electric current gives rise to magnetic fields b i can a...
TRANSCRIPT
Chapter 20.1 Induced EMF and magnetic flux
Electric current gives rise to magnetic fields
Can a magnetic field give rise to a current?
The answer is yes as discovered by Michael Faraday.
I
B
Michael Faraday1791-1867
B× Moving magnetic fields induce
currents
×B
Change the strength of B:Induced current
Induced EMF!
vB×
Chapter 20.1 Induced EMF and magnetic flux
Moving current loop induces currents
Fig. 20.2
Chapter 20.1 Induced EMF and magnetic flux
SI units: weber (Wb)
20.2&20.4 Faraday’s law of induction
Michael Faraday1791-1867
Faraday’s law:The induced emf, ɛ, in a closed wire is equal to the time rate of change of the magnetic flux, ΦB.
An induced emf, ɛ, always gives rise to a current whose magnetic field opposes the original change in magnetic flux.
Lenz’s law
Faraday’s Law and Lenz’s Law bothstate that a loop of wire will want its magnetic flux to remain constant.
2. A circular loop with a radius of 0.20 m is placed in a uniform magnetic field B=0.85 T. The normal to the loop makes an angle of 30o with the direction of B. The field increases to 0.95 T what is the change in the magnetic flux through the loop?
B=0.85T
B’=0.95T
30o
Change in flux?
' 'cos ' cosB B A BAθ θ∆Φ = −2'
' 30o
A A Rπθ θ
= =
= =
2
2
2
cos ( ' ) cos30( ' )
(0.2) cos30(0.95 0.85)
1.1 10
B
B
B
A B B R B B
x Wb
θ π
π−
∆Φ = − = −
∆Φ = −
∆Φ =
8. A circular coil with a radius of 20 cm is in a field of 0.2 T with the plane of the coil perpendicular to the field. If the coil is pulled out of the field in 0.30 s find the average emf during this interval
B
cos 0B BAN Nt t
θε ∆Φ −= =
∆ ∆
2 2
2
0.2 (0.2)0.3
8.4 10
B Rtx V
π πε
ε −
= =∆
=
N=cosθ=A=
11
πR2
B=0
20.3 Motional emf
x x x x x
x x x x x
x x x x x
x x x x x
A voltage is produced by a conductor moving ina magnetic field
B into the page
v
Charges in the conductor experience a force upward
F
L
The work done in movinga charge from bottom to top
W FL qvBL= =
F qvB=
The potential difference isWV vBLq
∆ = =
V∆
x x x x x
x x x x x
x x x x x
x x x x x
A voltage is produced by a conductor moving ina magnetic field
B into the page
v
Charges in the conductor experience a force upward
F
L W FL qvBL= =
F qvB=
The potential difference isWV vBLq
∆ = =
V∆
Voltage
velocity
20.3 Motional emf
20.3 Motional emf
x x x x x
x x x x x
x x x x x
x x x x x
B into the page
vFL
V∆
The potential difference can drive a current through a circuitThe emf arises from changing flux due to changing area according to Faraday’s Law
wire
R
I
Bx ABV vLB LBt t t
ε∆Φ∆ ∆∆ = = = = =
∆ ∆ ∆
BLvIR Rε
= =
Changing Magnetic Flux
x x x x x
x x x x x
x x x x x
x x x x x
B into the page
vF
LV∆
18. R= 6.0 Ω and L=1.2 m and B=2.5 T. a) What speed should the bar be moving to generate a current of 0.50A in the resistor? b) How much power is dissipated in R? c) Where does the power come from?
wire
R
I 0.5(6.0)2.5(1.2)
1.0 /
BLvIR RIRvBL
v m s
ε= =
= =
=
a)
b) 2 2(0.5) (6.0)1.5
P I RP W= ==
c) Work is done by theforce moving the bar
Lenz’s LawThe polarity of the induced emf is such that it induces a current whose magnetic field opposes the change in magnetic flux through the loop. i.e. the current flows to maintain the original flux through the loop.
NS
V
B Bin
B increasing in loop
Bin acts to oppose thechange in flux
Current direction thatproduces Bin is as shown (right hand rule)I
+
-
Emf has the polarity shown. ε drives current inexternal circuit.
ε
20.4 Lenz’s law revisited
Now reverse the motion of the magnet
NS
V
B
Bin
B decreasing in loop
Bin acts to oppose thechange in flux
Current direction thatproduces Bin is as shown (right hand rule)I
+
-
Emf has the polarity shown.
ε
The current reverses direction
20.4 Lenz’s law revisited
N S
Lenz’s Law and Reaction Forces
NS
V
B Bin
I
A force is exerted by the magnet on loop to produce the current
A force must be exerted by the current on the magnet to oppose the change
The current flowing in the direction shown induces a magnetic dipole in the current loopthat creates a force in the opposite direction
FmcFcm
20.4 Lenz’s law revisited
B
εThe flux through the loop
cosB BA θΦ =tθ ω=
ω = angular velocity (radians/s)
Normal to the plane
θ
Flux through a rotating loop in a B field20.5 Generators
B
t∆Φ∆
tBΦ
tB
t∆Φ∆
sinB BA tt
ω ω∆Φ= −
∆
cosB BA tωΦ =
BA
-BA
BAω
-BAω
BRelation between ΦB and
proportional to ω
20.5 Generators
The emf generated by a loop of N turns rotating at constantangular velocity ω is
sinNBA tε ω ω=
t
εNBAω
-NBAω
0
BNt
ε ∆Φ= −
∆
20.5 Generators
35. In a model ac generator, a 500 turn rectangular coil8.0 cmx 20 cm rotates at 120 rev/min in a uniform magneticfield of 0.60 T. a) What is the maximum emf induced in the coil?
sinNBA tε ω ω=The maximum value of ε
max
max(120 2 )(500)(0.6)(0.08 0.2) 60
60
NBAxx V
ε ωπε
=
= =
20.5 Generators
Rotational
Work
Alternating Current (AC) generator
20.5 Generators
Direct Current (DC) generator
20.5 Generators
DC motor
ε drives rotation
A generator is motor acting in reverse
I
20.5 Generators
20.6 Self-Inductance• a property of a circuit carrying a current
• a voltage is induced that opposes the change in current
• used to make devices called inductors
Self- inductance of a circuit a reverse emf is produceby the changing current
B
tε ∆Φ= −
∆
Self-inductance of a coil
I
Current increasing
BB increases,
changes magnetic flux inthe coil,
B A Bt t
∆Φ ∆=
∆ ∆
Bt
∆∆
Produces emf in coilB A BN N
t tε ∆Φ ∆= − = −
∆ ∆
The direction of the induced emf opposes the change in current.
ε+ -
20.6 Self-Inductance
A changing current in a coil induces an emf that opposes the change
I I
I increasing
ε +-
I decreasing
ε+ -
induced emfopposes I
induced emfsupports I
20.6 Self-Inductance
Inductance L is a measure of the self-induced emf
I
Current increasing
BNt
ε ∆Φ= −
∆
ILt
ε ∆= −
∆
The self-induced emf is
L is a property of the coil, Units of L , Henry (H) VsA
ε
proportionality constant is L
but B It t
∆Φ ∆∝
∆ ∆
20.6 Self-Inductance
Inductance of a solenoid with N turns and length ℓ, wound around an air core (assume the length is much larger than the diameter).
2
oNL Aµ=l
B oNBA IAµΦ = =l
Bo
N I At t
µ∆Φ ∆=
∆ ∆l2
Bo
N I IN A Lt t t
ε µ∆Φ ∆ ∆= − = − = −
∆ ∆ ∆l
lA
It
∆∆
Bt
∆∆
inductance proportional to N squared x area/length
20.6 Self-Inductance
An air wound solenoid of 100 turns has a length of 10 cm and a diameter of 1 cm. Find the inductance of the coil.
2 2 2
7 2 25
44 10 (100) (0.01) 1.0 10
0.1(4)
o oN N dL A
L x H
µ µ π
π π−−
= =
= =
l l
I
l= 10 cm
d=1 cm
20.6 Self-Inductance
20.7 RL circuits
The inductor prevents the rapid buildup of current
But at long time does not reduce the current,
ILt
ε ∆= −
∆
0It
∆=
∆at t=∞
t
oI I (1 e )−τ= −
LR
τ =Applications of Inductors:
Reduce rapid changes of current in circuits
Produce high voltages in automobile ignition.
20.8 Energy stored in a magnetic fieldEnergy is stored in a magnetic field of an inductor.
I I=Io
B=0 BoB increasing
ε
Work is done against ε to produce the B field.
This produces a change in the PE of the inductor
212LPE LI=
This stored PE can be used to do work