Oct 31, 2008
revised Nov 10, 2008
It is no disadvantage to check a paper written using a certain method by another valid method, but sometimes it is somewhat tiring.
2.1 Check of the behavior for r → 0: OK. The solution B ~ Jo is regular.
2.2 Check of the jump [B] at r=R:
From (6) we have (in my terminology, replacing NJ with j)
(6 −) B−(r) = i Γ j κR H1(1)(κR) Jo(κr) = i Γ j κR (J1(κR) + i N1(κR)) Jo(κr) ,
(6 +) B+(r) = i Γ j κR Ho(1)(κr) J1(κR) = i Γ j κR (Jo(κr) + i No(κr)) (J1(κR) ,
where Γ = 2π²/c and j = NJ. Hence we obtain the jump
(6 a) [B] = B+(R) − B−(R) = Γ j κR (Jo(κR) N1(κR) − J1(κR) No(κR)) .
Due to an identity of the Bessel/Neumann functions involving the Wronskian of the Bessel de. we have
(6 b) Jo(κR) N1(κR) − J1(κR) No(κR) = 2/πκR ,
[B] = B+(R) − B−(R) =
2/π Γ j =
4π/c j .
Possibly an appropriate constant factor is missing in (6).
We are here confronted with the problem that we are not given the field E by the author. However, for fields of the assumed cylindrical symmetry there is a simple method to calculate E from known B.
If B is given (see above) by
B(r) = C1 Ho(1)(κr) + C2 Ho(2)(κr)
then (see eqs. (3.14) and (4.1+) - (4.3+) of my web paper)
E(r) = c/i (C1 H1(1)(κr) + C2 H1(2)(κr))
We are interested in the jump [E] at r=R which is given by
[E] = c/i ([C1] H1(1)(κR) + [C2] H1(2)(κR)) .
where [Ck] = Ck+ − Ck− (k = 1, 2) .
From (6 +) we learn
(6 −') B−(r) = i/2 Γ j κR H1(1)(κR) (Ho(1)(κr) + Ho(2)(κr))
(6 +') B+(r) = i/2 Γ j κR 2J1(κR) (Ho(1)(κr) + 0 ),
[C1] = i/2 Γ j κR (2J1(κR) − H1(1)(κR)) = i/2 Γ j κR H1(2)(κR)
[C2] = − i/2 Γ j κR H1(1)(κR) .
By inserting these values in [E] we obtain
[E] = c/2 Γ j κR (H1(2)(κR) H1(1)(κR) − H1(1)(κR) H1(2)(κR)) = 0 .
This shows that the author's B-solution (6) via Maxwell yields an E-solution that passes continuously through the surface r = R.
Therefore the pair of functions (B(r), E(r)) off from the solenoids surface fulfils the homogeneous Maxwell equations, and E is continuous at r = R.
These conditions guarantee that the emf induced in a coil ∂Ω surrounding the solenoid fulfils
But perhaps, there might be a nice tiny flaw in the calculation above.
So check all calculations carefully! And perhaps the problems with the jump-condition of B will find an explanation too.