Page 186 - Electromagnetics

together form a closed surface. The outward flux of B through this surface is

B · dS = 0
S 0 +S 1 +S 2
so that

W =−I B · dS = I B · dS
S 0 S 1 +S 2
where ˆ n is outward from the closed surface. Finally, let 1,2 be the flux of B through S 1,2
in the direction determined by dl and the right-hand rule. Then
W =−I ( 2 − 1 ) =−I . (3.183)
Now suppose that the initial position of the filament is at infinity. We bring the filament
into a final position within the field B through a succession of small displacements,
each requiring work (3.183). By superposition over all displacements, the total work is
W =−I ( − ∞ ) where ∞ and are the fluxes through the filament in its initial and
final positions, respectively. However, since the source of the field is localized, we know
that B is zero at infinity. Therefore ∞ = 0 and

W =−I =−I B · ˆ n dS (3.184)
S
where ˆ n is determined from dl in the right-hand sense.
Now let us find the work required to position two current filaments in a field-free region
of space, starting with both filaments at infinity. Assume filament 1 carries current I 1
and filament 2 carries current I 2 , and that we hold these currents constant as we move
the filaments into position. We can think of assembling these filaments in two ways: by
placing filament 1 first, or by placing filament 2 first. In either case, placing the first
filament requires no work since (3.184)is zero. The work required to place the second
filament is W 1 =−I 1 1 if filament 2 is placed first, where 1 is the flux passing through
filament 1 in its final position, caused by the presence of filament 2. If filament 1 is
placed first, the work required is W 2 =−I 2 2 . Since the work cannot depend on which
loop is placed first, we have W 1 = W 2 = W where we can use either W =−I 1 1 or
W =−I 2 2 . It is even more convenient, as we shall see, to average these values and use
1
W =− (I 1 1 + I 2 2 ) . (3.185)
2
We must determine the energy required to keep the currents constant as we move the
filaments into position. When moving the first filament into place there is no induced
emf, since no applied field is yet present. However, when moving the second filament
into place we will change the flux linked by both the first and second loops. This change
of flux will induce an emf in each of the loops, and this will change the current. To keep
the current constant we must supply an opposing emf. Let dW emf /dt be the rate of work
required to keep the current constant. Then by (3.153)and (3.181)we have
dW emf d
=− J · E dV =−I E · dl =−I .
dt V dt
Integrating, we find the total work W required to keep the current constant in either
loop as the flux through the loop is changed by an amount :
W em f = I .

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