Electromagnetismo

ELECTRO | 9 de mayo
2010
|
HUGO ADRIAN FLORES CASTAÑEDA | PROBLEMARIO DE ELECTROMAGNETISMO |

ELECTROMAGNETISMO
PROBLEMAS SUPLEMENTARIOS

(-s u + v) * u = 0
-s u*u+v*u = 0
s=v*u/(u*u)
la proyección seria: w = v*u/(u*u) u
(4,10)*(2,3)= (4,10)*(2,3) /(2,3)*(2,3)= 3.2

Proye= a*b/ |a| =102 *3(10/2) = 1.5 (axaz)

|V|= V*V
Cos-1= u*v= |v| |u| Cos
u*v= -401 = 39
U= 1022^2= 104 X = 16904. = Cos-1 = -39/ 16904. = 161.5°
V= -42.5^2=16.25

|V|= V*V
Cos-1= u*v= |v| |u| Cos
u*v= (5.8*-6.93) (1.55* 4) = -34
U= 5.821.552= 36 X = 2304.8 = Cos-1 = -39/ 2304.8= 135°
V= -6.93242 =64

Módulo = √(4^2+3^2+2^2)
Módulo = √29
U = (4, 3, 2) / √29 =U = (4/√29, 3/√29, 2/√29)

a=(ax,ay,az)
b=(bx,by,bz)

Es igual a:
a.b=ax*bx+ay*by+az*bz
o
a.b=|a|*|b|*cosalfa
Por lo que: |a|=sqrt(ax^2+ay^2+az^2)

|b|=sqrt(bx^2+by^2+bz^2)

De esta forma:
A= 0, B=0

Ax= R = -3ay –Zaz = -3ay –Zaz . l R l Y2 + Z2 9 + Z2
R= -3ay -Zaz(0,3,Z)





R= (X1-X)ax + (Y1+ 5)ay +(Z1-Z)azAx= (X1-X)ax + (Y1+ 5)ay +(Z1-Z)az . (X1-X)2 + (Y1+ 5)2 + (Z1-Z)2
(0,0, h)

(x, -5, Z)


R= -Xax – Yay +(h+2)azAx= -Xax – Yay +(h+2)az . X2+Y2+(h+2)2
(0,0, h)

(x, y,-2)B x C = (Bxax + Byay + Bzaz) * (Cxax + Cyay + Czaz) = (BxCx + ByCy + BzCz)
A. B x C = (Axax + Ayay + Azaz) * (BxCx + ByCy + BzCz) = (AxBxCx + AyByCy + CzBzCz)

6 DIF DE 0 =90º A Y B SON PERP.2, 0, -1
3, 1, 0
-2, 6, -4

C=0 NO ES PENDICULAR A (B)

A= 2ax – ax= A.B= (2) (3) + (0) (1) + (-1) (0) =
C=0 NO ES PENDICULAR A (A)

B=3ax + ay= B.C= (3) (-2) + (1) (6) + (0) (4) =0
C= -2a +6ay -4ax A.C= (2) (-2) +(0) (6) + (-1) (-4) =0



1,-1, 0
0, 0, 2
-1, 3-0

A= ax-ay =
B=2az =
C= -ax +3ay -=

A.B= {(-1) (2)} - {(0) (0)} + {(0) (1)} – (1) (2)} + {(1) (0) – (-1) (0)} =
-4
A.B= (-2) + (-2) + (0) =
A.B=
A.B=-2, -2, 0
C=-1, 3. 0 {A.B} (C) = (-2) (0)} - {(0) (3)} + {(0) (-1)} – (-2) (0)} + {(-2) (3) – (2) (-1) =
-4
{A.B} (C) = 0 + 0 -4
{A.B} (C) =

A X B= 1 -1 0 A X B = (-1) (2)} - {(0) (0)} + {(6) (0)} – (1) (2)} + {(1) (0) – (-1) (0) =
0 0 2 A X B = -2 -2 O
-2 -2 0
-1 3 0...