divE =ρε(

rotE = 0

divB = 0

rotB ≠ 0

𝐅𝐋

S

dl

B

FJvD

B

F

IA B C

F1

F2

F3

F4Bn

BF2

F4

a

b

n

b

contattirotanti

q>0 + + + + +

h dIB E

q<0- - - -

h dIB E

vD

vD

divB = div 1µ(4π

�

�

Idl⃗×r:r;

=µ(4π

I1divdl⃗×r:r;

�

�

divdl⃗×r:r;

= divdl⃗×r⃗r<

= divı̂ ȷ̂ kAdx dy dzxr<

yr<

zr<

=

=𝜕 dy zr< − dz

yr<

𝜕x+𝜕 dz xr< − dx

zr<

𝜕y+𝜕 dx yr< − dy

xr<

𝜕z=

= dyz − dzy𝜕 1r<𝜕x

+ dzx − dxz𝜕 1r<𝜕y

+ dxy − dyx𝜕 1r<𝜕z

=

= dyz − dzy −3xrK

+ dzx − dxz −3yrK

+ dxy − dyx −3zrK

=

= −3rK

dyzx − dzyx + dzxy − dxzy + dxyz − dyxz = 0

perchédivB = 0