# INDUCTION

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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  ds  0ienclosed

 B  ds  0 (i1  i2 )

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
 0
B
2
Infinite Sheet of Charge


E
2 0

Infinite sheet of current

0
B
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   i0 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.   The magnetic field is doubled.
B.   The magnetic field is cut to a quarter.
C.   The magnetic field does not change.
D.   The magnetic field is quadrupled.
E.   The magnetic field is cut in half.
THE NUMBER OF TURNS IS DOUBLED
BUT THE LENGTH REMAINS THE SAME.
A.   The magnetic field is quadrupled.
B.   The magnetic field is cut to a quarter.
C.   The magnetic field is doubled.
D.   The magnetic field does not change.
E.   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 integration :

 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.   Vector
B.   Scalar
C.   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  dA  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
q
 E  EA 
0
d E 1 dq

dt    0 dt
Let
dq
 I d (in gap)
dt
Displacement 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 ..

?
?
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             d E
surface

Total Flux of the Electric
Field LEAVING a surfaceis
 E   E  n out dA
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       Magnetic field passing
through the loop is
CHANGING.
xxxxxxxxxxxxxxx    FLUX is changing.
xxxxxxxxxxxxxxx    There must be an emf
xxxxxxxxxxxxxxx     developed around the
xxxxxxxxxxxxxxx     loop.
   A current develops (as we
xxxxxxxxxxxxxxx         saw in demo)
xxxxxxxxxxxxxxx    Work has to be done to
xxxxxxxxxxxxxxx    move a charge
completely around the
loop.
 Again, for a current to
flow around the circuit,
xxxxxxxxxxxxxxx
xxxxxxxxxxxxxxx     there must be an emf.
xxxxxxxxxxxxxxx    (An emf is a voltage)
xxxxxxxxxxxxxxx    The voltage is found to
xxxxxxxxxxxxxxx     increase as the rate of
xxxxxxxxxxxxxxx     change of flux
xxxxxxxxxxxxxxx
increases.

xxxxxxxxxxxxxxx
xxxxxxxxxxxxxxx
xxxxxxxxxxxxxxx
xxxxxxxxxxxxxxx                    d
xxxxxxxxxxxxxxx            emf  
xxxxxxxxxxxxxxx                    dt

xxxxxxxxxxxxxxx   Using the right hand rule, we
xxxxxxxxxxxxxxx   would expect the direction
xxxxxxxxxxxxxxx   of the current to be in the
xxxxxxxxxxxxxxx   direction of the arrow shown.
xxxxxxxxxxxxxxx
xxxxxxxxxxxxxxx
xxxxxxxxxxxxxxx
MINUS SIGN)

The minus sign means
xxxxxxxxxxxxxxx                      that the current goes the
xxxxxxxxxxxxxxx                      other way.
xxxxxxxxxxxxxxx
xxxxxxxxxxxxxxx                      This current will produce
xxxxxxxxxxxxxxx                      a magnetic field that
xxxxxxxxxxxxxxx                      would be coming OUT of
xxxxxxxxxxxxxxx                      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?

xxxxxxxxxxxxxxx      Work/Unit Charge 
xxxxxxxxxxxxxxx
d
xxxxxxxxxxxxxxx     W / q  V   E  ds  
xxxxxxxxxxxxxxx                               dt
xxxxxxxxxxxxxxx
xxxxxxxxxxxxxxx
xxxxxxxxxxxxxxx

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   weber
   1 weber = 1 T-m2
WE FINALLY   STATED

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

   closed
surface
B  dA  0

No Monopoles
 B  dA  0
ANOTHER VIEW OF THAT DAMNED
MINUS SIGN AGAIN …..SUPPOSE
THAT B BEGINS TO INCREASE ITS
MAGNITUDE INTO THE PAGE
xxxxxxxxxxxxxxx      The Flux into the page
begins to increase.
xxxxxxxxxxxxxxx
   An emf is induced around a
xxxxxxxxxxxxxxx       loop
xxxxxxxxxxxxxxx      A current will flow
xxxxxxxxxxxxxxx      That current will create a
xxxxxxxxxxxxxxx       new magnetic field.
xxxxxxxxxxxxxxx      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
 0 ids
dBz  dB cosq  cosq
4 R 2  x 2 
R
cosq 
 x2 
R   2        1/ 2

q        0 ids               R
dBz 
4 R 2  x 2  R 2  x 2 1/ 2
ds  Rd
 0iR 2
Bz 
2R 2  x 2 
3/ 2
MORE WORK

In the small loop:

r 2 0iR 2
  Bz A  r 2 Bz 
2R x   2

2 3/ 2

For x  R (Far Away as prescribed)
r 2 0iR 2
         3                            dx/dt=v
2x
d 3r 2  0iR 2
emf            4
V
dt    2x
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

BLv
F  BiL  BL 
R
or
B 2 L2 v
F
R
v
 Fx  Fv
dW d
POWER 
dt  dt
B 2 L2 v
P          v
R
B 2 L2 v 2
P
R

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!

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 noEmf   .

   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   Ebattery  V (notation)
emf  
dt              d
 V  iR     0
dt
Ebattery  …
MOVING RIGHT ALONG V (notation)
d
 V  iR      0
dt
The flux is proportional 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
L/l               n 2 A
unit length
Depends only on geometry just like C and
is independent of current.
INDUCTIVE CIRCUIT
   Switch to “a”.
i            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.
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
Max Current Rate of
VR=iR      increase = max emf
~current
E
i   (1  e  Rt / L )
R
L
  (time constant)
R
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??
E=0A/d         We  move a charge dq
from the (-) plate to the
(+) one.
 The (-) plate becomes
more (-)
 The (+) plate becomes
+dq             more (+).
+q                   dW=Fd=dq x E x d
-q
THE CALC

          q
dW  (dq) Ed  (dq) d  (dq)      d
0        0 A
d             d q2
W
0 A  qdq   0 A 2
or
d         1  2 Ad 1   2      1
W        (A) 
2
 0 
2 0 A        2 0     2  0      2
energy    1
u             0 E           The energy is in
2

unit volume 2                the FIELD !!!

di
E  L  iR
dt
i :
di 2
iE  Li  i R
power
dt power
to      dissipated
circuit   by resistor
Must be dWL/dt
SO

dWL      di
 Li
dt      dt
1 2
WL  L  idi  Li
2
1               Energy
WC  CV     2
stored
2                in the
Capacitor
WHERE IS   THE ENERGY??

l
 B  ds   i  0 enclosed

0l  Bl   0 nil
B   0 ni
or
 0 Ni
B
l
 0 Ni
  BA                 A
l
REMEMBER   THE   INDUCTOR??

N
L
i
N  Number of turns in inductor
i  current.
Φ  Magnetic flux through 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 A
W  Ni              0 N i
2

2  l  2 0               l
1      A
W       N i 2     2 2

2 0
0
l
From before:
 0 Ni
B
l
1   A   1 2
W      Bl   2 2
B V (volume)
2 0    l 2 0
or
W   1 2                ENERGY IN THE
u       B                 FIELD TOO!
V 2 0
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.

```
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