ch29_Magnetic_Induct..

Comments

Transcription

ch29_Magnetic_Induct..
Chapter 29
Magnetic Induction
Dr. Ray Kwok
SJSU
Static Fields (stationary charges, steady current)
 r r Qt 
∫ E ⋅ da = ε  Gauss’s Law
o
 r r

 E ⋅ d l = 0  Conservative
∫ r

r
 B ⋅ da = 0  No magnetic charge
∫

r
r
 B ⋅ d l = µ I  Ampere’s Law
o t
∫
linear, homogeneous, isotropic
Qt = Q f + Qb =
Qf
εr
≤ Qf
Orsted’s Discovery
R
4/21/1820
Danmark
V
I
compass
S
Hans Christian Ørsted
(1777–1851) Danmark
cgs unit for B
r r
∫ B ⋅ d l = µoIt
André-Marie Ampère
(1775–1836) France
SI unit for current
Ampere’s Law, 9/18/1820,
after he learned about Orsted’s
discovery on 9/11/1820 !!
Faraday’s Experiment
With stationary magnet,
no current induced (1831)
Michael Faraday
(1791–1867) England
SI unit for capacitance
Joseph Henry
(1797–1878) USA
SI unit for inductance
Magnetic induction
Faraday’s Law
r r
dΦ
Vemf ≡ ∫ E ⋅ d l = −
dt
r r
Φ = ∫ B ⋅ da
oppose the “change” of
magnetic flux
x
Φ = ∫ Bldx = Blx
0
B
l
R
I
x
u
dΦ
dx
= Bl
= Blu = −Vemf
dt
dt
Vemf Blu
I=
=
R
R
direction given by Lenz’s Law
Lenz’s Law (1834)
“Back emf” to oppose the “change” of magnetic flux
B
R
I
u
x
Heinrich Friedrich Emil Lenz
(1804-1865) Italy
Example – moving magnet
Example - jumping ring
FB
FB B
I
Al ring
I
R
V
S
e.g. magnet in a copper tube
S
N
Bind
Eddy current - disc brake
Lenz’s Law
metal plate
r
r r
FB = IL × B
To reduce eddy current
transformer
Example – metal detector
Example – guitar pickup
Question
Find the direction of the current in the resistor R shown in Figure at
each the following steps: (a) at the instant the switch is closed, (b)
after the switch has been closed for several minutes, (c) when the
variable resistance r increases, (d) when the circuit containing R
moving to the right, away from the other circuit, and (e) at the instant
the switch is opened.
r
S
R
V
Right, 0, Left, Left, Left
Moving Conductor
FB
+
r rr
FB=qu×B
B
saturate when
u
FB
−
induced
emf
B
R
r r
FE + FB = 0 separate the charges
r
r r
qE + qu × B = 0 induced E
r
r r
E = −u × B similar to Hall Effect
V
= uB
l
V = uBl
induced current
+
I = |V/R| = B u/R
l
I
u
-
direction agrees w/ Lenz’s Law
emf is due to the FB,
no E field build up.
Example – rotating loop
θ = ωt
uniform B
d r r
Vemf = − ∫ B ⋅ da
dt
d
= − ( BA cos ωt ) = ωBA sin ωt
dt
AC generator
dΦ
Vemf = − N
dt
= NωBA sin ωt
DC generator
Electromotor
Project 1: Make one in two weeks.
To turn faster, should we
1.
2.
3.
4.
use thicker wire?
use more turns?
make bigger loop?
use stronger magnet?
Answer… (not counting friction)
dΦ
= NωBA sin ωt
dt
NωBA
| I |=
R
r
r r
FB = I l × B
r r
r r r
τ = r × FB = m × B torque
r
magnetic moment
m = NIAn̂
r r
I = moment of inertia
τ = Iα
Vemf = − N
l
resistivity
A'
m = ρV = ρA' l mass density
R =ρ
1. use thicker wire?
same.
e.g. half R, double mass, same α
2. use more turns?
same.
double N, double R, same I,
double mm, but double inertia, same α.
3. use bigger loop? (same N)
same.
bigger A, more Φ
somewhat higher R, but still more I ~A/R~ r
mm~A2/R~r3, inertia~r3, same α
4. use stronger magnet?
YES
more I, m, τ, α
Another important application
– telephony / loudspeaker
(?? before him)
Alexander Graham Bell
(1847 –1922) UK
(practical improvement)
Thomas Alva Edison
(1847 –1931) USA
Ideal Transformer (AC)
dΦ
dt
dΦ
V2 = − N 2
dt
V1 N1
=
Voltage transformer
V2 N 2
Φ is confined in the core (µ = ∞) V1 = − N1
P1 = P2
RL
V1I1 = V2 I 2
V1 I 2 N1
= =
V2 I1 N 2
2
2
V1 V2  N1   N1 

 = 
 R L
R in =
=
I1
I2  N 2   N 2 
Impedance transformer
Self-Inductance
back emf causes I lags V
NΦ
L≡
I
LI = NΦ
dΦ
dI
V = −N
= −L
dt
dt
Ideal Solenoid
ideal:
• large N tightly wound
• no end effect
• uniform internal B
• zero external B in the vicinity
Ampere’s Law:
µ o NI
= µ o nI
B=
l
r r
Φ = ∫ B ⋅ da =µ o nIA
NΦ Nµ o nIA
L=
=
= µ o n 2lA
I
I
LC Resonator (Lenz’s Law, later resonator…)
Transformer -
Primary / Secondary coils
same cross-section?
V1 V2
∂Φ
=
=−
N1 N 2
∂t
N 2 N1 M12
=
=
I1
I 2 Φ12
same equations
different cross-section?
Electrodynamics
 r r Qt 
 ∫ E ⋅ da = ε

o


dΦ 
 r r
∫ E ⋅ d l = −


dt 
 r r

 ∫ B ⋅ da = 0

 r r

 ∫ B ⋅ d l = µ o I t 
Gauss’s Law
Faraday’s Law
No magnetic charge
Ampere’s Law
Exercise
Rectangular loop of 20 turns is placed at 1 m
away from a long current-carrying wire as
shown.
I2
N = 20
I1(t) = 2 cos(60t) (A)
Total resistance of the wire in the loop is 4 Ω.
5m
Find I2.
I1(t)
3m
1m

Similar documents