Download Sources of the Magnetic Field

Induction and Inductance
Chapter 30
Magnetic Flux
   B  dA Like Gauss
Insert Magnet into Coil
Remove Coil from Field
Region
From The Demo ..
First experiment
Second experiment
Faraday’s Experiments
?
?
That’s Strange …..
These two coils are perpendicular to each other
Definition of TOTAL
ELECTRIC FLUX through a
surface:
   d
surface
Total Flux of the Electric
Field LEAVING a surface is
   E  n outdA
Magnetic Flux: 
THINK OF
MAGNETIC FLUX
as the
“AMOUNT of Magnetism”
passing through a surface.
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
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

weber

1 weber = 1 T-m2
We finally stated
FARADAY’s LAW
d
emf  
dt
From the equation
Lentz
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 Equations
(chapter 32 .. Just a Preview!)
Gauss
Faraday
Another View Of That hopeless
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!
Example of 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.
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!) …
#1 CHAPTER 30
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 …
 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.
#21 Figure 30-50 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. 29-27.) In the smaller loop, find (b) the induced emf and
(c) the direction of the induced current.
v
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 q
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 3/ 2
2R x
For x  R (Far Away as prescribed )

2
r 2  0iR 2
3
2x
d 3r 2  0iR 2
emf  

v
4
dt
2x
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 reestablish 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”
 0
di
L  iR  0
dt
i

R
e
t / 
VR=iR
~current
Max Current Rate of
increase = max emf
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??
For an answer
Return to the Big C
E=0A/d


+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
  L  iR
dt
i :
di 2
i  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
Coil
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
i
N
i
L
1
1 N 1
W  Li 2  i 2
 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
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.


10. A uniform magnetic field B increases in magnitude with
time t as given by Fig. 30-43b, where the vertical axis
scale is set by Bs=9 mT and the horizontal scale is set by
ts=3 s . A circular conducting loop of area A= 8x10-4 m2 lies
in the field, in the plane of the page. The amount of charge
q passing point A on the loop is given in Fig. 30-43c as a
function of t, with the vertical axis scale set by qs=3 mC
and the horizontal axis scale again set by ts=3 s. What is
the loop's resistance?
29. If 50.0 cm of copper wire (diameter=1mm ) is
formed into a circular loop and placed
perpendicular to a uniform magnetic field that is
increasing at the constant rate of 10.0 mT/s, at
what rate is thermal energy generated in the
loop?