Transcript 3.5 Electromagnetic Induction

Electromagnetic Induction
• emf is induced in a conductor placed in a magnetic field
whenever there is a change in magnetic field.
Faraday’s work
• Faraday suggested that an e.m.f. is induced in a
conductor when
1. there is a change in the number of lines ‘linking’ it,
2. it ‘cuts’ across field lines.
• As shown in the figure, if the coil moves towards the
magnet from X to Y, the number of magnetic field
lines linking it increases from three to five;
alternatively we can say it cuts two lines in moving
from X to Y.
• Hence, an e.m.f. is induced in the coil.
• The figure shows two stationary coils A and B.
• When a current flowing in coil A increases or
decreases, the magnetic flux linking coil B increases
and decreases respectively. Hence, an e.m.f. is
induced in coil B.
Magnetic flux
A
B
x
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x
x
x
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x
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• B: magnetic flux density (i.e. the number of magnetic
field lines per unit cross-section area)
• A: cross-section area,
• magnetic flux linking the area is the product BA
represents the number of field lines linking a surface
of cross-sectional area A.
•  = BA
• If B = 1 T and A = 1 m2,  is defined to be 1 (Tm2) or
weber (Wb).
Magnetic flux
• If the surface is not perpendicular to the field with the normal to
the surface making an angle q to the field, the magnetic flux
linking the area is  = BA cos q.
Normal
B
B q
 = BA
B-field ⊥ plane of coil
 = BA cos q
• If  is the flux through the cross-section area A of a coil of N
turns, the total flux through it, called the flux-linkage, is N since
the same flux  links each of the N turns.
Example 1
• A circular coil of 20 turns with diameter 10 cm is
placed in a region of uniform magnetic field of 1.5 T.
Find the flux-linkage if the plane of the coil
• (a) is perpendicular to the field,
• Solution:
Example 1
• A circular coil of 20 turns with diameter 10 cm is
placed in a region of uniform magnetic field of 1.5 T.
Find the flux-linkage if the plane of the coil
• (b) is along the field, and
• Solution:
Example 1
• A circular coil of 20 turns with diameter 10 cm is
placed in a region of uniform magnetic field of 1.5 T.
Find the flux-linkage if the plane of the coil
• (c) makes an angle of 30o to the field.
• Solution:
Faraday’s law
The induced e.m.f. is directly
proportional to the rate of change
of flux-linkage or rate of flux
cutting.
d
d
• Mathematically, e   N  or e  constant  N  .
dt
dt
• It is defined that 1 Wb is magnetic flux that induces in a
one-turn coil an e.m.f. of 1 volt when the flux is reduced
to zero in 1 s.
• By putting e = 1 V, dt = 1 s and d(N) = 1 Wb, we have
1 = constant x 1/1.
d
• Hence, e   N 
dt
Example 2
• (a) Suppose a 5000-turn coil of cross-section area 5 cm2
is at right angles to a flux density of 0.2 T, which is then
reduced steadily to zero in 10 s. Find the e.m.f. induced in
the coil.
• (b) Find the e.m.f. induced if the normal to the plane of
coil makes an angle of 60o with the field.
• Solution
Lenz’s law
Lenz's law
Induced I always flows to oppose the
movement which started it.
In both cases, magnet
moves against a force.
Work is done during
the motion & it is
transferred as electrical
energy.
• Lenz’s law is incorporated in the mathematic
expression of Faraday’s law by including a negative
sign to show that current due to the induced e.m.f.
produces an opposing flux change. So we have
d
e    N 
dt
Fleming's right-hand rule
For a wire cutting through a B-field...
motion or force F
magnetic
field B
induced
current I
Example 3
observer
• State the direction of induced current flowing through
coil B observed by the observer when the current
through coil A increases steadily.
• Solution:
Calculation of e.m.f.
B
v
l
• Consider a conducting rod of length l moving sideways
with constant velocity v through and at right angles to a
uniform magnetic field of flux density B.
• Area swept out per second by the rod per second = lv
• Flux cut per second = Blv
• e.m.f induced = rate of flux-cutting = flux cut per second
e = Blv
Alternative derivation
• Magnetic force = Bqv.
• An electric field is built up due to the
accumulation of charges.
• Electric force = qE
• Finally, equilibrium is reached when
magnetic force acting on electrons is
balanced by electric force.
• Hence, qE = Bqv ⇒ E = Bv
• An e.m.f. e is generated across the
conductor such that
e = El = Blv.
Example 4
• A metal aircraft of wing span l = 32 m is flying with speed
v = 190 ms-1 towards the earth’s magnetic north pole in a
region where the earth’s magnetic field BR = 4.3 x 10-5 T
and the angle of dip a = 65o.
• Calculate the e.m.f. induced across its wing tips.
• Solution:
Simple a.c. Generator
• According to the Faraday’s law of electromagnetic
induction,
Nd
d
e 
  N BA cos t  NBA  sin t
dt
dt
http://www.walter-fendt.de/ph11e/generator_e.htm
Simple d.c. Generator
e  NBA  sin t
Back e.m.f.
Sparks appear while opening a switch
• There is current flowing in the
coil of the electromagnet in
use.
• When the circuit is broken by
opening the switch, the current
starts to drop and the flux
linkage through the coil of the
electromagnet decreases
suddenly.
• By Faraday’s law, a large
induced e.m.f. would develop
across the coil of the
electromagnet so as to oppose
the change.
• Sparks occur due to the
discharge across the small
gap of the switch.
DC motors
• A d.c. motor consists of a coil
on an axle, carrying a d.c.
current in a magnetic field.
• The coil experiences a couple
as in a moving-coil
galvanometer which makes it
rotate.
• When its plane is perpendicular
to the field, a split-ring
commutator reverses the
current in the coil and ensures
that the couple continues to act
in the same direction thereby
maintaining the rotation.
Back emf in Motors
• When an electric motor is running, its armature
windings are cutting through the magnetic field
of the stator. Thus the motor is acting also as a
generator.
• According to Lenz's Law, the induced voltage in
the armature will oppose the applied voltage in
the stator.
• This induced voltage is called back emf.
Back emf and Power
Armature coils, R
Back emf, Eb
Driving source, V
V  e b  IR
Multiplying by I, then
VI  e b I  I R
2
eb I  VI  I R
2
• So the mechanical power developed in motor
 eb I
Variation of current as a motor
is started
I
I max 
V
R
Larger load
Zero load
t
0
• As the coil rotates, the angular speed as well as the
back emf increases and the current decreases until the
motor reaches a steady state.
The need for a starting resistance
in a motor
• When the motor is first switched on,  =0.
• The initial current, Io=V/R, very large if R is small.
• When the motor is running, the back emf increases,
so the current decrease to its working value.
• To prevent the armature burning out under a high
starting current, it is placed in series with a
rheostat, whose resistance is decreases as the
motor gathers speed.
Variation of current with the steady
angular speed of the coil in a motor
I
I max
V

R
V  NBA 
I
R
0
 max
V

NBA

• The maximum speed of the motor occurs when the current
in the motor is zero.
Eddy Current
• An eddy current is a
swirling current set up in a
conductor in response to a
changing magnetic field.
• When the magnetic flux
linkage through a
conductor changes, an
e.m.f. is induced in it.
• If the conductor is a lump
of metal. These are known
as Eddy Currents.
• Eddy currents may be quite large because of the
low resistance of the paths they follow.
Consider a metallic sheet moving away
from a magnetic field.
• By Lenz’s law, eddy
currents must flow in a
direction to oppose the
motion of the sheet.
• Hence, eddy currents
act as an effective brake
to its motion.
• The mechanical work
done is converted into
internal energy of the
sheet.
1
Applications
Smooth Braking Device
• The eddy currents induced in the copper plate
produce a strong braking effect on the plate which
stops oscillating quickly.
• If the copper plate is replaced by one with slits, the
induced eddy currents, which can only flow within the
narrow teeth between the slits, are greatly reduced.
This is because the resistance of the path which the
eddy currents follow is increased.
Braking effect in moving-coil
galvanometer
pointer
soft-iron
core
eddy
currents
• As the core swings in a magnetic field, eddy currents
are induced in it. Since the eddy currents flow in a
direction to oppose the motion, unwanted oscillations
are reduced.
• Ideal design of the core is to produce critical damping
in which oscillation is just avoided.
Metal Detector
• A pulsing current is applied to the coil, which then induces a
magnetic field shown in blue. When the magnetic field of the
coil moves across metal, such as the coin in this illustration, the
field induces electric currents (called eddy currents) in the coin.
• The eddy currents induce their own magnetic field, shown in red,
which generates an opposite current in the coil, which induces a
signal indicating the presence of metal.
• A metal detector can also be used to detect mines buried
underground.
Induction cooker
• The induction cooker uses coils of wire with high
frequency a.c. to produce large eddy currents in the
metal cooking pot placing above. The heating effect
of the eddy current cooks the food.
• Moreover, since eddy current is not induced in its
plastic case which is made up of non-metallic
material, the cooker is not hot to touch.
Transformer
• A transformer is a device for stepping up or down an
alternating voltage.
• For an ideal transformer,
– (i.e. zero resistance and no flux leakage)
d B
Vp   N p
dt
d B
Vs   N s
dt
Vs N s
 
Vp N p
Transformer Energy Losses
•
Heat Losses
1. Copper losses - Heating effect occurs in the
copper coils by the current in them.
2. Eddy current losses - Induced eddy currents flow
in the soft iron core due to the flux changes in the
metal.
•
Magnetic Losses
1. Hysteresis losses - The core dissipates energy on
repeated magnetization.
2. Flux leakage - Some magnetic flux does not pass
through the iron core.
Designing a transformer to
reduce power losses
• Thick copper wire of low resistance is used to reduce the
heating effect (I2R).
• The iron core is laminated, the high resistance between the
laminations reduces the eddy currents as well as the heat
produced.
• The core is made of very soft iron, which is very easily
magnetized and demagnetized.
• The core is designed for maximum linkage, common method
is to wind the secondary coil on the top of the primary coil and
the iron core must always form a closed loop of iron.
Transmission of Electrical Energy
• Wires must have a low resistance to reduce power loss.
• Electrical power must be transmitted at low currents to reduce
power loss.
• To carry the same power at low current we must use a high
voltage.
• To step up to a high voltage at the beginning of a transmission
line and to step down to a low voltage again at the end we
need transformers.
Direct Current Transmission
• Advantages
– a.c. produces alternating magnetic field which induces
current in nearby wires and so reduce transmitted
power; this is absent in d.c.
– It is possible to transmit d.c. at a higher average
voltage than a.c. since for d.c., the rms value equals
the peak; and breakdown of insulation or of air is
determined by the peak voltage.
• Disadvantage
– Changing voltage with d.c. is more difficult and
expensive.
Self Induction
• When a changing current passes
through a coil or solenoid, a
changing magnetic flux is produced
inside the coil, and this in turn
induces an emf.
• This emf opposes the change in
flux and is called self-induced emf.
• The self-induced emf will be against
the current if it is increasing.
• This phenomenon is called selfinduction.
Definitions of Self-inductance
(1)
• Definition used to find L
The magnetic flux linkage in a coil  the current flowing
through the coil.
  LI
Where L is the constant of proportionality for the coil.
L is numerically equal to the flux linkage of a circuit
when unit current flows through it.
L

I
Unit : Wb A-1 or H (henry)
Definitions of Self-inductance
(2)
• Definition that describes the behaviour of an
inductor in a circuit
d
dI
e 
 L
dt
dt
L
e
dI
dt
L is numerically equal to the emf induced in the circuit
when the current changes at the rate of 1 A in each second.
Inductors
• Coils designed to produce large self-induced
emfs are called inductors (or chokes).
• In d.c. circuit, they are used to slow the growth of
current.
• Circuit symbol
or
Inductance of a Solenoid
• Since the magnetic flux density due to a solenoid is
 o NI
B

• By the Faraday’s law of electromagnetic induction,
d
e  N
dt
d
d  o NIA
  N ( BA)   N (
)
dt
dt

2
2

N
A
 o N A dI
o
L



dt
Energy Stored in an Inductor
• The work done against the back emf in bringing
the current from zero to a steady value Io is
Io
W   eIdt
0

Io
0
dI
LI dt
dt
Io
  LIdI
0
1 2
 LI o
2
Current growth in an RL circuit
dI
V L
 IR
dt
I 
e
R
(1  e
 Rt
L
)
• At t = 0, the current is zero.
• So L dIdt  e
• As the current grows, the
p.d. across the resistor
increases. So the selfinduced emf (e - IR) falls;
hence the rate of growth of
current falls.
dI
0
• As t
dt
Decay of Current through an Inductor
• Time constant for RL
R
circuit  
L
I  Ioe
 Rt
L
• The time constant is the
time for current to
decrease to 1/e of its
original value.
• The time constant is a
measure of how quickly
the current grows or
decays.
emf across contacts at break
• To prevent sparking at the contacts of a switch in
an inductive circuit, a capacitor is often
connected across the switch.
The energy originally stored
in the magnetic field of the coil
is now stored in the electric
field of the capacitor.
+
-
1 2 1
LI  CV 2
2
2
Switch Design
• An example of using a protection diode with a relay coil.
+
-
• A blocking diode parallel to the inductive coil is used to
reduce the high back emf present across the contacts
when the switch opens.
Non-Inductive Coil
• To minimize the self-inductance, the coils of resistance boxes
are wound so as to set up extremely small magnetic fields.
• The wire is double-back on itself. Each part of the coil is then
travelled by the same current in opposite directions and so
the resultant magnetic field is negligible.