Page 168 - Electromagnetics

while the point forms are
∇× H(r) = J(r), (3.120)
∇· B(r) = 0. (3.121)

Note the interesting dichotomy between the electrostatic field equations and the magne-
tostatic field equations. Whereas the electrostatic field exhibits zero curl and a divergence
proportional to the source (charge), the magnetostatic field has zero divergence and a
curl proportional to the source (current). Because the vector relationship between the
magnetostatic field and its source is of a more complicated nature than the scalar rela-
tionship between the electrostatic field and its source, more effort is required to develop a
strong understanding of magnetic phenomena. Also, it must always be remembered that
although the equations describing the electrostatic and magnetostatic field sets decou-
ple, the phenomena themselves remain linked. Since current is moving charge, electrical
phenomena are associated with the establishment of the current that supports a magne-
tostatic field. We know, for example, that in order to have current in a wire an electric
field must be present to drive electrons through the wire.

The magnetic scalar potential. Under certain conditions the equations of magne-
tostatics have the same form as those of electrostatics. If J = 0 in a region V , the
magnetostatic equations are
∇× H(r) = 0, (3.122)
∇· B(r) = 0; (3.123)
compare with (3.5)–(3.6) when ρ = 0. Using (3.122)we can define a magnetic scalar
potential m :
H =−∇ m . (3.124)
The negative sign is chosen for consistency with (3.30). We can then define a magnetic
potential difference between two points as

P 2 P 2 P 2
V m21 =− H · dl =− −∇ m (r) · dl = d m (r) = m (r 2 ) − m (r 1 ).
P 1 P 1 P 1
Unlike the electrostatic potential difference, V m21 is not unique. Consider Figure 3.17,
which shows a plane passing through the cross-section of a wire carrying total current I.
Although there is no current within the region V (external to the wire), equation (3.118)
still gives

H · dl − H · dl = I.
2 3
Thus

H · dl = H · dl + I,
2 3

and the integral H · dl is not path-independent. However,

H · dl = H · dl
1 2
since no current passes through the surface bounded by 1 − 2 . So we can artificially
impose uniqueness by demanding that no path cross a cut such as that indicated by the
line L in the figure.

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