## 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 ~ 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)) .

(6 b)         Jo(κR) N1(κR) − J1(κR) No(κR) = 2/πκR ,

hence

(6 c)         [B] = B+(R) − B(R) = 2/π Γ j = /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 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 ),

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/dtΩ 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.