ТЭД - Лекция 3 2019

# ( )

3

∂D

× H =

+ j

E ( × H ) = E

∂D

+ E j

∂t

∂t

∂B

× E = −

H ( × E) = −H

∂B

∂t

∂t

∂D

∂B

E ( × H ) − H ( × E) = E

+ H

+ E j

∂t

∂t

∂D

∂B

(E × H ) = −E

− H

− E j

∂t

∂t

'( V, ) )*

∂

E

D + H

B

∫(E × H ) n

= −

jdV

∫

dV − ∫ E

S

∂t V

2

V

dW

= ∫

−

(E × H ) n dS + P

Π = E × H -

dt

S

!

4

$ . %

5

1

B = × A, A -

H =

× A

µ 0µ

∂A

∂A

× ( E +

) = 0

E = − ϕ −

∂t

∂t

εµ ∂

2

εµ ∂ϕ

2 A −

A

=

+ A − µ

µ j

0

c 2 ∂t 2

c 2 ∂t

εµ ∂ϕ

A +

= 0 - %

c

2

∂t

& & .

8

εµ ∂ 2 H

2 H −

= − × j

c

2

∂t 2

εµ ∂ 2 E

1

∂ j

2 E −

= −

ρ + µ 0µ

c 2

∂t

∂t

ε 0ε

" E H

2

u

−

1 ∂ 2u

= 0

z, t

v 2

∂t 2

∂

2

u

1

∂

2

u

1

∂

1

∂

−

= 0,

∂

−

∂

+

u(z, t ) = 0,

∂z

2

v

2

∂t

2

∂z

v ∂t ∂z

v ∂t

"

ξ = z − vt;η = z + vt,

z =

ξ + η

,

t =

η − ξ

2

2v

&

∂

=

∂z

∂

+

∂t

∂

=

1

∂

−

1

∂

,

∂

=

1

∂

+

1 ∂

2v ∂t

∂ξ ∂ξ ∂z

∂ξ ∂t

2 ∂z

2v ∂t

∂η 2 ∂z