Page 187 - Electromagnetics

So the total work required to keep I 1 constant as the loops are moved from infinity (where
the flux is zero)to their final positions is I 1 1 . Similarly, a total work I 2 2 is required
to keep I 2 constant during the same process. Adding these to (3.185), the work required
to position the loops, we obtain the complete assembly energy
1
W = (I 1 1 + I 2 2 )
2
for two filaments. The extension to N filaments is
N
1
W m = I n n . (3.186)
2
n=1
Consequently, the energy of a single current filament is
1
W m = I . (3.187)
2
We may interpret this as the “assembly energy” required to bring the single loop into
existence by bringing vanishingly small loops (magnetic dipoles)in from infinity. We
may also interpret it as the energy required to establish the current in this single filament
against the back emf. That is, if we establish I by slowly increasing the current from
zero in N small steps I = I/N, an energy n I will be required at each step. Since
n increases proportionally to I,wehave
N
I

W m = (n − 1)
N N
n=1
N
&
where is the flux when the current is fully established. Since (n−1) = N(N −1)/2
n=1
we obtain
1
W m = I (3.188)
2
as N →∞.
A volume current J can be treated as though it were composed of N current filaments.
Equations (3.128)and (3.186)give
N
1
W m = I n A · dl.
2
n=1 n
Since the total current is

N

I = J · dS = I n
CS n=1
where CS denotes the cross-section of the steady current, we have as N →∞
1
W m = A · J dV. (3.189)
2 V
Alternatively, using (3.135), we may write

1 J(r) · J(r )

W m = dV dV .
2 V V |r − r |

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