El campo magnético de una
corriente
El campo magnético producido
por una carga en movimiento
1820 Hans Christian
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.
El campo magnético producido
por una carga en movimiento
El campo magnético producido
por una carga en movimiento
2
r
qvsen
B
∝
φ
2ˆ
r
x
q
K
v
r
B
r
r
=
El campo magnético producido
por una carga en movimiento
2
ˆ
r
x
q
K
v
r
B
r
r
=
rˆ
3r
x
q
K
v
r
B
r
r
r
=
r
r
r
=
r
ˆ
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 04
r
vsen
q
B
φ
π
µ
=
El campo magnético producido
por una carga en movimiento
Campo magnético creado por corrientes eléctricas: Ley de Biot y Savart
El campo magnético de una
corriente
2 0ˆ
4
r
x
dq
d
B
v
r
r
r
π
µ
=
2 0ˆ
4
r
x
id
d
B
s
r
r
r
π
µ
=
3 04
r
x
id
d
B
s
r
r
r
r
π
µ
=
………. 2 0(
)(
)
4
r
sen
ds
i
dB
φ
π
µ
=
Sources of Electric field, magnetic field
From Coulomb's Law, a point charge dqgenerates electric field distance away from the source:
r
r
πε
=
r
2 01
4
dq
ˆ
d
r
E
r
r r r dE dq PFrom Biot-Savart's Law, a point current
generates magnetic field distance away from the source:
r
Ids
r
r
r dBr
Ids
µ
π
=
×
r
r
0 24
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
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
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 = ysec2θ dθ 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 −
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
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 circularExample: 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 rµ
µ
π
π
= × = r r 0 0 2 2 4 4 Id Ids ˆ ˆ d r r s B r kr
Ids
r Ids O Y X r rOff center points are not so easy to calculate.
I
ˆ
=
or
r
=
2
oµ
B
B
r
B
k
This is the field at the
center
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 = I1 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
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 = ysec2θ dθ 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 −
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 = I1ℓ B
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
Force on a Wire, the formula
The magnetic force is exerted on eachmoving 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 BIn 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 ACampo 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 = /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 circularfrom 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
oI
a
a
=
=
=
∫
2
2
o oB ds
B
π
µ
I
µ
I
B
π
is parallel with , so r B r dsWhen 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 IPlot the results
The field is proportional
to r inside the wire
The field varies as 1/r
outside the wire
Both equations are
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 Bµ
NI ⋅ = =∫
B s r r lI
I
o oN
B
=
µ
=
µ
n
l
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)