Oct 26, 2008
revised and supplemented on Nov 11, 2008
Let (r,θ,z) denote cylinder coordinates around the z-axis of an Euclidean 3-D space, the basic vectors of which are
x = xi +yj + zk, r = x − zk = xi +yj = r r°
and the radial and azimuthal unit vectors r° and r°×k respectively, where r = |r| = |x×k| = |r×k|.
1.2 Geometry of the solenoid
Axis = z-axis, radius R , length infinite.
The region r < R is called inner domain, the region r > R is called outer domain.
1.3 Physical data of the solenoid
σ = conductivity per z-length unit, λ = inductivity per z-length unit.
1.4 Physics around the solenoid using SI units
(1): The homogeneous Maxwell equations to be satisfied inside and outside the solenoid, i.e. separately for r < R and for r > R.
− No sources of E and B
(1.1) div E = 0, div B = 0 ,
− Faraday's law
(1.2) curl E = − Bt ,
− Ampère's law
(1.3) curl B = 1/c² Et (+ μj if r=R)
(2): The Maxwell equations imply the wave equations
(1.4) ΔE = 1/c² Ett , ΔB = 1/c² Btt for r ≠ R.
(3): All fields are oscillating ~ e−iωt at constant angular frequency ω (converts the wave eqs. to Helmholtz eqs.).
(4): E in azimuthal direction, B in direction of the axis of the solenoid, both spatially only depending on r. j flowing azimuthally at r=R.
(1.5) B = B k , E = E r° × k , j = j r° × k .
(5): Prescribed coupling conditions at r=R:
(a) E passing continously.
(b) B passing with a jump µj (according to Ampère's law), where j is the current density within the surface r=R.
(6): Alternating emf given by circulation of E around the solenoid: ∫r=R E·dr = 2πR Eo e−iωt.
Exercise: Evaluate div B und div E for B = B(r,t) k und E = E(r,t) r°×k .
All time derivatives and the angular frequency ω are vanishing.
The current density is time independent: jo = σE azimuthal on the surface of the
This implies the boundary conditions
(2.1) E = jo/σ for r → R from outside, and
(2.2) E → 0 for r → ∞ .
The unique solution of this outer domain problem is well-known:
(2.3) E = Eo R/r r°×k , where Eo = jo/σ.
B satifies div B = 0 and curl B = 0 and is | | k. Due to B = B k this yields grad B = 0, and hence B = const. Supposed B ≠ 0 we would have infinite flux of B in the outer domain. Thus, B=0 for r > R.
I the inner domain the constant B must be chosen such that the
jump condition (5b) of Sect.1,B(R+o)−B(R-o) = μ jo,
is fulfilled. Then r < R yields B(r) = B(R−o) = − μ jo
N.B. Orientation of j questionable, to be checked.
At the solenoid at r = R an azimuthal alternating current is assumed:
(3.1) j = jo e−iωt = jo r°×k e−iωt ,
where jo is a complex constant. For more about the correlation with the effective electrical field E see Sect. 6.
The effective electrical field E is assumed
(3.2) E = E(r) r°×k e−iωt
to be a solution of the wave equation (1.4-1) for r ≠ R. This yields the Helmholtz equation
(3.3) E'' + 1/r E' + (κ² − 1/r²) E = 0 where ' = d/dr and κ = ω/c.
This is a Bessel type differential equation with well-known solutions:
E(r) is a linear combination of the (complex-valued) Hankel functions H1(1), H1(2) of order 1:
(3.4) E(r) = c1H1(1)(κr) + c2H1(2)(κr)
Likewise the ansatz (see (1.5))
(3.5) B = B(r) e−iωt k
for the wave equation (1.4-2) yields the Helmholtz equation
(3.6) B'' + 1/r B' + κ² B = 0 ,
the solutions of which are Hankel functions of order 0:
(3.7) B(r) = C1Ho(1)(κr) + C2Ho(2)(κr)
There is a coupling between B(r) and E(r) by the Maxwell equations (1.2) and (1.3), s. Eq. (3.9) below.
The still free constants are to be determined from boundary and transmission conditions as will be shown in Sect. 4.
Knowing the field E we can calculate the field B by means of the Faraday law (1.2)
(3.8) Bt = − curl E
= grad [rE(r)] e−iωt × (r°×k) / r
= (rE)' grad r × e−iωt (r°×k) / r
= 1/r(rE)' e−iωt r° × (r°×k) = − 1/r(rE)' e−iωt k ,
using the fact that the field (r°×k) / r is irrotational. Hence
(3.10) Bt = 1/r (rE)' e−iωt k ,
(3.11) B = B(r) e−iωt k = i/ωr (rE)' e−iωt k ,
where a time-independent summand was ommitted.
This yields the following coupling between E(r) and B(r)
There exist quite similar couplings between the Hankel functions H1(s) and Ho(s):
(3.13) 1/s d/ds[sH1(s)] = Ho(s) .
which lead to couplings between the coefficients c und C in the Bessel representations (3.4) of E(r) and (3.7) of B(r):
(3.14) Ck = i/c ck (k = 1, 2)
mit der Ampère-Gleichung (1.3) im Innen- und Außenraum (j = 0):
(3.15) curl B = 1/c² Et
[1/r (rE)']' e−iωt grad r
= i/ω [1/r (rE)']' e−iωt r°×k
= i/ω (E'' + 1/r E' − 1/r² E) e−iωt r°×k
= − i/ω κ² E e−iωt r°×k
= − iω/c² E e−iωt r°×k .
(3.2) liefert andererseits
(3.17) 1/c² Et = − iω/c² E e−iωt r°×k
somit ist (3.15) in der Tat erfüllt.
The field functions for the inner (−) and the outer (+) domain are given by
(4.1+) B+(r) = C1+ Ho(1)(κr) + C2+ Ho(2)(κr)
(4.2+) E+(r) = c1+ H1(1)(κr) + c2+ H1(2)(κr) .
As a consequence of the Maxwell equations we have equation (3.14), which applied to the present situation yields
(4.3+) Ck+ = i/c ck+ .
The coupling between inner and outer domain is given by the prescribed jump of B at the current density j at the surface r=R:
Integration of B·dx = B dz along a closed path inside and outside the discontinuity surface r=R (where B is almost everywhere | | dx = kdz) due to the Ampère law (1.3) yields the following jump condition (since the identical factors eiωt of B and j can be reduced):
(4.4) [B] = B(R+0) − B(R−0) = μ jo .
The limits B(R+0) are given by (3.7):
(4.5) B(R+0) = C1+ Ho(1)(κR) + C2+ Ho(2)(κR) ,
hence (4.4) yields the 1st jump relation:
(4.7) Ck := Ck+ − Ck− (k=1,2).
The azimuthal field E, hence also E(r), must pass continuously through r=R:
(4.8) c1 H1(1)(κR) + c2 H1(2)(κR) = [E(R+0)] = 0
where ck := [ck+] (k=1,2). Due to the proportionality (4.3) we obtain the 2nd jump relation
Since the Hankel functions Ho(1)/(2), H1(1)/(2) constitute a fundamental system of the Bessel de. the determinant of the linear system (4.6+9) is well-known to fulfil the identity
(4.10) D := ( Ho(1) H1(2) − Ho(2) H1(1) ) (κR) = 4 i/πκR .
Therefore the linear system (4.7+10) can be solved explicitly:
Due to (4.7) the cofficient Ck are the jumps Ck+ − Ck−. Therefore we have to look for further conditions to be fulfilled by Ck+ in the inner and the outer domain respectively.
The outer domain: Here the first term in equ. (4.1+) represents an out-running waves ~ Hk(1)(κr) e−iωt while the second summand ~ Hk(2)(κr) e−iωt represents an incoming waves as can be seen by the asymptotic behavior of both summands (see Sect. 5, Discussion). Due to the causality principle no incoming waves can be admitted in the outer domain:
(4.12+) C2+ = 0 and therefore C2− = − C2 = − ¼ i μjo πκR H1(1)(κR).
The inner domain: Here the regularity of B is required. However, the (opposite) imaginary parts of the Hankels Ho(1)/(2) in (4.1−) have logarithmic singularities at r=0. Therefore the singularities have to be compensated by the condition
(4.12−) C1− = C2−.
So, by (4.12+) the coefficient C1− is determined as well to obtain:
These are the field functions for the inner domain. For the outer domain we know C2+ = 0 due to the causality principle, and C1+ = C1 + C1− where the right hand side is completely known now due to (4.11) and (4.12+).
Therefore the emf induced in the circular coil r = r1 > R is
(4.14) U(r1) = 2π r1 E(r1) e−iωt with E(r1) from (4.13+).
In addition and as response to a contradictory claim in the Abstract of the paper  we'll here calculate the emf induced in a coil ∂Ω: Due to the Stokes integral theorem we have
(4.15) ∫∂Ω E ·dx = ∫Ω curl E · do .
Here we can substitute curl E by −∂B/∂t due to the Faraday law that is fulfilled by construction of E and B (see (3.8)) to obtain
(I) The inner radiation is given by eq. (4.13−), E and B are regular also in the center.
(II) The E field passes continously through the solenoid surface r=R while the B field jumps by μj at that surface.
The behavior of the radiation for large distances r from the solenoid axis is given by the asymptotics of the Bessel/Hankel functions, see e.g. Asymptotics of the Bessel/Hankel functions The radiation is therefore decreasing spatially oscillating at least ~ r−½ .
Due to the Asymptotics of the Bessel/Hankel functions we have the far field of the AC solenoid to be linear combinations of the terms
(5.1) E : H1(1)(κr) ≈ (2/πr)½ ei[κ(r−ct)−3π/4] and H1(2)(κr) ≈ (2/πr)½ e−i[κ(r−ct)−3π/4] ,
(5.2) B : Ho(1)(κr) ≈ (2/πr)½ ei[κ(r−ct)−π/4] and Ho(2)(κr) ≈ (2/πr)½ e−i[κ(r−ct)−π/4] ,
Of course, if a time reversal transform t → −t is applied, then the above solution changes to an incoming wave. That wave would contradict the causality principle.
Which relation do we have between the electric field E along the solenoid surface and the current density j?.
Let the solenoid have an inductivity λ per z-length-unit in addition to the conductivity σ. Then the azimuthal field E = Eo e−iωt is only partially available for the ohm current density j, another part λ dj/dt is required for compensating the inductive counter voltage of the solenoid. The distribution is
(6.1) Eo e−iωt = j/σ + λ dj/dt .
The periodic solution j(t) of this de. is well known:
(6.2) j = jo e−iωt where jo = Eo eiφ / (λ²ω² + 1/σ²)½ and tan φ = ωλ σ .
This is the more detailed description of j = jo e−i(ωt−φ) in eq. (3.1), hence:
(6.3) j = |jo| r°×k e−i(ωt−φ) = Eo / (λ²ω² + 1/σ²)½ r°×k e−i(ωt−φ) .
 A. Sommerfeld, Elektrodynamik, § 21, Akad. Verlagsgesellschaft, Leipzig (1949).
 J. D. Jackson, Classical Electrodynamics, 3rd ed. , Sect.3.7 (p.111 ff.) and
Chap.5 (p.225.f.), John Wiley & Sons (1998).
 G. Rousseaux, R. Kofman, and O. Minazzoli, The Maxwell-Lodge effect:
significance of electromagnetic potentials in the classical theory,
Eur. Phys. J. D (2008), Vol.49, p.249-256 .
 K.E. Schmidt, The external magnetic field of a long solenoid,
G.W. Bruhn, Check of the web paper
"The external magnetic field of a long solenoid"