Page 236 - Engineering Electromagnetics, 8th Edition

218 ENGINEERING ELECTROMAGNETICS

only of x 1 , y 1 , and z 1 . Consequently, it may be factored out of the curl operation as
any other constant, leaving
1 J 1
H 2 = ∇ 2 × dν 1 (54)
4π vol R 12
The curl of the product of a scalar and a vector is given by an identity which may
be checked by expansion in rectangular coordinates or obtained from Appendix A.3,
∇× (SV) ≡ (∇S) × V + S(∇× V) (55)
This identity is used to expand the integrand of (54),
1
1 1
H 2 = ∇ 2 × J 1 + (∇ 2 × J 1 ) dν 1 (56)
4π vol R 12 R 12
The second term of this integrand is zero because ∇ 2 × J 1 indicates partial deriva-
tives of a function of x 1 , y 1 , and z 1 , taken with respect to the variables x 2 , y 2 , and z 2 ;
the first set of variables is not a function of the second set, and all partial derivatives
are zero.
The first term of the integrand may be determined by expressing R 12 in terms of
the coordinate values,

2 2 2
R 12 = (x 2 − x 1 ) + (y 2 − y 1 ) + (z 2 − z 1 )
and taking the gradient of its reciprocal. Problem 7.42 shows that the result is
1 R 12 a R12
∇ 2 =− =−
2
R 12 R 3 12 R 12
Substituting this result into (56), we have
1 a R12 × J 1
H 2 =− 2 dν 1
4π vol R 12
or

J 1 × a R12
H 2 = dν 1
vol 4πR 2 12
which is the equivalent of (3) in terms of current density. Replacing J 1 dν 1 by I 1 dL 1 ,
we may rewrite the volume integral as a closed line integral,

I 1 dL 1 × a R12
H 2 =
2
4πR 12
Equation (51) is therefore correct and agrees with the three definitions (3), (32),
and (46).
Next we will prove Amp`ere’s circuital law in point form,
∇× H = J (28)
Combining (28), (32), and (46), we obtain
B 1
∇× H =∇ × = ∇× ∇× A (57)
µ 0 µ 0