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Magnetic Induction
April 1, 2005
Happenings
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Short Quiz Today
New Topic: Magnetic Induction (Chapter 30)
Quiz NEXT Friday
Exam #3 – April 15th.
Should I do a class on Vectors and how we
use them??? [vote]
From The Demo ..
Faraday’s Experiments
?
?
Insert Magnet into Coil
Remove Coil from Field Region
That’s Strange …..
These two coils are perpendicular to each other
Remember Electric Flux?
Did you really think you were through
with this kind of concept???
We discussed the normal component of
the Electric Field vector
q
En  E n cos(q )  E  n
DEFINITION:
Element of Flux through a surface
DA
E
DF=ENORMAL x DA
(a scalar)
“Element” of Flux of a vector E
leaving a surface
DF  E  dA  E NORMAL  DA
also
DF  E  dA  E  ndA
n is a unit OUTWARD pointing vector.
This flux was LEAVING the closed
surface
q
Definition of TOTAL ELECTRIC
FLUX through a surface:
F   dF
surface
Total Flux of the Electric
Field LEAVING a surface is
F   E  n out dA
There is ANOTHER Kind of FLUX
F
THINK OF
MAGNETIC FLUX
as the
“AMOUNT of Magnetism”
passing through a surface.
Don’t quote me on this!!!
What is this thing called FLUX?
F B   B  dA
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Magnetic Flux is a Scalar
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The UNIT of FLUX is the

1 weber = 1 T-m2
weber
Consider a Loop
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Magnetic field passing
through the loop is
CHANGING.
FLUX is changing.
There is 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)
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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
dF
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 
dF
W / q  DV   E  ds  
dt
A magnetic field and an electric field are
intimately connected.)
So..
FARADAY’s LAW
dF
emf  DV   E  ds  
dt
Flux
F B   B  dA
The MAGNITUDE of the emf induced in a loop
is equal to the negative of the
rate at which the magnetic flux
through the loop changes with time.
Flux Can Change
F B   B  dA
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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 Equations
(Next Course .. Just a Preview!)
Gauss
Faraday
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.
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.
Take a deep breath!!
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
FB=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 …
F B  6t  7t
2
dF
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.
Figure 31-36 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. (Hint: See Eq. 30-29.)
In the smaller loop, find (b) the induced emf and (c) the direction of the induced current.
v
This combines some stuff and is a good
review for the final! Or Not.
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:
F  Bz A  r 2 Bz 
r 2 0iR 2
2R  x 
For x  R (Far Away as prescribed )
F
2
r 2 0iR 2
3
2x
dF 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?
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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...
dF
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!
Definition of Inductance L
NF B
L
i
UNIT of Inductance = 1 henry = 1 T- m2/A
FB is the flux near the center of the coil
Consider a Solenoid
l
 B  ds   i
0 enclosed
 Bl   0 nli
or
n turns per unit length
B   0 ni
So….
NF 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
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i
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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.
Consider a “uniform short” coil
For a single loop of the inductor
dF
emf  
dt
For N loops :
dF
emf( N )   N
dt
where F is the flux across " any section"
and is assumed to be uniform across ANY section.
From before
B   0 ni
so
NF  N ( BA)  N 0 niA
NF
N
L
 N 0 nA  l  0 nA
i
l
L   0 n 2 Al
Depends only on the geometry, just as
A
C  0
d
also depends only on geometry.
THE BIG INDUCTION
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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.
Trying to make
an equation out of this…..
We had
NF B  Li (definitio n of L)
We also know that :
dF total
d
emf  
  ( Li )
dt
dt
or
di
emf   L
dt
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
increase = max emf
E
(1  e Rt / L )
R
L
  (time constant)
R
i
IMPORTANT QUESTION
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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
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E=0A/d
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
+dq
+q
-q

We 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
F  BA 
 0 Ni
l
A
Remember the Inductor??
NF
L
i
N  Number of turns in inductor
i  current.
Φ  Magnetic flux throu gh one turn.
?????????????
So …
NF
i
NF
i
L
1
1 NF 1
W  Li 2  i 2
 NF i
2
2
i
2
 0 NiA
F
l
1   0 NiA 
1
2
2 2 A
W  Ni
0 N i

2  l  2 0
l
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
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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.