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INDUCTION
Spring ‘09
DIS HERE WEEK
 We
begin the study of magnetic induction
 There will be a quiz on Friday
 There is a new WebAssign.
 Exam #3 is Wednesday4/8

It will include the material covered through
April 6th.
 The

end is in sight …
Check the website for the Final Exam
Schedule
AMPERE’S LAW
B

d
s


i
0
enclosed

B

d
s


(
i

i
)
0
1
2

USE THE RIGHT HAND RULE IN THESE CALCULATIONS
LAST TIME: A SIMPLE EXAMPLE
FIELD AROUND A LONG STRAIGHT WIRE
 B  ds   i
0 enclosed
B  2r   0i
 0i
B
2r
THE FIGURE BELOW SHOWS A CROSS SECTION OF AN INFINITE CONDUCTING SHEET
CARRYING A CURRENT PER UNIT X-LENGTH OF L; THE CURRENT EMERGES
PERPENDICULARLY OUT OF THE PAGE. (A) USE THE BIOT–SAVART LAW AND
SYMMETRY TO SHOW THAT FOR ALL POINTS P ABOVE THE SHEET, AND ALL POINTS
P´ BELOW IT, THE MAGNETIC FIELD B IS PARALLEL TO THE SHEET AND DIRECTED AS
SHOWN. (B) USE AMPERE'S LAW TO FIND B AT ALL POINTS P AND P´.
FIRST PART
Vertical Components
Cancel
APPLY AMPERE
TOL CIRCUIT
B
Infinite Extent
B
  current per unit length
Current inside the loop is therefore :
i  L
THE “MATH”
B
Infinite Extent
B
 B  ds   i
0 enclosed
BL  BL   0 L
B
 0
2
Infinite Sheet of Charge

E
2 0
Infinite sheet of current
B
 0
2
COIL OR SOLENOID
Valve Application
Switches
THE MAGNETIC FIELD
A PHYSICAL SOLENOID/COIL MODEL
INSIDE THE SOLENOID
For an “INFINITE” (long) solenoid the previous
problem and SUPERPOSITION suggests that
the field OUTSIDE this solenoid is small!
MORE ON LONG SOLENOID
Field is ZERO far from coil!
Field looks “UNIFORM”
Field is ZERO far from coil
THE REAL THING…..
Finite Length
Weak Field
Stronger - Leakage
ANOTHER WAY
Far away
Ampere :
 B  ds   i
0 enclosed
0h  Bh   0 nih
B   0 ni
n=number of turns per unit length. nh=total number of turns.
THE LENGTH OF A SOLENOID IS DOUBLED BUT
THE NUMBER OF TURNS REMAINS THE SAME.
A.
B.
C.
D.
E.
The magnetic field is doubled.
The magnetic field is cut to a quarter.
The magnetic field does not change.
The magnetic field is quadrupled.
The magnetic field is cut in half.
THE NUMBER OF TURNS IS DOUBLED
BUT THE LENGTH REMAINS THE SAME.
A.
B.
C.
D.
E.
The magnetic field is quadrupled.
The magnetic field is cut to a quarter.
The magnetic field is doubled.
The magnetic field does not change.
The magnetic field is cut in half.
B  0 ni
APPLICATION
Creation of Uniform Magnetic Field Region
 Minimal field outside


except at the ends!
TWO COILS
“REAL” HELMHOLTZ COILS
Used for
experiments.
Can be aligned to
cancel
out the Earth’s
magnetic
field for critical
measurements.
THE TOROID
Slightly less
dense than
inner portion
THE TOROID
Ampere again. We need only worry
about the INNER coil contained in
the path of integratio n :
 B  ds  B  2r   Ni (N  total # turns)
0
so
0 Ni
B
2r
MAGNETIC FLUX
NEW CONCEPT – WELL SORTA
NEW
MAGNETIC FLUX
 B  dA  
M
 M   B  dA Like Gauss - Open Surface
MAGNETIC FLUX IS A
A.
B.
C.
Vector
Scalar
Tensor
MAGNETIC FLUX
 M   B  dA Like Gauss - Open Surface
For a CLOSED Surface we might expect this to be equal to
some constant times the
enclosed poles … but there ain’t no such thing!
B

d
A

0

CLOSED SURFACE
A PUZZLEMENT ..

closed
path
B  ds   0ienclosed
Let’s apply this to the gap of a capacitor.
CONSIDER THE POOR LITTLE CAPACITOR…
i
i
CHARGING OR DISCHARGING …. HOW CAN CURRENT
FLOW THROUGH THE GAP
In a FIELD description??
THROUGH WHICH SURFACE DO WE
MEASURE THE CURRENT FOR
AMPERE’S LAW?
I=0
IN THE GAP…
DISPLACEMENT CURRENT
The ELECTRIC FLUX through S2
 E  EA 
q
0
d E 1 dq

dt
 0 dt
Let
dq
 I d (in gap)
dt
Displaceme nt Current
dq
d E
Id 
 0
dt
dt
SO FAR ..
We found that currents create magnetic fields.
 Stationary charges do not.
 Static magnetic fields do not create currents.
 What about CHANGING magnetic fields??

FROM THE DEMO ..
FARADAY’S EXPERIMENTS
?
?
INSERT MAGNET INTO COIL
REMOVE COIL FROM FIELD REGION
Summary
THAT’S STRANGE …..
These two coils are perpendicular to each other
REMEMBER THE DEFINITION OF TOTAL
ELECTRIC FLUX THROUGH A CLOSED
SURFACE:
 E  
surface
d E
Total Flux of the Electric
Field LEAVING a surface is
 E   E  n outdA
MAGNETIC FLUX - REMINDER
Applies to an OPEN SURFACE only.
 “Quantity” of magnetism that goes through a surface.
 A Scalar

 B   B  dA
surface
CONSIDER A LOOP
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 Magnetic
field passing
through the loop is
CHANGING.
 FLUX is changing.
 There must be an emf
developed around the
loop.

A current develops (as we
saw in demo)
 Work
has to be done to
move a charge
completely around the
loop.
FARADAY’S LAW (MICHAEL FARADAY)
 Again,
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for a current to
flow around the circuit,
there must be an emf.
 (An emf is a voltage)
 The voltage is found to
increase as the rate of
change of flux
increases.
FARADAY’S LAW (MICHAEL FARADAY)
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Faraday' s Law
d
emf  
dt
We will get to the minus sign in a short time.
FARADAY’S LAW (THE MINUS SIGN)
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Using the right hand rule, we
would expect the direction
of the current to be in the
direction of the arrow shown.
FARADAY’S LAW (MORE ON THE
MINUS SIGN)
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The minus sign means
that the current goes the
other way.
This current will produce
a magnetic field that
would be coming OUT of
the page.
The Induced Current therefore creates a magnetic field that OPPOSES the attempt to
INCREASE the magnetic field! This is referred to as Lenz’s Law.
HOW MUCH WORK?
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Work/Unit Charge 
d
W / q  V   E  ds  
dt
A magnetic field and an electric field are
intimately connected.)
MAGNETIC FLUX
 B   B  dA
This is an integral over an OPEN Surface.
 Magnetic Flux is a Scalar


The UNIT of FLUX is the

1 weber = 1 T-m2
weber
WE FINALLY
STATED
FARADAY’s LAW
d
emf  V   E  ds  
dt
FROM THE EQUATION
Lentz
d
emf  V   E  ds  
dt
 B   B  dA
FLUX CAN CHANGE
 B   B  dA




If B changes
If the AREA of the loop changes
Changes cause emf s and currents and consequently
there are connections between E and B fields
These are expressed in Maxwells Equations
MAXWELL’S FOUR EQUATIONS
Ampere’s Law
Gauss

Faraday
No Monopoles
closed
surface
B  dA  0
 B  dA  0
ANOTHER VIEW OF THAT DAMNED
MINUS SIGN AGAIN …..SUPPOSE
THAT B BEGINS TO INCREASE ITS
MAGNITUDE INTO THE PAGE
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




The Flux into the page
begins to increase.
An emf is induced around a
loop
A current will flow
That current will create a
new magnetic field.
THAT new field will
change the magnetic flux.
THE STRANGE WORLD OF DR. LENTZ
LENZ’S LAW
Induced Magnetic Fields always FIGHT to stop
what you are trying to do!
i.e... Murphy’s Law for Magnets
EXAMPLE
OF
NASTY LENZ
The induced magnetic field opposes the
field that does the inducing!
DON’T HURT YOURSELF!
The current i induced in the loop has the direction
such that the current’s magnetic field Bi opposes the
change in the magnetic field B inducing the current.
Let’s do the
Lentz Warp
again !
AGAIN: LENZ’S LAW
An induced current has a direction
such that the magnetic field due to
the current opposes the change in
the magnetic flux that induces the
current. (The result of the
negative sign!) …
OR
The toast will always fall buttered side down!
AN EXAMPLE
The field in the diagram
creates a flux given by
B=6t2+7t in milliWebers
and t is in seconds.
(a) What is the emf when
t=2 seconds?
(b) What is the direction
of the current in the
resistor R?
THIS IS
AN EASY ONE
…
 B  6t  7t
2
d
emf 
 12t  7
dt
at t  2 seconds
emf  24  7  31mV
Direction? B is out of the screen and increasing.
Current will produce a field INTO the paper
(LENZ). Therefore current goes clockwise and R
to left in the resistor.
The diagram shows two parallel loops of wire having a common
axis. The smaller loop (radius r) is above the larger loop (radius
R) by a distance x >> R. Consequently, the magnetic field due
to the current i in the larger loop is nearly constant throughout
the smaller loop. Suppose that x is increasing at the constant rate
of dx/dt = v. (a) Determine the magnetic flux through the area
bounded by the smaller loop as a function of x. In the smaller
loop, find (b) the induced emf and (c) the direction of the
induced current.
v
DOING IT
B is assumed to be
constant through the
center of the small
loop and caused by
the large one.
q
THE CALCULATION OF BZ
dBz  dB cos q  cos q
cos q 
R
R
 0 ids
4 R 2  x 2 

1/ 2
 x2
q
 0 ids
R
dBz 
4 R 2  x 2 R 2  x 2
ds  Rd 
2

Bz 


 0iR 2
2 R x
2

2 3/ 2

1/ 2
MORE WORK
In the small loop:
  Bz A  r 2 Bz 
r 2 0iR 2
2R  x 
For x  R (Far Away as prescribed )

2
r 2 0iR 2
3
2x
d 3r 2 0iR 2
V

emf  
4
2x
dt
2 3/ 2
dx/dt=v
WHICH WAY IS CURRENT IN SMALL
LOOP EXPECTED TO FLOW??
B
q
WHAT HAPPENS HERE?
 Begin
to move handle
as shown.
 Flux through the loop
decreases.
 Current is induced
which opposed this
decrease – current
tries to re-establish
the B field.
MOVING THE BAR
Flux  BA  BLx
Dropping the minus sign...
d
dx
emf 
 BL
 BLv
dt
dt
emf BLv
i

R
R
MOVING THE BAR TAKES WORK
F  BiL  BL 
BLv
R
or
v
B 2 L2 v
F
R
dW d
POWER 
 Fx   Fv
dt
dt
B 2 L2 v
P
v
R
B 2 L2 v 2
P
R
WHAT ABOUT A SOLID LOOP??
Energy is LOST
BRAKING SYSTEM
METAL
Pull
Back to Circuits for a bit ….
DEFINITION
Current in loop produces a magnetic field
in the coil and consequently a magnetic flux.
If we attempt to change the current, an emf
will be induced in the loops which will tend to
oppose the change in current.
This this acts like a “resistor” for changes in current!
REMEMBER FARADAY’S LAW
d
emf  V   E  ds  
dt
Lentz
LOOK AT THE FOLLOWING CIRCUIT:
Switch is open
 NO current flows in the circuit.
 All is at peace!

CLOSE THE
CIRCUIT…
After the circuit has been close for a long time, the
current settles down.
 Since the current is constant, the flux through the coil
is constant and there is no
.

Emf

Current is simply E/R (Ohm’s Law)
CLOSE THE
CIRCUIT…
 When
switch is first closed, current begins to flow
rapidly.
 The flux through the inductor changes rapidly.
 An emf is created in the coil that opposes the
increase in current.
 The net potential difference across the resistor is
the battery emf opposed by the emf of the coil.
CLOSE THE
CIRCUIT…
d
emf  
dt
Ebattery  V (notation)
d
 V  iR 
0
dt
MOVING RIGHT ALONG
Ebattery  V…
(notation)
d
 V  iR 
0
dt
The flux is proportion al to the current
as well as to the number of turns, N.
For a solonoid,
  i  Li  N B
d
di
L
dt
dt
di
 V  iR  L  0
dt
DEFINITION OF INDUCTANCE L
N B
L
i
UNIT of Inductance = 1 henry = 1 T- m2/A
B is the flux near the center of one of the coils
making the inductor
CONSIDER A SOLENOID
l
 B  ds   i
0 enclosed
 Bl   0 nli
or
n turns per unit length
B   0 ni
SO….
N B nlBA nl 0 niA
L


i
i
i
or
L   0 n 2 Al
or
inductance
2
L/l 
 n A
unit length
Depends only on geometry just like C and
is independent of current.
INDUCTIVE CIRCUIT

i




Switch to “a”.
Inductor seems like a
short so current rises
quickly.
Field increases in L and
reverse emf is generated.
Eventually, i maxes out and
back emf ceases.
Steady State Current
after this.
THE BIG INDUCTION
 As
we begin to increase the current in the coil
 The current in the first coil produces a magnetic
field in the second coil
 Which tries to create a current which will reduce
the field it is experiences
 And so resists the increase in current.
BACK TO
THE REAL WORLD…
Switch to “a”
i
sum of voltage drops  0 :
di
 E  iR  L  0
dt
same form as the
capacitor equation
q
dq
E R
0
C
dt
SOLUTION
E
 Rt / L
i  (1  e
)
R
time constant

L

R
SWITCH POSITION “B”
E0
di
L  iR  0
dt
E t / 
i e
R
VR=iR
~current
Max Current Rate of
increase = max emf
E
(1  e Rt / L )
R
L
  (time constant)
R
i
IMPORTANT QUESTION
Switch closes.
 No emf
 Current flows for a while
 It flows through R
 Energy is conserved (i2R)

WHERE DOES THE ENERGY COME FROM??
FOR AN ANSWER
RETURN TO THE BIG C
 We
E=0A/d
+dq
+q
-q
move a charge dq
from the (-) plate to the
(+) one.
 The (-) plate becomes
more (-)
 The (+) plate becomes
more (+).
 dW=Fd=dq x E x d
THE CALC

q
dW  (dq) Ed  (dq) d  (dq)
d
0
0 A
d
d q2
W
qdq 

0 A
0 A 2
or
1  2 Ad 1   2 
1
2


W
(A) 
  0  2  Ad   0 E Ad
2 0 A
2 0
2  0 
2
d
2
energy
1
2
u
 0 E
unit volum e 2
The energy is in
the FIELD !!!
WHAT ABOUT POWER??
di
E  L  iR
dt
i :
di 2
iE  Li  i R
dt
power
to
circuit
Must be dWL/dt
power
dissipated
by resistor
SO
dWL
di
 Li
dt
dt
1 2
WL  L  idi  Li
2
1
2
WC  CV
2
Energy
stored
in the
Capacitor
WHERE IS
THE ENERGY??
l
 B  ds   i
0 enclosed
0l  Bl   0 nil
B   0 ni
or
B
 0 Ni
l
  BA 
 0 Ni
l
A
REMEMBER
THE
INDUCTOR??
N
L
i
N  Number of turns in inductor
i  current.
Φ  Magnetic flux throu gh one turn.
SO …
N
L
i
N
i
L
1 2 1 2 N 1
W  Li  i
 N i
2
2
i
2
 0 NiA

l
1   0 NiA 
1
2
2 2 A
W  Ni
0 N i

2  l  2 0
l
1
A
W
 N i
2 0
l
2
0
2 2
From before :
B
 0 Ni
l
1
A
1 2
W
Bl

B V (volume)
2 0
l 2 0
2 2
or
W
1 2
u

B
V 2 0
ENERGY IN THE
FIELD TOO!
IMPORTANT CONCLUSION
A region of space that contains either a magnetic
or an electric field contains electromagnetic
energy.
 The energy density of either is proportional to the
square of the field strength.
