Expert: Philip A. Stahl - 2/15/2008

Question
Can you give advice on solving this problem?

Determine the magnetic field of a cylindrically symmetric configuration as a function of distance from the axis, r:
B(r) = B_z(r)z^ + B_phi(r)phi^  and assuming a force-free field configuration of: curl B = alpha B where we assume that alpha(r) = alpha_0.

Answer
Hello again,

This problem is actually relatively straightforward.

The first thing to note is the force-free configuration and in this case: curl B(r) = alpha(r) B(r) = alpha_0 B(r)

note that when 'alpha' is denoted below it will be taken to mean 'alpha_0', when B is used it is taken to be B(r) to avoid excessive notational clutter.

Now, you have to start first with the curl:

In matrix format:

Curl B =

(1/r r^     phi^           z^)

(@/@r    @/@phi          @/@z)

(0        rB_phi         B_z)


where ‘@’ denotes the partial derivative symbol. Note that your B(r) is a function of only *two* components, in phi^ and z^ - there is no r^ component. Thus, the above matrix takes this into account. Indeed, 1/r r^ -> 0.

Taking each coordinate curl remaining:


Curl B_phi =

(-@ B_z/@r) phi^


Curl B_z =

[1/r (@/@r(rB_phi)]z^


Two ways to work it:

1)Straight forward use of partial derivatives which reduce to normal derivatives

(*Why* are the curl B_r terms omitted?)

A)   alpha (B_phi) =  (-@ B_z/@r) phi^

B)   alpha(B_z) =  [1/r (@/@r(rB_phi)]z^

->

(B_phi) =  1/ alpha (-@ B_z/@r) phi^


substitute. above into equation (B):

alpha(B_z) =  [1/r (@/@r(r[(1/ alpha (-@ B_z/@r) phi^]]z^


x (-alpha)

-alpha^2 (B_z) =  -[1/r (@/@r(r[ (@ B_z/@r)]]z^

->

1/r  d/dr (r d B_z  / dr  )  - alpha^2 B_z        =   0

This equation has classical special function solutions!

2)Using vector identities:

The force-free field eqn. is:

Curl (B) = alpha (B)

Further working, by taking the curl of both sides, yields:

curl curl B  = curl alpha (B)

Which, via vector identity is:

DIV DIV B – DIV^2 B =  alpha  curl B

Into the above, substitute Curl (B) = alpha (B)  and DIV B = 0, to obtain:

DIV 2(B) = (alpha)^2 B

where 'DIV'  (“divergence”) is another vector operator.

From this, we can obtain:


DIV^ 2(B) - (alpha)^2 B = 0

In cylindrical geometry, one has:

DIV^2  =  1/r [@/ @r  ( r * @/ @r)]

and you ought to be able to satisfy yourself – with further working – that the end result is the same as in part (1). I suggest that to finish the working you get hold of the CRC Mathematical Handbook and look up 'special functions' and specifically "Bessel functions".

Hope this helps!