El campo magnético de una corriente

Texto completo

(1)

El campo magnético de una

corriente

(2)

El campo magnético producido

por una carga en movimiento



1820 Hans Christian

(3)

El campo magnético producido

por una carga en movimiento



1876 Henry Rowland



Campo magnético

producido por su disco

en rotación fue apenas

0.00001 del campo de

la Tierra.

(4)

El campo magnético producido

por una carga en movimiento

(5)

El campo magnético producido

por una carga en movimiento

2

r

qvsen

B

φ

2

ˆ

r

x

q

K

v

r

B

r

r

=

(6)

El campo magnético producido

por una carga en movimiento

2

ˆ

r

x

q

K

v

r

B

r

r

=

3

r

x

q

K

v

r

B

r

r

r

=

r

r

r

=

r

ˆ

(7)

µ0: permeabilidad del vacío 2 7 7 0 4 10 Tm/ A 4 10 N / A − − = = π π µ

El campo magnético producido

por una carga en movimiento

3 0

4

r

x

q

v

r

B

r

r

r

π

µ

=

2 0

4

r

vsen

q

B

φ

π

µ

=

(8)

El campo magnético producido

por una carga en movimiento

(9)

Campo magnético creado por corrientes eléctricas: Ley de Biot y Savart

El campo magnético de una

corriente

2 0

ˆ

4

r

x

dq

d

B

v

r

r

r

π

µ

=

2 0

ˆ

4

r

x

id

d

B

s

r

r

r

π

µ

=

3 0

4

r

x

id

d

B

s

r

r

r

r

π

µ

=

………. 2 0

(

)(

)

4

r

sen

ds

i

dB

φ

π

µ

=

(10)

Sources of Electric field, magnetic field

From Coulomb's Law, a point charge dq

generates electric field distance away from the source:

r

r

πε

=

r

2 0

1

4

dq

ˆ

d

r

E

r

r r r dE dq P

From Biot-Savart's Law, a point current

generates magnetic field distance away from the source:

r

Ids

r

r

r dB

r

Ids

µ

π

=

×

r

r

0 2

4

Id

ˆ

d

r

s

B

r

Difference:

1. A current segment , not a point charge dq, hence a vector.

2. Cross product of two vectors, is determined by the right-hand rule, not .

r

Ids

r dB

ˆr

(11)

Campo magnético total B

=

B

B

r

r

d

=

=

=

0

3

2

0

4

ˆ

4

r

x

id

r

x

id

d

B

s

r

s

r

B

r

v

v

r

r

π

µ

π

µ

(12)

2 0 ˆ 4 r Id dB = l×r π µ φ π µ sen r Idx dB 0 2 4 = Líneas de campo magnético creadas por

un hilo conductor θ π µ cos 4 2 0 r Idx dB = θ ytg x= dx = ysecdθ θ cos r y = θ θ π µ d y I dB cos 4 0 = ( 1 2) 0 4π θ θ µ sen sen y I B = +

Para un conductor muy largo:

2 2 1 π θ θ = = y I B π µ 2 0 =

Campo magnético debido a una corriente en

un conductor rectilíneo

) (π φ

φ = sen

(13)

For a long, straight conductor the

magnetic field goes in circles

 The magnetic field lines are

circles concentric with the wire

 The field lines lie in planes

perpendicular to to wire

 The magnitude of the field is

constant on any circle of radius a

 A different and more

convenient right-hand rule for determining the direction of the field is shown

B

(14)

Campo magnético creado por una espira circular 2 0 ˆ 4 r Id dB = l×r π µ 2 2 0 2 0 4 ˆ 4 x R Idl r Id d + = × = π µ π µ l r B 2 2 2 2 0 2 2 4 R x R R x Idl R x R dB dBsen dBx + + =         + = = π µ θ

(

)

(

)

+ = + = = dl R x IR dl R x IR dB Bx x 3/2 2 2 0 2 / 3 2 2 0 4 4 π µ π µ

(

2 2

)

3/2 0 (2 ) 4 x R R IR Bx + =

π

π

µ

Líneas de campo creado por una espira circular Líneas de campo creado por una espira circular

(15)
(16)

Example: at the center of a

circular loop of wire with current I



From Biot-Savart Law, the

field at O from is

B

r

I I I = 2

= 2 2 = 4 4 2 o o o full circle µ µ µ B ds πr πr πr r

µ

µ

π

π

= × = r r 0 0 2 2 4 4 Id Ids ˆ ˆ d r r s B r k

r

Ids

r Ids O Y X r r

Off center points are not so easy to calculate.

I

ˆ

=

or

r

=

2

o

µ

B

B

r

B

k



This is the field at the

center

(17)

Example problems

A long, straight wire carries current I. A right-angle bend is made in the middle of the wire. The bend forms an arc of a circle of radius r. Determine the magnetic filed at the center of the arc.

Formula to use: Biot-Savart’s Law, or more specifically the results from the discussed two examples: I I sin = 2 = 0 4 4 π o o µ µ B θ πr πr

For the straight section

For the arc = I

= I = I

1 4 2 2 2 4 4 4 8 o o o circle µ µ πr µ B ds πr πr r

The final answer: = I + I + I = I  + 

  2 1 4 8 4 4 2 o o o o µ µ µ µ B πr r πr r π magnitude

(18)

2 0 ˆ 4 r Id dB = l×r π µ φ π µ sen r Idx dB 0 2 4 = Líneas de campo magnético creadas por

un hilo conductor θ π µ cos 4 2 0 r Idx dB = θ ytg x= dx = ysecdθ θ cos r y = θ θ π µ d y I dB cos 4 0 = ( 1 2) 0 4π θ θ µ sen sen y I B = +

Para un conductor muy largo:

2 2 1 π θ θ = = y I B π µ 2 0 =

Campo magnético debido a una corriente en

un conductor rectilíneo

) (π φ

φ = sen

(19)
(20)
(21)
(22)

Magnetic Force Between Two Parallel

Conductors

 Two parallel wires each carry

steady currents

 The field due to the current in

wire 2 exerts a force on wire 1 of

F1 = I1B

2

 Substituting the equation for B2

gives



Check with right-hand

rule:

 same direction currents attract

each other

 opposite directions currents

repel each other

2 B r I I a = 1 2 l 1 2 o µ F π

The force per unit length on the wire is I I

a = l 1 2 2 o B µ F π

(23)

Force on a Wire, the formula

 The magnetic force is exerted on each

moving charge in the wire



 The total force is the product of the force

on one charge and the number of charges in the wire

 d q = × F v B r r r

(

)

= × = × r r r r r d d q nAL qnA L F v B v B

 In terms of the current, this becomes

 I is the current

 is a vector that points in the direction of the

current

 Its magnitude is the length L of the segment

B

= ×

I

F

L B

r

r

r

L r d I = nq ⋅ r r Q v A

(24)

Campo magnético creado por un solenoide

(

)

(

2 2

)

3/2 2 0 2 / 3 2 2 2 0 2 4 ) 2 ( 4 x R nIdx R R x R di dBx + = + = π π µ π π µ Líneas de campo debidas a dos espiras que transportan la misma corriente en el mismo sentido

(

)

+ = b a x R x dx nI R B 3/2 2 2 2 0 2 4π π µ       + + + = 2 2 2 2 0 2 1 R a a R b b nI Bx µ nIdx di = L N n = /

(25)

Campo magnético creado por una espira circular 2 0 ˆ 4 r Id dB = l×r π µ 2 2 0 2 0 4 ˆ 4 x R Idl r Id d + = × = π µ π µ l r B 2 2 2 2 0 2 2 4 R x R R x Idl R x R dB dBsen dBx + + =         + = = π µ θ

(

)

(

)

+ = + = = dl R x IR dl R x IR dB Bx x 3/2 2 2 0 2 / 3 2 2 0 4 4 π µ π µ

(

2 2

)

3/2 0 (2 ) 4 x R R IR Bx + =

π

π

µ

Líneas de campo creado por una espira circular Líneas de campo creado por una espira circular

(26)
(27)
(28)
(29)
(30)

from a long, straight conductor

re-calculated using Ampere’s Law

B

r

Choose the Gauss’s Surface, oops, not again!

Choose the Ampere’s loop, as a circle with radius a.

Ampere’s Law says

=

B

r

d

s

r

µ

o

I

a

a

=

=

=

2

2

o o

B ds

B

π

µ

I

µ

I

B

π

is parallel with , so r B r ds

(31)

When the wire has a size: with radius

R



Outside of the wire, r > R



Inside the wire, we need

I

’,

the current inside the

ampere’s circle

2 2 o o µ d B πr µ B πr ⋅ = = → =

B s r r I ( ) I 2 2 2 2 2 o o r d B πr µ R µ B r πR ⋅ = = → =   =  

B s r r ( ) I ' I ' I I

(32)

Plot the results



The field is proportional

to r inside the wire



The field varies as 1/r

outside the wire



Both equations are

(33)

Ideal (infinitely long) Solenoid



An ideal solenoid is

approached when:

 the turns are closely spaced

 the length is much greater than

the radius of the turns



Apply Ampere’s Law to loop 2:

⋅ = ⋅ = =

r r

r r

l 1 1 path path d d B ds B B s B s o d B

µ

NI ⋅ = =

B s r r l

I

I

o o

N

B

=

µ

=

µ

n

l

(34)

Magnetic Field of a Toroid



The toroid has N turns of

wire



Find the field at a point at

distance r from the center

of the toroid (loop 1)



There is no field outside

the coil (see loop 2)

2

2

o o

d

B

π

r

µ

N

µ

N

B

π

r

=

=

=

B

s

r

r

(

)

I

I

Figure

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Referencias

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