Electromagnetic Induction

Chapter 31A - Electromagnetic
Induction
AA PowerPoint
PowerPoint Presentation
Presentation by
by

Paul
Paul E.
E. Tippens,
Tippens, Professor
Professor of
of Physics
Physics
Southern
Southern Polytechnic
Polytechnic State
State University
University
©

2007

Objectives: After completing this
module, you should be able to:

• Calculate the magnitude and direction of
the induced current or emf in a conductor
moving with respect to a given B-field.
• Calculate the magnetic flux through an
area in a given B-field.
• Apply Lenz’s law and the right-hand rule
to determine directions of induced emf.
• Describe the operation and use of ac and
dc generators or motors.

Induced Current
When aa conductor
conductor moves
moves
When
across flux
flux lines,
lines, magnetic
magnetic
across
forces on
on the
the free
free electrons
electrons
forces
induce an
an electric
electric current.
current.
induce
Right-hand force
force rule
rule shows
shows
Right-hand
current outward
outward for
for down
down and
and
current
inward for
for up
up motion.
motion. (Verify)
(Verify)
inward
Down

B
Down

I

Up
Up

F

v

B

I

v

F
B

Induced EMF: Observations
Faraday’s observations:

B Flux lines  in Wb

• Relative motion induces emf.
• Direction of emf depends on
direction of motion.

N turns; velocityv

• Emf is proportional to rate at
which lines are cut (v).

Faraday’s Law:

• Emf is proportional to the
number of turns N.


E = -N
t

The negative sign means that E opposes its cause.

Magnetic Flux Density
• Magnetic flux lines
 are continuous
and closed.


B
A

A



• Direction is that
of the B vector at
any point.

Magnetic Flux
density:

When
When area
area AA isis
perpendicular
perpendicular to
to flux:
flux:


B  ;  = BA
A

The unit of flux density is the weber per square meter.

Calculating Flux When Area is
Not Perpendicular to Field
The flux penetrating the
area A when the normal
vector n makes an angle
of  with the B-field is:

  BA cos

n
A





B

The angle  is the complement of the angle  that the
plane of the area makes with B field. (Cos  = Sin )

Example 1: A current loop has an area of 40 cm2
and is placed in a 3-T B-field at the given angles.
Find the flux  through the loop in each case.
x
x
x
x

x x
x x
A
x x
x x

x
x
x
x

A = 40 cm2

n
n



n

(a) = 00 (b) = 900 (c) = 600

(a) = BA cos 00 = (3 T)(0.004 m2)(1);

12.0 mWb

(b) = BA cos 900 = (3 T)(0.004 m2)(0);

0 mWb

(c) = BA cos 600 = (3 T)(0.004 m2)(0.5); 6.00 mWb

Application of Faraday’s Law
Faraday’s Law:


E = -N
t

A change in flux can
occur by a change in area or
by a change in the B-field:
 = B A

Rotating loop = B A
n
n
n

 = A B

Loop at rest = A B

Example 2: A coil has 200 turns of area 30 cm2.
It flips from vertical to horizontal position in a
time of 0.03 s. What is the induced emf if the
constant B-field is 4 mT?
N = 200 turns
n

A = 30 cm2 – 0 = 30 cm2
 = B A = (3 mT)(30 cm2)
 = (0.004 T)(0.0030 m2)
 = 1.2 x 10-5 Wb


1.2 x 10-5 Wb
E  N
 (200)
0.03 s
t

N


B

S

B = 4 mT; 00 to 900
EE== -0.080
-0.080 VV

The negative sign indicates the polarity of the voltage.

Lenz’s Law
Lenz
’s law:
Lenz’s
law: An
An induced
induced current
current will
will be
be in
in such
such aa direction
direction
as
as to
to produce
produce aa magnetic
magnetic field
field that
that will
will oppose
oppose the
the
motion
motion of
of the
the magnetic
magnetic field
field that
that isis producing
producing it.
it.
Induced B

Induced B
Left motion

I

N

S

Flux increasing to left induces
loop flux to the right.

I

Right motion
N

S

Flux decreasing by right move
induces loop flux to the left.

Example 3: Use Lenz’s law to determine direction
of induced current through R if switch is closed
for circuit below (B increasing).
Close switch. Then what is
direction of induced current?

R

The rising
rising current
current in
in right
right circuit
circuit causes
causes flux
flux to
to increase
increase
The
to the
the left,
left, inducing
inducing current
current in
in left
left circuit
circuit that
that must
must
to
produce aa rightward
rightward field
field to
to oppose
oppose motion.
motion. Hence
Hence
produce
current II through
through resistor
resistor RR isis to
to the
the right
right as
as shown.
shown.
current

Directions
Directions of
of
Forces
Forces and
and EMFs
EMFs
An
An emf
emf EEisis induced
induced by
by
moving
moving wire
wire at
at velocity
velocity vv
in
in constant
constant BB field.
field. Note
Note
direction
direction of
of I.I.
From
’s law,
From Lenz
Lenz’s
law, we
we see
see
that
that aa reverse
reverse field
field (out)
(out) isis
created.
created. This
This field
field causes
causes
aa leftward
leftward force
force on
on the
the wire
wire
that
that offers
offers resistance
resistance to
to the
the
motion.
-hand
motion. Use
Use right
right-hand
force
force rule
rule to
to show
show this.
this.

x x x x x x x x x
x x x x x xI x x
x x x x x x x x x
x x I x x x x v x vx
x x x x x Lx x x x
x x x x x x x x
x x x x x x x x x

x

I

B
v

Induced
emf

x x x
x x
x x x
x x
x x x
x x
x x x

I
v

Lenz’s law

B

Motional EMF in a Wire
Force F on charge q in wire:

F = qvB; Work = FL = qvBL

Work qvBL
E=

q
q

EMF: E = BLv

x x x x x xI x
x x x x x x
x x x x x x x
x Ix x x x x
x x x x L
x xv x
x x x x x x
x x x x x x x

B

v sin 



B
v

x

If wire of length L moves with
velocity v an angle  with B:

E = BLv sin 

F

v

Induced Emf E

Example 4: A 0.20-m length of wire moves
at a constant speed of 5 m/s in at 1400 with
a 0.4-T B-Field. What is the magnitude and
direction of the induced emf in the wire?

E = BLv sin 
E = (0.4 T)(0.20 m)(5 m/s) sin140

v
0

E = -0.257 V
Using
-hand rule
, point
Using right
right-hand
rule,
point fingers
fingers
to
to right,
right, thumb
thumb along
along velocity,
velocity, and
and
hand
hand pushes
pushes in
in direction
direction of
of induced
induced
emf
—to the
emf—to
the north
north in
in the
the diagram.
diagram.

North


B

South

v

South

I

North

B

The AC Generator
• An alternating AC current is Rotating Loop in B-field
produced by rotating a loop
B
in a constant B-field.
v
• Current on left is outward
by right-hand rule.

I

I

v

B

• The right segment has an
inward current.
• When loop is vertical, the
current is zero.

I in R is right, zero, left, and then zero as loop rotates.

Operation of AC Generator

I=0

I=0

Calculating Induced EMF
Rectangular
loop a x b
Each segment a
has constant
velocity v.

a
b

.

n

Both segments a moving with v
at angle  with B gives emf:

E = Bav sin  ; v   r    b 2 

ET  BA sin 

B



B

Area A = ab

ET  2 Ba  b 2  sin 

n


b/2 x v

v = r


r = b/2
v sin 

n
B

v

x 

x

.

x

Sinusoidal Current of Generator
.
+E

-E
The emf varies sinusoidally with max and min emf
For N turns, the EMF is:

E  NBA sin 

Example 5: An ac generator has 12 turns of
wire of area 0.08 m2. The loop rotates in a
magnetic field of 0.3 T at a frequency of 60
Hz. Find the maximum induced emf.
 = 2f = 2(60 Hz) = 377 rad/s

.

Emf is maximum when  = 900.

Emax = NBA ; Since sin   1
Emax = (12)(0.3 T)(.08 m 2 )(377 rad/s)

n


B

x
f = 60 Hz

The maximum emf generated is therefore:

Emax = 109 V

IfIf the
’s law
the resistance
resistance isis known,
known, then
then Ohm
Ohm’s
law ((VV == IR
IR)) can
can
be
be applied
applied to
to find
find the
the maximum
maximum induced
induced current.
current.

The DC Generator
The simple ac generator
can be converted to a dc
generator by using a single
split-ring commutator to
reverse connections twice
per revolution.

E

t

Commutator

DC Generator

For
For the
the dc
dc generator:
generator: The
The emf
emf fluctuates
fluctuates in
in magnitude,
magnitude,
but
but always
always has
has the
the same
same direction
direction (polarity).
(polarity).

The Electric Motor
In
, aa current
In aa simple
simple electric
electric motor
motor,
current loop
loop experiences
experiences aa
torque
torque which
which produces
produces rotational
rotational motion.
motion. Such
Such motion
motion
induces
induces aa back
back emf
emf to
to oppose
oppose the
the motion.
motion.
Applied voltage – back emf
= net voltage

VV –– EEbb == IR
IR
Since
Since back
back emf
emf EEbb increases
increases with
with
rotational
, the
rotational frequency
frequency,
the starting
starting
current
current isis high
high and
and the
the operating
operating
current
 sin
current isis low:
low: EEbb== NBA
NBA
sin 

Eb

I
V
Electric Motor

Armature and Field Windings
In the commercial motor,
many coils of wire around
the armature will produce
a smooth torque. (Note
directions of I in wires.)
Series-Wound Motor: The
field and armature wiring
are connected in series.

Motor

Shunt-Wound Motor: The field windings and the
armature windings are connected in parallel.

Example 6: A series-wound dc motor has an
internal resistance of 3 . The 120-V supply
line draws 4 A when at full speed. What is the
emf in the motor and the starting current?
VV –– EEbb == IR
IR

Recall that:

Eb

I

120 V – Eb = (4 A)(3 

V

The back emf
in motor:

Eb = 108 V

The starting current Is is found by noting that Eb = 0
in beginning (armature has not started rotating).
120 V – 0 = Is (3 

Is = 40 A

Summary
Faraday’s Law:


E = -N
t

A change in flux can
occur by a change in area or
by a change in the B-field:
 = B A

 = A B

Calculating flux through an area in a B-field:


B  ;  = BA
A

  BA cos

Summary (Cont.)
Lenz
’s law:
Lenz’s
law: An
An induced
induced current
current will
will be
be in
in such
such aa direction
direction
as
as to
to produce
produce aa magnetic
magnetic field
field that
that will
will oppose
oppose the
the
motion
motion of
of the
the magnetic
magnetic field
field that
that isis producing
producing it.
it.
Induced B

Induced B
Left motion

I

N

S

Flux increasing to left induces
loop flux to the right.

I

Right motion
N

S

Flux decreasing by right move
induces loop flux to the left.

Summary (Cont.)
An emf is induced by a wire
moving with a velocity v at an
angle  with a B-field.

E = BLv sin 

B

v sin 



v

Induced Emf E

In
In general
general for
for aa coil
coil of
of NN turns
turns of
of area
area AA rotating
rotating
with
with aa frequency
frequency in
in aa B-field,
B-field, the
the generated
generated emf
emf
isis given
given by
by the
the following
following relationship:
relationship:
For N turns, the EMF is:

E  NBA sin 

Summary (Cont.)
The
The ac
ac generator
generator isis
shown
shown to
to the
the right.
right. The
The
dc
dc generator
generator and
and aa dc
dc
motor
motor are
are shown
shown below:
below:

V
DC Generator

Electric Motor

Summary (Cont.)
The
The rotor
rotor generates
generates aa back
back
emf
emf in
in the
the operation
operation of
of aa
motor
motor that
that reduces
reduces the
the
applied
applied voltage.
voltage. The
The
following
following relationship
relationship exists:
exists:
Applied voltage – back emf
= net voltage

VV –– EEbb == IR
IR

Motor

CONCLUSION: Chapter 31A
Electromagnetic Induction