = π [δ_{m1}e^{i(ϕ−α)} + δ_{m,−1}e^{−i(ϕ−α)}]

we find

M(d) = 2πa^{2} ∫_{0}^{2π} dα cos(ϕ − α) ∫_{0}^{∞} dkJ_{1}(ka)J_{1}(kρ)

We must now perform the integral over α, keeping in mind that

ρ cos ϕ = d + a cos α ρ sin ϕ = a sin α

cos(ϕ − α) = \frac{1}{ρ} (a + d cos α)

For this purpose we use another formula from problem 3.13:

J_{0}(kρ) = J_{0}(k\sqrt{d^{2} + a^{2} + 2ad cos α}) = ∑_{m=−∞}^{∞} e^{im(π−α)}J_{m}(ka)J_{m}(kd)  (10.6)

which gives by differentiation

\frac{a + d cos α}{ρ}J_{1}(kρ) = − ∑_{m=−∞}^{∞} e^{im(π−α)}J′_{m}(ka)J_{m}(kd)

We obtain

M(d) = −2πa^{2} ∫_{0}^{2π} dα ∑_{m=−∞}^{∞} e^{im(π−α)} ∫_{0}^{∞} dk J_{1}(ka)J′_{m}(ka)J_{m}(kd)

and finally

M(d) = 4π^{2}a^{2} ∫_{0}^{∞} dk [J_{1}(ka)]^{2} J_{0}(kd)

F = II′ \frac{∂M}{∂d} = −4π^{2}a^{2}II′ ∫_{0}^{∞} kdk [J_{1}(ka)]^{2} J_{1}(kd)

\frac{ad sin α}{ρ}J_{1}(kρ) = − ∑_{m=−∞}^{∞} ime^{im(π−α)}J_{m}(ka)J′_{m}(kd)

It is convenient to perform first an integration by parts, using

−\frac{∂}{∂k} J_{0}(ka) = aJ_{1}(ka) \frac{∂}{∂k} J_{1}(kρ) = \frac{ρ}{2} [J_{0}(kρ) − J_{2}(kρ)]

to obtain

∫_{0}^{∞} dkJ_{1}(ka)J_{1}(kρ) = \frac{ρ}{2a} ∫_{0}^{∞} dkJ_{0}(ka) [J_{0}(kρ) − J_{2}(kρ)]

\frac{1}{|R + a − a′|} = ∑_{l=0}^{∞} ∑_{m=−l}^{l} e^{im(ϕ−α′)}P_{l}^{m}(0)P_{l}^{m}(0) \frac{a}{ρ}

First, we need the result of Problem 5.4(b) for the vector potential of a loop of radius a

A_{ϕ}(ρ, z) = \frac{2πIa}{c} ∫_{0}^{∞} dk e^{−k|z|}J_{1}(ka)J_{1}(kρ)

This follows from Jackson’s eq. (5.36)

A_{ϕ}(ρ, z) = \frac{Ia}{c} ∫_{0}^{2π} \frac{cos ϕ dϕ}{\sqrt{ρ^{2} + a^{2} + z^{2} − 2aρ cos ϕ}}

by using the Green function formula from Problem 3.14

\frac{1}{\sqrt{ρ^{2} + a^{2} + z^{2} − 2aρ cos ϕ}} = ∑_{m=−∞}^{∞} e^{imϕ} ∫_{0}^{∞} dk e^{−k|z|}J_{m}(ka)J_{m}(kρ)  (10.7)

Second, we need a formula for J_{1}(kρ) that is the analog of the formula

J_{0}(k\sqrt{d^{2} + a^{2} − 2ad cos ϕ}) = ∑_{m=−∞}^{∞} e^{imϕ}J_{m}(ka)J_{m}(kd)  (10.8)

again from Problem 3.14. To prove this formula, we follow three steps:

(i) Put a = 0 in (10.7) and obtain

\frac{1}{\sqrt{ρ^{2} + z^{2}}} = ∫_{0}^{∞} dk e^{−k|z|}J_{0}(kρ)

(ii) Put ρ^{2} = d^{2} + a^{2} − 2ad cos ϕ in the above and obtain

\frac{1}{\sqrt{d^{2} + a^{2} − 2ad cos ϕ + z^{2}}} = ∫_{0}^{∞} dk e^{−k|z|}J_{0}(k\sqrt{d^{2} + a^{2} − 2ad cos ϕ})

(iii) Comparing this result with eq. (10.7) for ρ = d, obtain the Laplace transform of (10.8). Now do the following:

(i) Take ∂/∂a of (10.7) and then put a = 0. Using J′_{1}(0) = \frac{1}{2} and J_{−1} = −J_{1}obtain

\frac{ρ}{(ρ^{2} + z^{2})^{3/2}} = ∫_{0}^{∞} kdk e^{−k|z|}J_{1}(kρ)

(ii) Put ρ^{2} = d^{2} + a^{2} − 2ad cos ϕ in the above and obtain

\begin{matrix}√ & \overline{d^{2} + a^{2} − 2ad cos ϕ} \\ (d^{2} + a^{2} + z^{2} − 2ad cos ϕ) & 3/2 \end{matrix} = ∫_{0}^{∞} kdk e^{−k|z|}J_{1}(k\sqrt{d^{2} + a^{2} − 2ad cos ϕ})

(iii) Compare this result with the derivative of eq. (10.7) for ρ = d, obtain the Laplace transform of (10.8).

we can also write

M(d) = −4π^{2} ∑_{n=1}^{∞} \frac{a^{2n+2}}{d^{2n+1}} |P_{2n}(0)|^{3} \frac{n}{n + 1}