Check of the web paper
"The external magnetic field of a long solenoid"

Gerhard W. Bruhn, Darmstadt University of Technology

Oct 31, 2008
revised Nov 10, 2008

1. Preliminary remarks

1) The paper under consideration reports (non-trivial) results without proofs. It is not self-explaining with very sparse information.
2) The author does not use SI units, probably cgs units.
3) The paper gives no E field data, only B-data, therefore the Maxwells cannot be checked directly.
4) Eq. (6) gives the B-field inside and outside the solenoid. This will allow to check the boundary conditions for B.

It is no disadvantage to check a paper written using a certain method by another valid method, but sometimes it is somewhat tiring.

2. The boundary conditions of B

2.1 Check of the behavior for r → 0: OK. The solution B ~ J_o 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) J_o(κr) = i Γ j κR (J₁(κR) + i N₁(κR)) J_o(κr) ,

(6 +) B⁺(r) = i Γ j κR H_o⁽¹⁾(κr) J₁(κR) = i Γ j κR (J_o(κr) + i N_o(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 (J_o(κR) N₁(κR) − J₁(κR) N_o(κR)) .

Due to an identity of the Bessel/Neumann functions involving the Wronskian of the Bessel de. we have

(6 b) J_o(κR) N₁(κR) − J₁(κR) N_o(κR) = ²/_πκR ,

hence

(6 c) [B] = B⁺(R) − B⁻(R) = ²/_π Γ j = ^4π/_c j .

Possibly an appropriate constant factor is missing in (6).

3. The boundary conditions of E

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 H_o⁽¹⁾(κr) + C₂ H_o⁽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 [C_k] = C_k⁺ − C_k⁻ (k = 1, 2) .

From (6 +) we learn

(6 −') B⁻(r) = ⁱ/₂ Γ j κR H₁⁽¹⁾(κR) (H_o⁽¹⁾(κr) + H_o⁽²⁾(κr))

(6 +') B⁺(r) = ⁱ/₂ Γ j κR 2J₁(κR) (H_o⁽¹⁾(κr) + 0 ),

hence

[C₁] = ⁱ/₂ Γ j κR (2J₁(κR) − H₁⁽¹⁾(κR)) = ⁱ/₂ Γ j κR H₁⁽²⁾(κR)

and

[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

∫_∂Ω E·dx = ∫_Ω curlE · do = − ∫_Ω ^∂B/_∂t · do = − ^d/_dt ∫_Ω B·do .

But perhaps, there might be a nice tiny flaw in the calculation above.

Devil never sleeps!

So check all calculations carefully! And perhaps the problems with the jump-condition of B will find an explanation too.