Transcript Short Version : 27. Electromagnetic Induction

Short Version :
27. Electromagnetic Induction
27.1. Induced Currents
4 results from Faraday / Henry (1831)
v = 0, I = 0
v > 0, I > 0
1. Current induced in coil by
moving magnet bar.
v >> 0, I >> 0
v < 0, I < 0
2. Moving the coil instead of the
magnet gives the same result.
3. An induced current also results when a
current-carrying circuit replaces the magnet.
4. A current is also induced when the
current in an adjacent circuit changes.
 changing B induces currents (electromagnetic induction)
27.2. Faraday’s Law
•
Magnetic flux
•
Flux & Induced EMF
Magnetic flux
Magnetic flux:
 B   B  dA
Reminder:
 B  dA  0
For a uniform B on a flat surface:
B  B  A  BAcos
Move magnet right
 more lines thru loop
Example 27.1. Solenoid
A solenoid of circular cross section has radius R,
consists of n turns per unit length, and carries current I.
Find the magnetic flux through each turn of the solenoid.
I
B
B out of plane
2
  B A  0 n I  R
Example 27.2. Nonuniform Field
A long, straight wire carries current I.
A rectangular wire loop of dimensions l by w lies in a plane containing
the wire, with its closest edge a distance a from the wire, and its
dimension l parallel to the wire.
Find the magnetic flux through the loop.
 B   B  dA

Area element for integration

aw
a
0 I l
aw
ln
2
a
0 I
 Il
l dr  0
2
2 r

aw
a
dr
r
Flux & Induced EMF
Faraday’s law of induction:
The induced emf in a circuit is proportional to the rate of change of
magnetic flux through any surface bounded by that circuit.

C
Edl  E  
d B
dt

d
B  dA

S
dt
Note: dB/dt can be due to
•
changing B caused by
• relative motion between circuit & magnet,
• changing current in adjacent circuit,
•
•
changing area of circuit,
C
changing
orientation between B & circuit.
C
C is CCW about S.
Example 27.3. Changing B
A wire loop of radius 10 cm has resistance 2.0 .
The plane of the loop is perpendicular to a uniform B that’s increasing at 0.10 T/s.
Find the magnitude of the induced current in the loop.
 B  BA
 B  r2
d B d
  B  r2 
dt
dt
S
C
E 
I
  r2
dB
dt
2
3
d B
   0.1 m   0.10 T / s   3.14 10 V
dt
E
3.14  103 V
I

 1.57  103 A
R
2.0 
CCW
Example 27.4. Changing Area
Two parallel conducting rails a distance l apart are connected at one end by a resistance R.
A conducting bar completes the circuit, joining the two rails electrically but free to slide along.
The whole circuit is perpendicular to a uniform B, as show in figure.
Find the current when the bar is pulled to the right with constant speed v.
Let x = 0 be at the left end of rail.
B  B A
C
S
E 
I
I
x
Bl x
d B
 B l v
dt
Blv
E

R
R
CCW
27.3. Induction & Energy
m
I
Direction of emf is to oppose
magnet’s motion.
RH rule: thumb // m.
Loop ~ magnet with N to left.
Magnet moving right
I
m
RH rule: thumb // m.
Loop ~ magnet with S to left.
Magnet moving left
Lenz’s law :
Direction of induced emf is
such that B created by the
induced current opposes the
changes in  that created
the current.
Motional EMF & Len’s Law
Motional emf: induced emf due to
motion of conductor in B.
Square loop of sides L & resistance R pulled with
constant speed v out of uniform B.
Force on e:
F   e v  B
Force on current carrying wire:
Fmag  I L  B
Fmag , net  Fapplied
downward force
 upward I
(CW)
B  B L x
S
C
x
I
d B
BLv
dt
E  B L v
<0
I
E2
 B L v

P  IE 
R
R
 Fv
d B
0
dt
dx
 v
dt
E
BLv

R
R
CW
2
F v I LBv
Work done is used to heat up circuit ( E conservation ).
 B L v

R
2
Application. Electric Generators
World electricity generation ~ 2TW.
Rotating loop changes
 & induces emf.
Rotating slip rings.
Electric
load
Sinusoidal
AC output
Stationary
brushes
Rotating
conducting
loop
Work required due
to Lenz’s law.
Hand-cranked generator ~ 100W
Example 27.5. Designing a Generator
An electric generator consists of a 100-turn circular coil 50 cm in diameter.
It’s rotated at f = 60 rev/s to produce standard 60 Hz alternating current.
Find B needed for a peak output voltage of 170 V,
which is the actual peak in standard 120 V household wiring.
2
1 turn
Loop
rotation
d 
 BA cos  B    cos  2 f t 
2
2
d B
 N B 
E 
dt
d  d
cos  2 f t 
 
2
dt
 
2
d 
 2 N B  2 f   sin  2 f t 
2
E peak
d 
 2N B  f  
2
2
2
 0.5 m 
170 V  2 100 B   60 / s  

 2 
2
B  23 103 T
2
EM induction is basis of magnetic recording ( audio, video, computer disks, …).
Iron
Coil
Card motion
Modern hard disks:
Giant magnetoresistance.
Magnetic strip
Information stored in
magnetization pattern
Swiping a credit card.
Patterns of magnetization on the strip induce currents in the coil.
Eddy Currents
Eddy current: current in solid conductor induced by changing .
Usage: non-frictional brakes for rotating saw blades, train wheels, …
Application: Metal Detectors
Nothing between coils
Induced
Current
Current
detector
AC
Strong I
Transmitter
coil
Receiver
coil
Weak I :
alarm.
Metal between coils
Closed & Open Circuits
B of induced I points out of page
Setting n // Bin
RH rule gives CCW I
 C is CCW &  < 0
Bin  

Bin 
C
+
_
B
E>0
d  /d t < 0
E>0
 I is CCW
Inductance:
27.4. Inductance
Mutual Inductance:
Changing current in one circuit
induces an emf in the other.
Large inductance:
two coils are wound on same iron
core.
Applications:
Transformers, ignition coil, battery chargers, …
Self-Inductance:
Changing current induces emf in own circuit
& opposes further changes.
Applications:
Inductors  frequency generator / detector …
B  Lself I
[ L ] = T  m2 / A = Henry
Example 27.6. Solenoid
A long solenoid of cross section area A and length l has n turns per unit length.
Find its self-inductance.
B of solenoid:
B  0 n I
1turn  BA  0 n I A
  n l 1turn  0 n2l I A
L

 0 n 2 l A
I
+E direction = V  along I.
B  L I
E 
d I /d t < 0
d B
dI
 L
dt
dt
back emf
Rapid switching of inductive devices
can destroy delicate electronic
devices.
Inductor
 (lower voltage)
 (lower voltage)
dI
0
dt
E
E
dI
E  L
0
dt
+ (higher voltage)
+ (higher voltage)
Example 27.7. Dangerous Inductor
A 5.0-A current is flowing in a 2.0-H inductor.
The current is then reduced steadily to zero over 1.0 ms.
Find the magnitude & direction of the inductor emf during this time.
+ here ( E > 0 ) helps keep I flowing
E  L
dI
0
dt
5.0 A 

E    2.0 H   
3 
 1.0  10 s 
 10000V
Inductors in Circuits
Current through inductor can’t change instantaneously.
Switch open: I = 0
Switch just closed :
I = 0, dI/dt  0; | EL | = E0
L ~ open circuit.
Long after switch closing:
I  0, dI/dt = 0; EL = 0;
L ~ wire.
E0  I R  EL  0
+
_
I   V = IR 
But rate is 
E0  I R  L
dI
0
dt
dI
d 2I
R
L 2 0
dt
dt
EL < 0 ; | EL | 
R
dE
EL  L  0
L
dt
EL  E0 e R t / L
I
EL t  0  E0
1
E0  EL   E0 1  e  R t / L 
R
R
Inductive time constant = L / R
c.f. capacitive time constant = RC
Example 27.8. Firing Up a Electromagnet
A large electromagnet used for lifting scrap iron has self-inductance L = 56 H.
It’s connected to a constant 440-V power source;
the total resistance of the circuit is 2.8 .
Find the time it takes for the current to reach 75% of its final value.
I
E0
R t / L
1  e  R t / L   I  1  e


R
  2.8   t 
0.75  1  exp  
56 H 


t  20 ln 0.25  28 s
Switch at B, battery’s shorted out.
I  exponentially.
Switch at A, I .
E0  I R  L
dI
0
dt
0
E0
I
1  e R t / L 

E
R
 0
R
I R  L
t 0
t
dI
0
dt
I  I0 e
Short times: IL can’t change instantaneously.
Long times: EL = 0 ; inductor  wire.
R t / L
 I0
t 0
0
t 
Making the Connection
Verify that the current in just after the switch is reopened has the value indicated
Immediately after the switch is closed:
L ~ open circuit.
Long time after the switch is closed:
Immediately after 2nd switch
opening :
L ~ short circuit.
I in L ~ continues.
I
E0
R1  R2
I R2  0
I R2  
E0
R1
27.5. Magnetic Energy
Any B contains energy.
This eruption of a huge prominence from the sun’s
surface releases energy stored in magnetic fields.
Magnetic Energy in an Inductor
RL circuit:
E0  I R  EL  0
I E0  I 2 R  L I
Power from
battery
Power
dissipated
I E0  I 2 R  I EL  0

dI
0
dt
PL  L I
Power taken
by inductor
dI
dt
Energy stored in inductor:
U   P dt   L I
dI
dt
dt

1
L I2
2
U  I  0  0
Example 27.10. MRI Disaster
Superconducting electromagnets like solenoids in MRI scanners store a lot of magnetic energy.
Loss of coolant is dangerous since current quickly decays due to resistance.
A particular MRI solenoid carries 2.4 kA and has a 0.53 H inductance.
When it loses superconductivity, its resistance goes abruptly to 31 m. Find
(a)the stored magnetic energy, and
(b) the rate of energy release at the instance the superconductivity is lost.
U
2
1
1
L I 2   0.53 H   2.4  103 A  1.5  106 J
2
2
3
3
P  I 2 R   2.4  10 A  31  10    1.8 105 W
2
In practice, Cu / Ag are incorporated into the superconducting wires to reduce R.
Magnetic Energy Density
Solenoid with length l & cross-section area A :
1
1
U  L I 2  0 n 2 A l I 2
2
2
Magnetic Energy Density :
uB 
c.f. electric energy density :
uE 

1
2 0
1
2 0
B2
1
0 E 2
2
L  0 n2 A l
B2 A l
B  0 n I
(Eg. 27.6)
27.6. Induced Electric Fields
EMF acts to separate charges:
Battery:
Motional emf:
Stationary loop in changing B :
 Edl  E  
Static E begins / ends on charge.
d B
dt
chemical reaction
F = v  B.
induced E

d
B  dA

dt
Induced E forms loop.
Faraday’s law
Example 27.11. Solenoid
A long solenoid has circular cross section of radius R.
The solenoid current is increasing, & as a result so is B in solenoid.
The field strength is given by B = b t, where b is a constant.
Find the induced E outside the solenoid, a distance r from the axis.
Loop for Faraday’s law
Symmetry  E lines are circles.
 E  d l  2 r E

S
S out  C ccw
d B
dt
b R2
E
2r

d
b t  R2 

dt
CCW
 b  R2
Conservative & Nonconservative Electric Fields
E
W against E
For stationary charges (electrostatics) :
 Edl  0
E is conservative
Induced fields (electromagnetics) :
E does W
 Edl  0
E is non-conservative
GOT IT? 27.7
The figure shows three resistors in series surrounding an infinitely long solenoid with a changing
magnetic field;
the resulting induced electric field drives a current counterclockwise, as shown.
Two identical voltmeters are shown connected to the same points A and B.
What does each read?
Explain any apparent contradiction.
Hint: this is a challenging question!
VA  VB = 2IR
VA  VB = IR
E  3R I
Diamagnetism
Classical model of diamagnetism (not quite right)
B = 0: net = 0
B0
This e slows down.
net  0
This e speeds up.
Superconductor is a perfect
diamagnet (Meissner effect).