Transcript Document

MAGNETIC INDUCTION
MAGNETUIC FLUX:
FARADAY’S LAW, INDUCED EMF:
MOTIONAL EMF:
THE AC GENERATOR:
THE TRANSFORMER:
INDUCTANCE:
STORED ENERGY:
Written by Dr. John Dayton
MAGNETUIC FLUX:
Magnetic flux is defined as the portion per unit area of the
magnetic field penetrating perpendicularly through a surface.
  B cos q  A
q = angle between B and A
A
q
B
The net magnetic flux through a closed surface is zero.
EXAMPLE: A uniform magnetic field, B, exists in space oriented
horizontally to the right. At a certain point is a cube positioned so that
its left face is perpendicular to the magnetic field. Each face of the
cube has an area A. What is the magnetic flux through each face of
the cube and the net magnetic flux through the cube?
The flux through the left side is 1=-BA, A is outward and so is antiparallel to B.
The flux through the right side is 2=+BA, A is parallel to B.
The flux through the other sides is 0, A is perpendicular to B.
The net flux is the sum of the flux through each side, net = 0.
The magnetic flux through any closed surface is always zero.
FARADAY’S LAW, INDUCED EMF:
If a conducting loop or coil has through its encircled area a changing
magnetic flux, then there will be an induced emf in the loop or coil.

E  N
t
N = number of loops
of wire
In general:
 B
A

cos q  A  B cos q 
  sin(q ) BA
t
t
t
In most problems the changing flux will depend on only one term.
q

t
INDUCED ELECTRIC CURRENT:
E
I
R
To determine the direction of the conventional current flow around the
conducting loop first determine the direction of the area vector of the
encircled area. Looking down on the area from above, traversing the
perimeter in a counter-clockwise sense is considered positive and the
area vector would be pointing towards you. If /t is positive with
this area vector, the current flow is in the negative direction, or
clockwise. This is due to the “-” sign in Faraday’s law.
The induced emf is not localized as it is in a battery. It is spread out
around the entire conducting path. The current flows towards
decreasing potential. Every complete circuit it gains a U from the emf
and looses U to the resistance, maintaining energy conservation.
DETERMINING THE DIRECTION OF AN INDUCED ELECTRIC CURRENT
A
1
4
+ direction
conducting loop with enclosed area
B
determine the positive direction around
loop in relation to the area vector
2
A
B decreasing
magnetic field and flux through loop
5
A
3
choose a direction for the area vector
current
If B is decreasing,  is negative and E
is oriented in the positive direction.
If B is increasing  is positive and E is
oriented in the negative direction.
EXAMPLE: A solenoid of length 10cm made with 1500
turns of wire carries a current of 12A that is steadily
decreasing at the rate of 3A/s. Refer to the diagram for the
initial direction of this current. Inside the solenoid is a coils
whose center lies on the axis of the solenoid. The plane of
the coil makes a 30O angle with the axis of the solenoid. The
coil is made with 400 turns of wire, is circular with a radius
of 3cm, and has a net resistance of 0.3 Ohms.
10cm
30o
BS
mc
Calculate the following:
 the induced emf in the coil:
IS
E   N c   t    N c  o N s ls  I s t   rc2 cos 60o   .032V


 the induced current in the coil:  t   Bs t  Ac cos q BA 
 t   o N s ls  I s t   Ac cos q BA 
I c  E R  0.107 A
 the magnetic dipole moment of the coil:
mc  Nc Ic rC2  0.121A  m2
mc
 the torque on the coil when Is is 2A:
Bs  o N s I s ls  .038T
 m
o
B
sin
6
0
 .004 N  m
c s
Ic

EXAMPLE:
Sliding rod. Two parallel, horizontal, frictionless, conducting tracks
are connected together at their left end by a wire with a resistor of resistance R. The
separation between the rods is l. A conducting rod rests on the rod and is free to move
along the rods. A uniform magnetic field, B, fills space and is directed vertically
upward. The rod is pulled to the right with a constant velocity v. A conducting loop is
form on the left with the rod on one side. Calculate the following:
B
I
R
FM
I
FA
v
Rod
x
Changing Area of Loop:
Induced emf in loop
Induced Current:
Power Loss:
Magnetic Force on Rod:
Applied Force on Rod:
Power Applied to Rod:
Answers on next slide.
Tracks
l
EXAMPLE:
Sliding rod. Two parallel, horizontal, frictionless, conducting tracks
are connected together at their left end by a wire with a resistor of resistance R. The
separation between the rods is l. A conducting rod rests on the rod and is free to move
along the rods. A uniform magnetic field, B, fills space and is directed vertically
upward. The rod is pulled to the right with a constant velocity v. A conducting loop is
form on the left with the rod on one side. Calculate the following:
B
I
FA
v
R
FM
I
Tracks
l
Rod
x
Changing Area of Loop:
Induced emf in loop
Induced Current:
Power Loss:
Magnetic Force on Rod:
Applied Force on Rod:
Power Applied to Rod:
A  lx (see diagram)
E   t  BA t  Bl x t  Blv
I  Blv R
(see diagram)
PR  I 2 R  B 2l 2v 2 R
FM  IlB sin q   B2l 2v R
(see diagram)
FA  B 2l 2v R to maintain constant v (see diagram)
PA  FAv  B 2l 2v 2 R  PR
MOTIONAL EMF:
A conducting rod is moved through a magnetic field. Mobile electrons
experience a force FB that pulls them to one side of the rod. Thus one
end of the rod becomes electrically negative and the other becomes
electrically positive. An electric field builds between the ends of the
rod. The electric field exerts a force FE on the electrons. Eventually FE
equals FB and no further charge moves.
B
FE  FB
eE  evB
E
e  evB
l
E  Bvl
l
__
-
FE
+
rod
See next slide for more details.
-
FB
electron
E
v
MOTIONAL EMF:
B Uniform Magnetic Field
E
I
FB
+
FE
v
Mobile Electrons
Moving Rod
Click to see each of the following:
Effective current of electrons
Magnetic force on electron
Charge separation in rod
Orientation of electric field
Electric force on electrons
EXAMPLE:
A metal bar of length 75cm is rolling down
an incline at a steady speed of 3m/s. A uniform magnetic field of
0.5G exists in space perpendicular to the surface of the incline.
What is the potential difference between the ends of the rod?
E  Bvl  .5  104 T   3 ms  .75m   1.125  104V
THE AC GENERATOR:
A
q
B
Rotating Loop
The induced emf causes an alternating

E  N
 NBA sin q  current in the loop. The interaction
t
between this current and the magnetic
Emax  Eo  NBA
field is a torque that opposes the
rotation of the loop. Thus an external
E  Eo sin q   Eo sin t 
source of energy will be required to
keep the loop in rotation.
EXAMPLE:
A coil is made to rotate inside a
solenoid. The solenoid consists of 2500 turns of wire,
has a length of 30cm, and carries a current of 6A.
The coil is made with 500 turns of wire, is circular
with a radius of 2cm, and has a net resistance of .75
Ohms. If the coil is rotating at 500 revolutions per
minute, what is the maximum current in the coil?
Eo
Io 
Need E0
R
IS
Eo  N C BS A C 
Need BS , AC and 
BS 
o N S I S
lS
4  10


7 Tm
A
  2500 6 A  .0628T
.3m
AC   r   .02m   .00126m
2
C
2

2
 500rev 
1
  2 f  2 

52.360
s

60
s


2.072V
Io 
 2.763 A
.75
Eo   500 .0628T  .00126m 2  52.360 s 1   2.072V
THE TRANSFORMER:
Primary Coil
A transformer
consists of a primary
coil (input voltage),
a secondary coil
(output voltage), and
an iron core the
coils are wrapped
around.
EP NP
Iron Core
Secondary
Coil
NS
ES


Ideally: I PEP  I S ES
EP  N P
ES  N S
t
t
EP
NP
I

I

I
EP
E
N
S
P
P
ES
NS
 S
ES  EP S
NP NS
NP
Magnetic field lines (in blue) from the primary coil are directed
through the iron core to the secondary coil. The changing flux they
cause in the secondary coil induces the emf in the secondary coil. This
will not work if the current in the primary coil is constant.
EXAMPLE:
The primary coil of a transformer has 2000 turns
of wire. (a) If the maximum emf in the primary coil is 24V and 75V is
required at the secondary coil, how many turns of wire are required in
the secondary coil? (b) If the maximum current in the primary coil is
5A, what is the ideal maximum current from the secondary coil?
EP
ES

NP NS
I PEP  I S ES

ES N P  75V  2000 
NS 

 6250
EP
24V
I PEP  5 A 24V 
 IS 

 1.6 A
ES
75V
A transformer whose secondary coil voltage is greater than that of the
primary is called a step up transformer. If the secondary voltage is less
than the primary voltage, it is called a step down transformer.
INDUCTANCE:
V
V
EO
Inductor
EO
When the switch is closed Back emf:
the voltage builds to Eo.
The reason it does not

B
I 

E  N
 N
A   N  0 n  A
instantly become Eo is
t
t
t 

because the change in
N 2 I
I
current produces a
E   0
A
 L
l
t
t
changing magnetic field
2
N
and an induced emf that
L  0
A
opposes Eo. This emf is
l
called a back-emf and the
process is known as selfL is Inductance and has SI unit of Henry, H.
induction.
STORED ENERGY:
As current begins to flow through an inductor a magnetic field is
created and increases in strength. Within the field energy is stored. In
a short time t while the current is increasing an amount of energy U
is added to the field.
U  Pt  I Et
 I 
U  I  L  t  ILI
 t 
As the current increases more and more energy is stored until the
current stops increasing at Ifinal. The total energy stored is:
U   U   ILI
U  L  I I  12 LI 2
U  12 LI 2
STORED ENERGY:
Energy density may also
be calculated:
2


N
A 2
2
0
1
1
U  2 LI  2 
I
 l 
2
2 2


N
I  l 
o
1
U  2
 A

2
 l
  0 
 Al 
U B  
 0 
U
U 1 B2
u

2
Vol Al
0
1
2
u
1
2
2
B2
0
EXAMPLE: A circuit consists of a 20 resistor, 24V battery and an
inductor in series. The inductor is a solenoid made of 100 turns of wire
wrapped around a cylinder 2cm long with a radius of 5mm. Calculate
the inductance of the solenoid and the energy density of the field
2
2
7 Tm
inside.
2
4


10
100

.005
m
 

o N A 
A 
L

 4.93  105 H
l
.02m
V 24V
I 
 1.2 A
R 20
U  LI 
2
1
2
1
2
 4.93  10
5
H  1.2 A  3.55  105 J
2
U
U
3.55  105 J
J
u



22.6
2
2
m3
Vol l r
.02m   .005m 
B
o NI
l

7 Tm
4


10

A  100 1.2 A
.02m
.00754T 

B
1
1
u2
2
 22.6 mJ3
7 Tm
o
4  10 A
2
2
also
 .00754T