"The external magnetic field of a long solenoid"

Oct 31, 2008

revised Nov 10, 2008

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.

**2.1 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 H_{1}^{(1)}(κR) J_{o}(κr)
=
i Γ j κR (J_{1}(κR) + i N_{1}(κR)) J_{o}(κr) ,

(6 +)
B^{+}(r) =
i Γ j κR H_{o}^{(1)}(κr) J_{1}(κR)
=
i Γ j κR (J_{o}(κr) + i N_{o}(κr))
(J_{1}(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) N_{1}(κR) −
J_{1}(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) N_{1}(κR) −
J_{1}(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).

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) = C_{1} H_{o}^{(1)}(κr) +
C_{2} H_{o}^{(2)}(κr)

then (see eqs. (3.14) and (4.1__+__) - (4.3__+__) of
my web paper)

E(r) = ^{c}/_{i} (C_{1} H_{1}^{(1)}(κr) +
C_{2} H_{1}^{(2)}(κr))

We are interested in the jump [E] at r=R which is given by

[E] = ^{c}/_{i} ([C_{1}] H_{1}^{(1)}(κR) +
[C_{2}] H_{1}^{(2)}(κR)) .

where [C_{k}] = C_{k}^{+} − C_{k}^{−}
(k = 1, 2) .

From (6 __+__) we learn

(6 −')
B^{−}(r) =
^{i}/_{2} Γ j κR H_{1}^{(1)}(κR)
(H_{o}^{(1)}(κr) + H_{o}^{(2)}(κr))

(6 +')
B^{+}(r) =
^{i}/_{2} Γ j κR 2J_{1}(κR)
(H_{o}^{(1)}(κr) + 0 ),

hence

[C_{1}] = ^{i}/_{2}
Γ j κR (2J_{1}(κR) − H_{1}^{(1)}(κR)) =
^{i}/_{2} Γ j κR H_{1}^{(2)}(κR)

and

[C_{2}] = − ^{i}/_{2}
Γ j κR H_{1}^{(1)}(κR) .

By inserting these values in [E] we obtain

[E] = ^{c}/_{2} Γ j κR
(H_{1}^{(2)}(κR) H_{1}^{(1)}(κR) −
H_{1}^{(1)}(κR) H_{1}^{(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.