c20-electromagnetic-induction-student2 (1).ppt

7/27/2019 C20-Electromagnetic-Induction-Student2 (1).ppt

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UNIT 20 :

ELECTROMAGNETIC INDUCTION

20.1 Magnetic flux

20.2 Induced emf

20.3 Self-inductance20.4 Mutual inductance

20.5 Energy stored in inductor

Electromagnetic induction is the product ionof an elec tr ical po tent ial di f ference (induced

em f) acro ss a conduc tor s i tuated in a

chang ing m agnet ic f ield .


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20.1 MAGNETIC FLUX ,

is defined as the scalar product between

the magnetic flux de

nsity, Band the vectorof the surface area, A .

BAAB cos

B

A

area, A

BA

A

B

0= = 0

= 90

Unit:T.m2 or Wb


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Example 20.1.1A small surface of area 10 mm2 inside a uniform magnetic field of strength

0.10 T is inclined at an angle to the direction of the field. Determine the

magnetic flux through the surface if

i) = 0,ii) = 30

iii) = 90

Solution :

BAAB cos


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20.2 INDUCED EMF

An electric current produces a magnetic field.

(chapter 19)If electr ic cu rrents produce a magnet ic f ield,

is i t possible that a magnet ic f ield can

produce an electr ic current ?

Scientists (American Joseph Henry and the

Englishman Michael Faraday) independently

found that is possible. Henry actually made the discovery first, but

Faraday published his results earlier and

investigated the subject in more detail.


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20.2 INDUCED EMF

The diagram below shows the apparatus used

by Faraday in his attempt to produce anelectric current from a magnetic field.

Faradays experiment to induce an emf


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20.2 INDUCED EMF

In this experiment, Faraday hoped by using a

strong enough battery, a steady current in X

would produce a current in a second coil Y but

failed.

Faraday saw the galvanometer in circuit Y

deflect strongly at the moment he closed theswitch in circuit X.

And the galvanometer deflected strongly in

the opposite direction when he opened theswitch.

A steady current in X had produced no

currentin Y.


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Only when the current in X was starting or

stopping was a current produced in Y.

Faraday concluded that although a steady

magnetic f ieldproduces no current, a

chang ing magnet icfield can produce an

electr ic cu rrent. Such a current is called an induced current.

We therefore say that an induced current is

produced by a chang ing magnet ic f ield. The corresponding emf required to cause

this current is called an induced em f.

20.2 INDUCED EMF


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Induced emfis an electromotive force

resulting from the motion of a conductor

through a magnetic field , or from a change inthe magnetic flux that threads a conductor.

a) A current is induced when a magnet is

moved toward a coil/loop.

b) The induced current is opposite when the

magnet is moved away from the coil/loop.

c) No current is induced if the magnet does

not move relative to the coil/loop.

Faraday did further experiments on

electromagnetic induction, as thisphenomenon is called.( refer diagram )

20.2 INDUCED EMF


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10Micheal Faradays experiment

20.2 INDUCED EMF


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20.2 INDUCED EMF


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Direction of the induced current depends on :

i ) the direction of the magnets motion and

ii) the direction of the magnetic field.

Magnitude of the induced current depends on :

i ) the speed of motion (v,Iind)

ii) the number of turns of the coil (N,Iind)

iii)the strength of the magnetic field (B,Iind)

From the observations, Michael Faraday

found that,

the current/emf is induced in a coil/loop or complete

circuit whenever there is a change in the magnetic flux

through the areasurrounded by the coil

20.2 INDUCED EMF


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Faradays law and Lenzs law

the magnitude of the induced e.m.f.

is propo rt ional to the rate of change

of the magnetic flux

Faradays law

Lenzs law

an induced electric currentalways f lows in such a direct ion

that i t opposes the change

produc ing i t.

20.2 INDUCED EMF


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dtdB

dtdBor

timeofchange:dt e.m.f.induced:fluxmagneticofchange:Bd

The (-) sign indicates that the direct ion ofinduced e.m .f. always opposesthe change of

magnetic flux producing it (Lenzs law).

These two laws are summed up in the

relationship,

20.2 INDUCED EMF


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The concept ofFaraday's Law is that any change

in the magnetic environment of a coil of wire will

cause a voltage (emf) to be "induced" in the coil.

No matter how the change is produced, the

voltage will be generated.

The change could be produced by

a) changing the magnetic f ield strength ,

b) mov ing a magnet toward or away from thecoi l ,

c) mov ing the coi l in to o r out o f the magnet ic

f ield,

d) rotat ing the co i l relat ive to the magnet, etc.

20.2 INDUCED EMF
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html


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(A) Induced emf in coil 20.2 INDUCED EMF

(A) I d d f i il


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Notes

i ) the magnitude of induced emf,

initialfinal

initialfinal

ttN

dt

dN

dt

d

or

ii) the flux through the coil can change in anyof 3 ways,

a) B , b) A , c) BAAB cos==

dt

dBNA

dt

dANB

dt

dB


(A) I d d f i il


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Notes

iii)

RIIR induced

If the coil is connected in series to aresistor of resistance Rand the induced

e.m.fexist in the coil as shown in figure

below.dtdB||

and

R inducedI

inducedI+-



L ' L (b d i f )


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Lenz's Law (based on censervation of energy)

When an emf is generated by a change in magnetic

flux according to Faraday's Law, the polarity of the

induced emf(next slide) is such that it produces acurrent whose magnetic field opposes the change

which produces it.

The induced magnetic field inside any loop of wirealways acts to keep the magnetic flux in the loop

constant.

In the examples below, if the B field is increasing, theinduced field acts in opposition to it.

If it is decreasing, the induced field acts in the

direction of the applied field to try to keep it constant.

(A) I d d f i il


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7.2 INDUCED EMF

The polar ity o f the induced em f

Induced current is directed out of the positive

terminal, through the attached device (resistance)

and into the negative terminal.


(A) Induced emf in coil



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Example 20.2.1

A coil of wire 8 cm in diameter has 50 turnsand is placed in a Bfield of 1.8 T. If the B

field is reduced to 0.6 T in 0.002 s , calculate

the induced emf.




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Solution Faradays law and Lenzs law

d = 8 cm, N = 50 turns, B from 1.8 T to 0.6 T

in 0.002 s

V151

t

BBdN

t

BBNA

t

BABAN

tN

dt

dN

initialfinal

initialfinalinitialfinal

initialfinalB

2

2

dt

dBNA



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Example 20.2.2

An elastic circular loop in the plane of thepaper lies in a 0.75 T magnetic field pointing

into the paper. If the loops diamater changes

from 20.0 cm to 6.0 cm in 0.50 s,

a)What is the direction of the induced current,

b)What is the magnitude of the average

induced emf, and

c)If the loops resistance is 2.5 , what is the

average induced current during the 0.50 s ?




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Solution:


B=0.75 T, di= 20.0 cm, df= 6.0 cm, t = 0.50 s

a) Direction of the induced current,

b) Magnitude of the average induced emf,

c) R = 2.5 ,

dt

dANB


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Example 20.2.3 Faradays law and Lenzs law

A circular shaped coil 3.05 cm in radius,

containing 40 turns and have a resistance of3.55 is placed perpendicular to a magnetic

field of flux density of 1.25 x 10-2 T. If the

magnetic flux density is increased to 0.450 T

in time of 0.250 s, calculate the induced

current flows in the coil.



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How to determine the direction of induced

current.- Lenzs law

inducedI

inducedI

Direction of induced current induced-current right hand

rule.

N

Case A

-+

Thumb induced magnetic field

Fingers - induced current



F d l d L l


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How to determine the direction of induced current.- Lenzs law


Consider a bar magnet that is moved

towards a solenoid.

As the north pole of the magnet approaches

the solenoid, the amount of magnetic field

passing through the solenoid increases ,thus increasing the magnetic flux through

the solenoid.

The increasing flux induces an emf

(current) in the solenoid and galvanometer

indicates that a current is flowing.

Case A

F d l d L l


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The direction of the induced current is

such as to generate a magnetic field in thedirection that opposes the change in the

magnetic flux, so the direction of the

induced field must be in the direction thatmake the solenoid right end becomes a

north pole.

This opposes the motion of the bar magnetand obey the Lenzs law.

Case A

F d l d L l


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(a) When the magnet is moved toward the stationary

conducting loop, a current is induced in the

direction shown.

(b) This induced currentproduces its own magnetic

field(Binduced) directed to the left that counteracts

the increasing external flux.

Case B

Bexternal Binduced

F d l d L l


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(c) When the magnet is moved away from the

stationary conducting loop, a current is induced

in the direction shown.

Case B

(d) This induced currentproduces a magnetic field

(Binduced) directed to the right and so counteracts

the decreasing external flux.

Bexternal Binduced

Faradays law and Lenzs law(A) Induced emf in coil


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Example 20.2.4

Calculate the current

through a 37 resistor

connected to a single

turn circular loop 10 cmin diameter, assuming

that the magnetic field

through the loop is

increasing at a rate of0.050 T/s. State the

direction of the current.

Faradays law and Lenzs lawFaradays law and Lenzs law




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Example 20.2.4

R = 37 , d = 10 cm

dB/dt = 0.050 T/s.

dtdBd

dtdBA

2

2||

A10x1.0637

10x3.93 5--4

R

I

Direction ofIinduced : from b to a.


I induced

I inducedSN


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(B) Induced emf of a straight conductor

vdtdx

Consider a

straight conductorof length lis

moved at a speed

vto the right on a

U-shapedconductor in a

uniform magnetic

field Bthat points

out the paper.

This conductor travels a distance dx =vdtin a time

dt.

dA


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vdtdx

The area of the

loop increases byan amount

dA

lvdtdA

ldxdA

According toFaradays law,the e.m.f. isinduced in theconductor andits magnitude isgiven by


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sinBlv

Blv

dt

lvdtB

dtdAB

dtd

= angle between v andB

= 90 o

This induced emf is called motional induced emf.

BvL

(B) I d d f f t i ht d t


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dA

vdtdx

FappliedFB

As the conductor is moved to the right (Fapplied to

the right) with speed v, the magnetic flux through

the loop increases.

A current is induced

in the loop.

The induced currentflows in the direction

that tends to oppose

this change.

In order to oppose this change, the current through

the conductor must produce a magnetic force (F=BIL)

directed to the left.

Faradays law and Lenzs law(B)Induced emf of a straight conductor


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Faraday s law and Lenz s law

The direction of the induced current due toinduced e.m.f. flows in the linear conductor can be

determine by using Flemings right hand rule(based on lenzs law).

(B)Induced emf of a straight conductor

P

vdtdx

FappliedFB

dAQ

The induced currentflows from P to Q.

B

inducedinduced

orI

)(motion

Thumb direction ofMotion

First finger direction ofField

Second finger direction ofInduced current

or Induced e.m.f.

Only for the

straight

conductor.

Fapplied


Polarity


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dA

vdtdx

When the conductor is moved to the right (Fappliedto the right) with speed v, the electrons in the rod

move with the same speed.

Therefore, each

feels a forceF=Bqv,

which acts upward in

the figure.

If the rod were not in contact with the U-shaped

conductor, electrons would collect at the upper end

of the rod, leaving the lower end positive. There

must thus be an induced emf.

Polarity

Induced emf of a straight conductorExample 20 2 5


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Induced emf of a straight conductorExample 20.2.5

Suppose the length in figure above is 0.10 m,

the velocity z is 2.5 m/s, the total resistance of

the loop is 0.030 and B is 0.60 T. Calculate

a) the induced emf

b) the induced current

c) the force acting on the rodd) the power dissipated in the loop

d)

c)

b)

||a)

2

RIP

BILF

RI

Blv

dissipated



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A 0.2-m length of wire moves at a constant

velocity of 4 m/s in a direction that is 40 o with

respect to a magnetic flux density of 0.5 T.

Calculate the induced emf.

sinBlv



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In figure above, a rod with length l= 0.400 m

moves in a magnetic flux with magnitudeB =

1.20 T. The emf induced in the moving rod is

3.60 V.

a) Calculate the speed of the rod.

b) If the total resistance is 0.900 ,

calculate the induced current.

c) What force does the field exert on the

rod as a result of this current?

7.50 m/s , 4.00 A , 1.92 N to the left


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Fig 31-CO, p.967

(C) I d d f i t ti il


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(C) Induced emf in a rotating coil

An ac generator / dynamo

(transforms mechanical energy into electric energy)



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An ac generator / dynamo

(transforms mechanical energy into electric energy)



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Consider a coil ofN

turns each of areaA and

is being rotated about ahorizontal axis in its own

plane at right angle to a

uniform magnetic field of

flux density B.

As the coil rotates with

the angular speed , the

orientation of the loopchanges with time.

A




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The emf induced in the loop is given by

Faradays law,

NABt

tNAB

tNAB

BAdt

tABd

N

tABABdt

dN

maxsin

sin

sin

cos

coscos

oo ,

constantareand,

,

The emf induced in the loop varies

sinusoidally in time.

NAB sin




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The alternating emf induced in the loop plotted

as a function of time.

0 0 0 0 0

max max max max

A


Example 20 2 8 Induced emf in a rotating coil


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Example 20.2.8

The armature of a simple ac generator

consists of 100 turns of wire, each having anarea of 0.2 m2 . The armature is turned with

a frequency of 60 rev/s in a constant

magnetic field of flux density 10-3 T.

Calculate the maximum emf generated.

NABmax



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Example 20.2.9

0

(V)

t

28 V

-28 V

0.21 s

0.42 s

0.84 s0.63 s

The drawing shows a plot of the output emf of agenerator as a function of time t. The coil of this

device has a cross-sectional area per turn of

0.020 m2 and contains 150 turns. Calculate

a)The frequency of the generator in hertz.b)The angular speed in rad/s

c)The magnitude of the magnetic field.

2.4 Hz , 15 rad/s , 0.62 T



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Example 20.2.10

An amarture in ac generator consists of 500

turns, each of area 60 cm2 . The amarture is

rotated at a frequency of 3600 rpm in a

uniform 2 mT magnetic field. Calculate

a) the frequency of the alternating emf

b) the maximum emf generatedc) the instantaneous emf at time when the

plane of the coil makes an angle of 60o

with the magnetic field ?

380 rad/s, 1.13 V, 2.26 V

20 3 SELF-INDUCTANCE


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20.3 SELF-INDUCTANCE

Self- induct ionis defined as the process o f

producing an indu ced e.m .f . in the coi l dueto a change of cu rrent f lowing through the

same coi l .

Consider a current is present in the circuit

above.

NS

S RII



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This current produces a magnet field in the

coil that causes a magnetic flux through the

same coil.

This coil is said to have sel f - inductance(inductance).

NS

S RII

This flux changes when

the current changes.

An emf is induced inthis coil called a self-

induced emf.

A coil that has inductanceis called an

i nductor.



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The symbol for an inductor is

if air-cored, and if it has

a core of magnetic material.

By Lenzs law, the induced current opposes

the change that cause it.

If the current is increasing, the direction of

the induced field and emf are oppositeto that

of the current, to try to decrease the current.

If the current is decreasing, the direction of

the induced field and emf are in the same

direction as the current, to try to increase the

current.



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(a) A current in the coil produces a magnetic field

directed to the left.

(b) If the current increases, the increasing magneticflux creates an induced emf having the polarity

shown by the dashed battery.

(c) The polarity of the induced emf reverses if

the current decreases.

Iinduced Iinduced



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IB

The magnetic flux in a coil is proportional

to the current

LIB coiltheofinductance-self:L

From the Faradays law,

dt

d B

dt

LId

dt

dIL

. (1)

. (2)



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dt

dIL

dt

dIL

Self-inductance, L is defined as the ratio

of the self induced e.m .f. to the rate of

change of current in the coi l .

20.3 SELF-INDUCTANCE(1) = (2) dId


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(1) (2)

dt

dIL

dt

dB

If the coil has N turns, hence

dt

dN

dt

dIL B

BdNdILBNLI

I

NL B - scalar quantity

- unit is henry (H).

1-21-

AT m1Wb A1H1



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I

NL B

The value of the self-inductance depends on

a) the size and shape of the coil

b) the number of turn (N)c) the permeability of the medium in the

coil ().

Self-inductance does not depend on current.



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I

NL B

Self-inductance of a Loop and Solenoid

l

NI

B

0

0BAB

cos

l

AN

L

20 AlnL

2

0

From And

By substituting we get,

orFor the medium-core solenoid :

l

AN

L

2

0r

l

ANL

2

0ror

where typermeabilirelative:rmediumoftypermeabili:

solenoidtheofarea:A

nl

N

20.3 SELF-INDUCTANCEExample 20.3.1


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Example 20.3.1

If the current in a 230 mH coil changes

steadily from 20.0 mA to 28.0 mA in 140 ms,

what is the induced emf ?

dt

dIL

Example 20.3.2

Suppose you wish to make a solenoid

whose self-inductance is 1.4 mH. Theinductor is to have a cross-sectional area of

1.2 x 10 -3 m2 and a length of 0.052 m. How

many turns of wire needed ? 220 turns

(Given 0 = 4 x 10-7 H m-1)

l

ANL

20



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Example 20.3.3

The current in a coil of wire is initially zero

but increases at a constant rate; after 10.0 s

it is 50.0 A. The changing current induces

an emf of 45.0 V in the coil.

a) Calculate the self inductance of the coil.

b) Calculate the total magnetic flux throughthe coil when the current is 50.0 A.

dtdILa)

b)I

NL B



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Example 20.3.4

A 40.0 mA current is carried by a uniformly

wound air-core solenoid with 450 turns, a 15.0

mm diameter and 12.0 cm length. Calculate

a) the magnetic field inside the solenoid.

b) the magnetic flux through each turn.

c) the inductance of the solenoid.

a)

b)

l

NIB 0

0BAB cos

I

NL

Bc)

l

ANL

20or

(Given 0 = 4 x 10-7 H m-1)

20.4 MUTUAL INDUCTANCE20.4 MUTUAL INDUCTANCE


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Mutual Inductance for two coaxial solenoids

lI1

I1

N1N2

A

Ac generator Consider a long solenoid with length land cross

sectional areaA is closely wound with N1 turns of wire.

A second solenoid with N2 turns surrounds it at its

centre as shown in figure above.

20.4 MUTUAL INDUCTANCEMutual Inductance for two coaxial solenoids

Th fi t l id i th t d t


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The first solenoid is the one connected to an ac

generator, which sends an alternating current I1

through it.

The currentI1 produces a magnetic field lines

inside it and this field lines also pass through the

solenoid 2 as shown in figure.

If the currentI1changes with time, the magneticflux through the solenoids 1 and 2 will change with

time simultaneously. Due to the change of magnetic flux through the

solenoid 2, an e.m.f. is induced in solenoid 2.

At the same time, the self-induction occurs in the

solenoid 1 since the magnetic flux through it changes.

This process is known as mutual induction.

Mutual Inductance for two coaxial solenoids 20.4 MUTUAL INDUCTANCE


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Mutual inductance, M

If the currentI1 in solenoid 1 is continously changing,then the flux it produces will also change continously.

The changing magnetic flux from the solenoid 1

induces an emf in the solenoid 2.

The induced emf in the solenoid 2 is proportional to

the rate of change of the currentI1 in solenoid 1.

dt

dI12

dt

dIM 1212 .. (1)

Mutual induction is defined as the process of

producing an induced e.m.f.in one circuit/coil due tothe change of current in another circuit/coil.


Mutual inductance M


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dt

dIM 2121

Also the induced emf in the solenoid 1 is proportional

to the rate of change of the currentI2

in solenoid 2.

dt

dI21

inductancemutual

alityproportionofconstanta

MMM 1221

The mutual inductance of the two solenoids is the

same if current flows in the solenoid 2 and flux linksthe solenoid 1, causing an induced emf when a

change in flux linkage occurs.


Mutual inductance M


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Rearrange,

Mis defined as the ratio of the induced emf in one

solenoid/coil/ to the rate of change of current in theother solenoid/coil.

dtdI

dtdI

M2

1

1

2

UnitM: Henry (H)

From Faradays law,

dt

dN

dt

d

N

222

.. (2)


Mutual inductance M

20.4 MUTUAL INDUCTANCE


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(1) = (2)

dt

dN

dt

dIM 22

121

22121 dNdIM

22121 NIM

1

2221

I

NM

SinceM12

=M21

=M, equation above can be

written as

2

11

1

22

I

N

I

NM


Mutual inductance M



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1

2221

I

NM From

l

AINAB o 1112

and

l

ANNMMM

lI

AINN

I

NM

o

o

121221

1

112

1

2221

He mutual inductance of the solenoid 2 is,

2

11

1

22

I

N

I

NM

l

ANNM o 12

Mutual inductance M



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2

11

1

22

I

N

I

NM l

ANNM

o 12

dt

dIM 2121

Example 20 5 120.4 MUTUAL INDUCTANCE


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The primary coil of a solenoid of radius 2.0 cm

has 500 turns and length of 24 cm. If thesecondary coil with 80 turns surrounds theprimary coil at its centre, calculate

a. the mutual inductance of the coils

b. the magnitude of induced e.m.f. insecondary coil if the current in primary coilchanges at the rate 4.8 A s-1.

Example 20.5.1



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rp = 2.0 cm , Np =500 , lp 24 cm Ns = 80

dIs/dt= 4.8 A s-1

Solution 20.5.1

p

pso

lANNM

dtdIM

p

s

lANNM o 12

dtdIM 2121 a) b)


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20.4 MUTUAL INDUCTANCETransformer


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laminated iron

core

primary coilsecondary coil

NP

turnsNS

turns

Vp(input)

Vs

(output)

A transformer is a device forincreasing ordecreasing an ac voltage.

The operation of transformer is based on the

principle ofmutual induction and self-induction.

Symbol

in circuit

20.4 MUTUAL INDUCTANCETransformer


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Two types of transformer

a) step-up transformer(Ns > Np)

b) step-down transformer(Np > Ns).

There are three assential parts;

(1) a primary coil connected to an ac source

(2) secondary coil(3) soft iron core

When ac voltage is applied to the input coil

(primary coil), the alternating current produces

an alternating magnetic flux that is concentrated

in the iron core, without any leakage of flux

outside the core.

20.5 ENERGY STORED IN INDUCTOR


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The functions of an inductor are ;

- to control current- to keep energy in the form of magnetic field

An inductor carrying current has energy

stored in it. It is because a generator does work to

establish a current in an inductor.

Suppose an inductor is connected to a

generator whose terminal voltage can bevaried continously from zero to some final

value.



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As the voltage is increased, the currentIin the

circuit rises continously from zero to its final value.

While the cu rrent is r is ing, an emf (back

emf)is induced in the inductor.

Because of this, the generator that supplies the

current must maintain a po tent ial di f ferencebetween its terminals while the current is rising

(changing), and therefore it must supp ly energy to

the induc tor.

Thus, the generator must do workto push thecharges through the inductor against this induced

emf.


T d thi h t b li d b th


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To do this, power has to be supplied by the

generator to the inductor.

LIdIPdt

dt

dILIP

IP

The total work done while the current is changed

from zero to its final value is given by

2

0 0

2

1LIU

dILdUU I

This work is stored asenergy in the inducto r

.

PdtdU

dt

dU

dt

dWP

tWP

UW

F l i l id th lf i d t i



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For a long air-core solenoid, the self-inductance is

l

ANL

2

0

l

AIN

2

1U

22

02LI

2

1U

Therefore the energy s tored in the so leno idis given by

2LI2

1U

How much energy is stored in a 0.085-Hinductor that carries a current of 2.5 A ?

Example 20.5.1


Example20 5 2


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Example20.5.2

A steady current of 2.5 A in a coil of 500 turns

causes a flux of 1.4 x 10-4 Wb to link (passthrough) the loops of the coil. Calculate

a) the average back emf induced in the coil if

the current is stopped in 0.08 sb) the inductance of the coil and the energy

stored in the coil (inductor).

dtdB||

dtdIL

2LI1

U

c20-electromagnetic-induction-student2 (1).ppt

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