Appendix
199
∂B z
∂t
= −
∂
∂x
(B z v x − B x v z ) −
∂
∂y
B z v y
.
(A.14.2)
Using (6.8) and (6.40), the second term on the right-hand side of (A.14.1) is written
as
B
∂
∂z
(v x cosθ − v z sinθ ) = −
B
sinθ
·
∂v z
∂z
.
(A.14.3)
This equation is reduced to
∂B
∂t
sinθ + Bcosθ
∂θ
∂t
= −B
∂
∂y
v y sinθ
−
B
sinθ
·
∂v z
∂z
.
(A.14.4)
The first term on the right-hand side of (A.14.2) is written as
−B
∂
∂x
(v x cosθ − v z sinθ ) = −
B
cosθ
·
∂v x
∂x
,
(A.14.5)
and this equation is reduced to
∂B
∂t
cosθ − Bsinθ
∂θ
∂t
= −B
∂
∂y
v y cosθ
−
B
cosθ
·
∂v x
∂x
.
(A.14.6)
Summing (A.14.4) multiplied by sinθ and (A.14.6) multiplied by cosθ , we have
∂B
∂t
= −B
∂v y
∂y
.
(A.14.7)
Subtracting (A.14.6) multiplied by sinθ from (A.14.4) multiplied by cosθ , we have
∂θ
∂t
= α f v y +
1
sinθ cosθ
·
∂v x
∂x
.
(A.14.8)
Thus, (6.41) and (6.42) are derived.
A.15 Proof of (6.45)
From the relationships of μ 0 H e = Bcosθ 0 and μ 0 H I = Bsinθ 0 , we have
∂θ 0
∂t
=
∂θ
∂t
=
μ 0
B
cosθ 0
∂H I
∂t
,
∂B
∂t
= μ 0 sinθ 0
∂H I
∂t
.
(A.15.1)
Using (6.49), (6.45) is written as