Comprehensive Guide to Electromagnetic Induction and Inductance Principles

Oh electricity and magnetism are interrelated and are also

interconvertible let us look at a few examples in this setup an electric current

passing through a conductor produces a magnetic field which can be observed through the

deflection of a magnetic compass needle placed near the conductor this proves that moving electric charges

produce magnetic fields electric motors work on this principle however in this Center we see that an

electric current is induced in a closed coil when subjected to a changing magnetic field the phenomenon in which

an electric current is generated by varying magnetic fields is called electromagnetic induction

electric generators work on this principle we will now discuss and learn about

three experiments by faculty and Henry relating to electromagnetism in the first experimental setup a coil

is connected to a galvanometer when a bar magnet with it's not fall facing the koi is moved to words or away

from the koi the galvanometer needle dislikes to the right and left side of zero reading respectively showing the

presence of a current in the coil now let us see what happens if the South Pole of the magnet faces the coin

the galvanometer needle dislikes to the left side of zero reading as the magnet approaches the coil and the Flex to the

right when the magnet moves away from the coil join the presence of a current in the

Kali the galvanometer needle deflects only as long as the bar magnet is in motion

once the bar magnet comes to rest the galvanometer needle settles down at zero reading indicating that there is no

current in the coil from all these observations we can conclude that whenever there is relative

motion between a bar magnet and a coil an electric current is induced in the coil

in the second experimental setup there are two coils coil one connected to a galvanometer coil to connect it to a

pantry due to the steady current in coil to a steady magnetic field is set up around

the coil - and this magnetic field is also linked to the coil one when coin to escape stationary and coin

one is moved towards coin too a current is induced in coil 1 and the galvanometer needle dislikes to the left

of 0 when going to escape stationary and kiwanis moved away from coil to

a current is induced in coil 1 and the galvanometer needle deflects to the right of 0

if we keep calling one stationary and move going to towards coil one a current is induced in coil 1 and a

deflection is observed in the galvanometer needle to the left of 0 now if we keep going one stationary

and move going to away from coit one a current is induced in coil 1 and a deflection is observed in the

galvanometer needle to the right of 0 from all these observations we can conclude that whenever there is relative

motion between a current carrying coil and a closed coil in which a galvanometer is connected a current is

induced in the closed coil in the third experiment the tab key is provided in the coil - second

here we can observe the deflection of the galvanometer needle even when the two coins are stationary

this deflection is observed only and those instants when the tab key is either switched on or off

this happens because of the change in magnetic field during switching on and off

when a ferromagnetic material like an iron truant is pleased coaxial e along the two coins

the effect of the magnetic field linked to the coin - increases due to the nature of the ferromagnetic material as

it allows more number of magnetic lines of force to link within the area of the cony

hence the deflection of the galvanometer needle increases indicating an increase in the induced current

before attempting to understand Faraday's law of electromagnetic induction we should lay capitulate the

terms used in electromagnetic induction these terms are representation of the surface area by a

vector e uniform magnetic field B and magnetic flux phi b

the area of a plane surface is represented by a vector a drawn along the outward normal to the plane surface

in the case of a curved surface this area is split into very small areas and each is approximated to a planar

area in a uniform magnetic field be the number of magnetic field lines cutting

across an area placed perpendicular to the fleet lines is same for each unit area of the surface

if a plaintiff area a is placed in a uniform magnetic field B with its area vector e making an angle theta with the

magnetic field then the magnetic flux 5b through the area can be written as Phi B is equal to

B dot a which is equal to B a cos theta the equation 1 magnetic flux Phi B can be varied by

changing any one or more of the terms B e and theta the SI unit of magnetic flux is Weber or

Tesla meters squared magnetic flux phi b is a scalar quantity when the magnetic field has different

magnitudes and directions at the various parts of a curved surface the magnetic flux through the curved surface is

computed by extending equation one phi b is equal to b 1 dot d a 1 plus B 2 dot d a 2 plus and so on which is equal

to Sigma all bi dot Dai let this be equation 2 so far we have recalculated the terms

used in electromagnetic induction learn about Faraday's experiments where we use these terms

party conducted three experiments from which he concluded that an EMF is induced in a coil when the magnetic flux

cutting across the coil changes with time in the first experiment

Baddeley observed that the relative motion between a magnet and a coil changes the magnetic flux across the

coin this change in magnetic flux with time induces an EMF in the coil

similarly in the second experiment he observed that the relative motion between the two coins one carrying a

current changes the magnetic flux across the coins this change in the magnetic flux with

time induces an EMF in the other coil in his third experiment with two coins he found that if the tab key is pressed

the current in coil two and the resulting magnetic field Rises from zero to maximum value in a short time

this results in a change in the magnetic flux with time across coil one inducing an EMF in it if the key is kept

pressed there is a steady magnetic field across coil one thereby there is no EMF induced in it and the current drops to

zero when the tab key is released the current and pointer decreases from a maximum

value to zero in a short time resulting in the magnetic flux across coil one to drop suddenly thereby inducing an EMF in

it from the observations of these three experiments Faraday concluded that the

time rate of change of magnetic flux through a closed coil induces an EMF through it and stated the Faraday's law

of electromagnetic induction the magnitude of the induced EMF in a circuit is equal to the time rate of

change of magnetic flux through the coil we can put this law in the form of a mathematical equation e is equal to

minus D Phi B divided by DT let this be equation three in this equation e is the induced EMF

the negative sign indicates the direction of EMF and hence the current in the closed-loop

when n number of turns are closely wound to form the coin the change of flux with time for each coin is the same and we

can write for the total induced EMF as e is equal to minus n into D Phi B divided by DT

let this be equation for in this equation e is the total induced e-m-f

the negative sign indicates the direction of EMF and hence the current in the closed loop therefore the total

induced EMF can be increased by increasing the number of turns n of the coil

before attempting to learn about lenses nor let us first recapitulate about the right hand grip rule applied to a

current carrying coil consider a right hand fist with the thumb extending out if the fingers are wrapped in a circle

pointing in the direction of the current through the coil then the thumb points towards the end of the coil indicating

the North Pole Lenz's law helps us in identifying the polarity of the induced EMF in a coin

Lenz's law states that the polarity of induced EMF in a closed loop is such that it tends to produce a

current which opposes the change in the magnetic flux that produces it we can put this statement in simple

terms if a current is induced by an increasing flux it will weaken the original flux

if a current is induced by a decreasing flux it was strengthened the original flux

the magnitude of the induced EMF is given by Saturday's law which is e is equal to minus D Phi B divided by DT

e is the induced EMF in the coin and Phi B is the magnetic flux the negative sign in this equation

represents the effect as stated in lenses law let us look at Saturday's experiments

and apply Lenz's law to them to identify the polarity of the induced EMF and the direction of induced current in the coin

in this experiment the coil which is connected to a galvanometer is at rest

a bar magnet with it's not full facing the coil is moved towards the coin this causes the magnetic flux through

the coil to increase a current is induced in the koi which opposes this increase in flux

the induced current is in the counterclockwise direction looking from the bar magnet site

this current produces a knock polarity towards the North Pole of the magnet and opposes the motion of the magnet thereby

opposing the increase in magnetic flux through the cony a force is required to move the magnet against this repelling

force the same magnet when moved away from the stationary coil causes the magnetic flux

through the coil to decrease a current is induced in the coil which opposes this decrease in flux

the induced current is in a clockwise direction looking from the bar magnet site

this current produces a south polarity towards the north pole of the receding magnet

this south pole attracts the north pole of the bar magnet thereby resisting the motion of the magnet

when the South Pole of the magnet is towards the coin and it is moved towards the coin a sound polarity is induced

near the South Pole of the magnet when the same magnet is moved away from the coil a north polarity is induced

near the South Pole in all these experiments we can also use an open circuit in the place of a closed

loop in this case to an EMS is induced but there is no induced current in the

coin the direction of this induced e-m-f can be found using lenses law

here is an easy way to understand the direction of induced currents and EMF we have so far taken lenses la for

granted suppose that the induced current is in the direction opposite to the one given

in the figure in that case the South Pole due to the induced current will be near the

approaching North Pole of the magnet causing the bar magnet to be attracted towards the coil

a gentle push on the magnet will start the process of motion and its velocity and kinetic energy will continuously

increase without expending any energy if this is true we can construct a perpetual motion machine with a suitable

arrangement this violates the law of conservation of energy and hence cannot happen

so we can conclude that lenses law is correct in all the experiments discussed a force

is required to move the magnet bar either towards or away from the coil the product of the force and the

distance moved gives the work done on the magnet work is done by spending some mechanical

energy by the person conducting the experiment the mechanical energy is converted to electrical energy in the

coin the electrical energy is converted into heat energy by heating up the coil which

is dissipated into the atmosphere from these observations we can say that the energy is only converted from one

form to another or the energy is concerned we have studied about induced EMF in a

circuit due to the rate of change of magnetic flux to the circuit this happens due to the relative motion

between the coil c1 and the bar magnet we know that the induced EMF E is equal to minus D Phi by DT where D Phi by DT

is the time rate of change of the magnetic flux and the minus sign is in accordance with the Lenz's law

let this be equation 1 now let us study about induced EMF produced due to the motion of a

conductor in a magnetic field consider PQRS to be a rectangular metallic loop having a galvanometer G

connected between R and s the part PQ can slide forwards or backwards in between the parts PS and

our cue of the loop let us now place PQRS in a uniform magnetic field such that its plane is

perpendicular to the magnetic field for a time independent magnetic field the magnetic induction B remains

constant both in magnitude as well as direction let the length of PQ be L and that of

our Q DX let the conductor PQ move towards the ends s start with a constant velocity V

we observed that the conductor PQ the direction of the magnetic induction B and the velocity V of the conductor

are mutually perpendicular to each other we see that the conductor initially is at a distance X from the RS end

as the conductor moves towards the RS end it is displaced from its original position and its velocity V can be

written as DX by DT since the plane of the loop is normal that is perpendicular to the magnetic

field maximum magnetic flux passes through the loop

this magnetic flux Phi be passing through the loop is equal to the product of the magnetic induction B and the area

a of the enclosed loop product of L and X represents the area swept B

hence we can now rewrite the magnetic flux 5b as equal to BL x since a equal to LX

as the conductor moves towards the RS end we see that X is varying and hence the area enclosed by the loop also

changes with the time therefore we see that magnetic flux phi b also changes with time as the magnetic

flux depends on the area enclosed hence according to principle of electromagnetic induction an induced EMF

is developed across PQ this induced EMF E is equal to minus D Phi B by DT as given by equation 1

substituting for phi b we get e is equal to minus d by dt of b LX as B and L are constant and only X is

the variable we can write e is equal to minus BL into DX by DT as DX by DT represents the velocity V of

the conductor P Q we can write electromotive force e is equal to minus B and V let this be equation two

[Music] the induced EMF E is also called the motional EMF in this case

the concept of motional EMF can also be explained by using the Lawrences force acting on the free charge carriers of

conductor PQ we know that when a charge Q enters perpendicular to a uniform magnetic

field of induction B then the Lawrences force on the charge FB is equal to QV b as the conductor moves with a speed V

free electrons in the conductor also moves with speed V in the magnetic field B in addition to the random velocity

they have within the metallic conductor PQ the average magnetic force due to the

random velocity is zero but the average magnetic force on each free electron due to velocity V is FB

equal to QV B where Q is the charge on the electron equal to minus 1 point 6 into 10 power minus 19 coulomb

from Fleming's left hand rule we can say that in this situation the magnetic force on a positive charge moving with

velocity V in this magnetic field we'll be along PQ in accordance with the same room we can

see that the force on the electrons FB will be directed along QP and we see that the three electrons in

the conductor move towards B as a result of this movement the negative charge builds up that P and

a positive charge appears at Q an electric field is developed within PQ such that the electric field exerts a

force F e equal to QE on each electron in electrostatic equilibrium the force F II will be equal to FB and

this build of charge stops since f n becomes equal to FB we have QE equal to QV be simplifying we get e

equal to VB due to the charge buildup a potential difference is developed across PQ which

is the induced EMF II we know that the potential difference e is equal to e into L substituting the value for the

electric field strength e equal to VB we get induced EMF e equal to b lv since Lenz's law is obeyed here also we can

write this induced EMF as e equal to minus b lv this induced EMF E is the motional EMF due to the induced EMF

an induced current flows in the closed loop PQRS and the current can be detected by the galvanometer

this induced current flows along the path QRS B this equation for the motional EMF

derived in this manner is the same as the equation to which we derived earlier now consider the case when the conductor

PQ is stationary in the magnetic field but the magnetic field varies with time that is its magnitude or direction or

both magnitude and Direction are changing with time since the conductor is at rest its velocity V is equal to

zero and as seen earlier the magnetic force F be equal to QV B is zero the moment the magnetic field starts

waving an induced current flows in the loop which can be detected by the galvanometer

since the magnetic field force FB is zero the electrons are forced to move only by the electric field and hence we

conclude that an electric field appears instantaneously this electric field is produced by the

changing magnetic field and is called the induced electric field this electric field causes an EMF to be

developed in the loop and hence an induced current flows in the loop from these observations we can say that

just like the moving charges in the conductor can produce magnetic fields and exert a magnetic force on a magnet

in its field a waving magnetic field can also exert a force on stationary charges

so if we move about magnet towards a stationary charge the charge experiences a force due to

the varying magnetic field this is the fundamental significance of Faraday's discovery

consider a rectangular conducting loop PQRS placed in a uniform magnetic field of strength be the arm PQ of the loop

can be slit over its arms PS and Q are arms PS and Q are of the loop a part of the rails sm + r/n connected to s are

the length of SR is L the length of the arm PS at an instant is x let the electrical resistance of DQ be our

let us assume that the resistances of the rails SM RN and the link SR is negligible compared to the resistance of

arm PQ hence the overall resistance of the rectangular loop is our the flux through the loop at any instant

phi b is equal to the product of the strength of the magnetic field b and the area enclosed e

here area of the loop E is equal to the product of the breadth and length of the loop L and X in this case the

instantaneous flux Phi B is equal to B LX let this be equation 1

from Saturdays law we get induced EMF E is equal to minus D Phi B by DT the negative symbol is in accordance

with Lenz law on substituting phi b from equation 1 and simplifying we arrive at the

equation e is equal to minus d BL X by DT the strength of the magnetic field B and the length of the arm s out of the

loop L are constant the equation transforms to e is equal to minus VL DX by DT

but DX by DT is the instantaneous velocity V of the arm PQ of the loop which is directed towards the sr part of

the loop as shown hence e is equal to minus B LV let this be equation 2

this EMF is the motional EMF as discussed earlier as the resistance of the rectangular

loop is our the magnitude of the induced current I is equal to the magnitude of e by r

the current is directed from P to Q in the arm substituting the value of e from

Equation two we arrive at the equation i is equal to b lv by r at this be equation 3 as the current

carrying on PQ is in the magnetic field there exists a force on the arm the direction of this force acting on

the arm PQ is given by Fleming's left hand rule according to the rule that come the four

finger and the middle finger of the left hand are stretched such that they are mutually perpendicular

if the full finger and the middle finger indicate the directions of the magnetic field and direction of the current

respectively the pump indicates the direction of the force

as per the rule the direction of the force on PQ is opposite to the direction of the motion of the arm PQ

vectorially the force on the arm PQ is given by F is equal to I L cross B and this force arises due to the drift

velocity of charges responsible for the current along the arm and the consequent Laurence's force acting on them

hence the nough to love the force F is equal to IL b sine-theta

where Peter is the angle between the directions of the current and the magnetic field

yeah the angle between the arm PQ and the magnetic field is 90 degrees hence the equation for the magnitude of

the force transforms to f is equal to i lb sine 90 as sign 90 is equal to 1 this can be

rewritten as f is equal to i lb let this be equation for on substituting equation 3 in equation 4

and simplifying we get S is equal to B Square L square V by R let this be equation five

this force arises due to the drift velocity of the charges through the arm PQ that are responsible for the current

and the Lawrences falls on them if we need to maintain uniform velocity for the arm PQ we need to apply a

constant force F in the direction of the velocity the applied force must be equal in

magnitude and opposite in direction to the magnetic force does the power required to push the arm

PQ along the rails is given by the equation P is equal to f3 on substituting the value of F from

equation 5 and simplifying we get the equation b is equal to b square l square v square by r

let this be equation six notice that the work done in moving the arm PQ is mechanical in nature

this mechanical energy is converted into heat energy when the work is done this energy refer to as June heat is

given by P G is equal to I square R on substituting the value of I from Equation 3 and simplifying we get PJ is

equal to B Square L square v square by r let this be equation seven notice that the right hand sides of the equations

six and seven are equal thereby confirming the law of conservation of energy does mechanical energy which was

needed to move the arm PQ is converted into the induced EMF and then to thermal energy

thus we conclude that the induced EMF in a conductor when moved in a magnetic field is in accordance with law of

conservation of energy that is now study the relationship between the flow of charge through the

arm PQ and the slugs from Saturdays law we have learned that the magnitude of the induced e-m-f

modulus e is equal to change in magnetic flux Delta Phi B by delta T let this be equation 8 however the EMF

induced in the circuit is equal to IR but rate of charge flow is current hence current can be replaced by delta q

by delta t is the equation nine the left hand signs of equations eight

and nine are equal equate the right hand sights to arrive at the equation Delta Q is equal to

Delta Phi B by R we use an apparatus that allows a copper plate to swing back and forth draw a

magnetic field like a pendulum the plane in which the copperplate

swings is bad linked to the plane of the plate and normal to the direction of the magnetic field

as the copperplate enters the field the changing magnetic flux induces an EMF in the plate which in turn causes the free

electrons in the plate to move reducing circulating eddy currents note that the magnetic field is directed from

the North Pole to the South Pole of the magnet that provides the field when viewed from the north pole side of

the magnet the induced eddy current is counterclockwise as the plate enters the field

as the plate enters the magnetic field the external magnetic flux through the plate increases

hence by Lenz's long the induced current must provide effective magnetic poles on the plate that are repelled by the poles

of the magnet does here a north pole is induced over the surface of the plate facing the

North Pole of the magnet as the plate leaves the magnetic field the external magnetic flux through the plate

decreases hence again a spur lenses law the induced

current must provide a magnetic South Pole on the surface of the plate facing the North Pole of the magnet

this is quite opposite to the case when the plate enters the magnetic field the induced eddy current always produces

a retarding force FB when the plate enters or leaves the field this retarding force slows down the swinging

of the plate and eventually brings it to rest does the oscillations of the plate are

damped when rectangular slots are cut in the plate eddy currents and the

corresponding retarding force are greatly reduced this reduction in the retarding force is because the

rectangular cuts in the plate prevent the formation of large current loops hence the pendulum plate with holes or

slots reduces electromagnetic damping and the plate swings more freely let us now explore the uses of eddy

currents eddy currents are produced when a metal plate is subjected to a changing

magnetic field heats it hence these currents are unwanted in certain situations

reduction in the surface area of the plate helps introducing eddy currents in the metallic cores of transformers

electric motors and other such devices in which a coil is wound over a metallic core

eddy currents are minimized by using laminations of metal to make a metal call instead of using a single metal

piece the laminations are separated by an insulating material like lacquer

the plane of the laminations is arranged parallel to the magnetic field such that it cuts across the eddy current paths

this arrangement reduces the strength of the eddy currents since the dissipation of electric energy

into heat depends on the square of the strength of the electric current heat loss is substantially reduced

let us now study the applications of eddy currents strong electromagnets are situated above

the rails in some electrically pants trains when the electromagnets are activated

any currents induced in the wheels oppose the motion of the Train as there are no mechanical linkages

taking effect of the Train is smoked certain galvanometers have a fixed core made of a non-magnetic metallic material

when the coil oscillates the eddy currents generated in the core opposed the motion and bring the coil to rest

quickly induction furnace can be used to produce high temperatures and can be utilized to

prepare alloys by melting the Constituent metals a high-frequency alternating current is

passed through a coil which surrounds the metals to be melted the eddy currents generated in the

metals produce high temperatures sufficient to melt it the shiny metal disc in the analog type

electric parameter rotates due to eddy currents electric currents are induced in the

disk by magnetic fields produced by the sinusoidal varying currents in the coin consider a circuit consisting of a coin

having n number of turns connected to a source of EMF in which current can be delete with time

when the source current is in the direction shown the magnetic field directed as shown is set up inside the

COI when the source garden changes with time netic slugs through the coil also

changes and induces an EMF in the coil this phenomenon is called self induction the flux linkage through a coil of

interns is proportional to the current through the coil and is expressed as n Phi B is directly proportional to AI

replacing the proportionality symbol with a constant we get n Phi B is equal to L times the current ie where the

proportionality constant L is called the self inductance of the coil let this be equation 1

[Music] Elle is also called the coefficient of self induction of the coin

when the current in the circuit is varied the flux linked with the Corney changes

and an EMF is induced in the coil this means D by DT of n Phi be equals to L di by DT let this be equation 2

from Faraday's law photo coil of interns the self induced EMF E is equal to minus D by DT of n Phi B

let this be equation three comparing equations two and three we get the equation e is equal to minus L di by

DT that this be equation for [Music]

it is important to know that from lenses law the polarity of this induced EMF is such that it opposes the change in the

magnetic field from the source current [Music] even Saturdays equation concerns that

the self induced e-m-f opposes any change of current in the coil does this self-induced EMS is also

called back EMS [Music] the self induced e-m-f plays the role of

inertia it is the electromagnetic analog to mass in mechanics hence work needs to be performed to

establish a current in a circuit and it is stored as magnetic potential energy for a current ie at an instant in a

circuit the rate of work done is DW by DT is equal to modulus of e times I let this be equation 5

on substituting the magnitude of e from equation four in equation size and simplifying we get DW by DT is equal to

Li di by DT let this be equation six [Music]

the total amount of work done can be established by integrating equation 6 on both sides thus we get integral DW is

equal to integral over 0 to AI Li di [Music] dance the work done or energy consumed

in establishing a current in the circuit is w is equal to half Li square this expression is similar to the

expression for kinetic energy of a particle of mass M moving with velocity V which is given by E is equal to half

MV square this clearly shows that self-inductance l in electromagnetic induction is

analogous to mass m in mechanics thus self-inductance is the measure of electrical inertia it opposes the growth

and decay of the current in a circuit let us now calculate the self-inductance of a long solenoid let the length of the

solenoid be L the current through it bi the number of terms be in and its area of

cross-section be a let the number of turns per unit length of the solenoid be n which is equal to n

by L the magnetic field induced in the solenoid due to the current I is B is

equal to MU knot n I let this be equation seven here we neglect the edge effects of the

solenoid thus assuming the magnetic field in the solenoid to be uniform throughout

the magnetic flux is equal to the product of the magnetic field strength and the area

thus phi b is equal to b a let this be equation eight [Music]

using equation 7 and equation 8 we get the total flux linked with the solenoids is n CB is equal to n L into mu not ni

into a simplifying the expression we get n Phi B is equal to MU naught n square a Li

let this be equation 9 we have already established that M Phi B is equal to L times the current ie

barring this expression with equation nine and simplifying we get the value of L

as L is equal to mu naught n square e n let this be equation 10 this space in the coins of the solenoid

is filled with the material of magnetic permeability nu R then L is equal to mu R mu knot n square

al let this be equation 11 note that the self-inductance of the solenoid depends on the geometry and the

permeability of the medium an AC generator converts mechanical energy into electrical energy

in its simplest form an AC generator consists of a rectangular coil mounted on a rotor shaft

the coil is also referred to as the armature the coned is placed in a uniform

magnetic field the axis of the coil is normal to the direction of the magnetic field

the koi all the armature is mechanically rotated in the uniform magnetic field by some external means

let the number of turns in the coil be n the ends of the coil are connected to an external circuit by means of slip rings

and brushes the coin rotates inducing an EMF and current in its sense

the current induced in the koi is fed into external circuits through the slip rings that act as terminals to the

generator let us now find the expression for the EMF generated by an AC generator

consider the area vector e for the coil the area vector is normal to the plane of the coy

initially let the area vector lie in the direction of the magnetic field this implies that plane of the coil is

normal to the direction of the magnetic field let Peter be the angle between the area

vector E and the magnetic field B had to give an instant of time T when the coil is rotated at a constant

angular velocity Omega the value of Peter at the instant of time tea

is given by the equation beta is equal to Omega multiplied by time T let this be equation 1

the effective area of the coil which is linked with the magnetic field is e cos theta

from Equation one it is clear that the effective area of the coil is a time dependent function

in other words the rotation of the koi causes the magnetic flux through it to change

thereby inducing an EMF in the coil the flux induced in the coil and any instant of time is given by the equation

Phi B is equal to B a cos theta let this be equation 2 on substituting equation 1 in equation 2

and simplifying we get the equation Phi B is equal to B a cos Omega T

let this be equation three from Saturdays law the induced EMF for a coil of n turns rotating in a magnetic

field is given by E is equal to minus nd Phi B by DT the negative sign is in accordance with

lenses law on substituting the value of Phi B from equation three and simplifying we get e

is equal to minus n D by DT of B a cos Omega T strength of the magnetic field B and the

area of the coil a are constants the equation now transforms to e is equal to minus NBA D by DT of cos Omega

T from differential calculus we have D by DT of course Omega T is

equal to minus Omega sine Omega T hence the instantaneous value of the EMF e is equal to NBA Omega sine Omega T

let this be equation for note that NBA Omega is the maximum value of the EMF that occurs when the value of

sine Omega T is plus or minus one that is when theta is 90 degrees or 270 degrees this is because the

change of flux is the maximum at these two angles of the coin as the value of sine function varies

from plus 1 to minus 1 the sign all polarity of the EMF changes with time

accordingly the direction of the induced current changes periodically for this reason this current is called

an alternating current let us denote the maximum EMF MBA Omega as a note

then the instantaneous EMF is equal to e naught sine omega-t let this be equation five

we know that Omega is equal to two pi nu when you is the frequency of revolution of the generators coil

substituting the value of omega in equation 5 we get is equal to e naught sine 2 by nu T

let this be equation six dance equation six gives the expression for

the instantaneous value of the EMF earlier we have learned Oh an electric current is induced by the relative

motion between a closed coil and a magnet [Music]

electric currents can also be induced in a closed coil by causing a change in the magnetic flux produced by its relative

motion with another current carrying coil in its vicinity or by a flux change produced by the same coin

let us first understand the experimental setup that helps us demonstrate this phenomenon

the coil on the left is called a primary coil and is connected to a battery let the current in this coil be i1

on the right is called the secondary coil and it is connected to a galvanometer

the primary and the secondary coils are placed coaxial and an iron core runs along the axes of

the coils let B be the magnetic field strength due to the current in the primary coil

when the secondary coil is moved towards the primary coil the magnetic flux through the secondary coil increases

as a consequence of the increasing flux in the secondary coil the current induced in it increases as

it moves towards the primary coin let the current induced in this coid be I

when the secondary coil moves away from the primary coil the magnetic flux through the secondary coil reduces as a

consequence of the decreasing magnetic flux the current in the secondary coil decreases

the magnetic flux in the secondary coil phi b is directly proportional to the current

i in it the variations of the current in the secondary coil are directly proportional

to the variations in the magnetic flux does the rate of change of flux is directly proportional to the rate of

change of current let the number of terms in the secondary coil be in

hence the net flux change in the secondary coil and scipy is directly proportional to I

the term M Phi B is called flux linkage [Music] the constant of proportionality em

between the flux linkage and the current is called the mutual inductance of the coil

inductance is a scalar quantity and has the dimensions M L Square t power minus 2 e power minus 2

the SI unit of inductance is Henry named after the scientist Joseph Henry and is denoted by age

[Music] one Henry is equal to 1 Tesla meter square per ampere

so far we have explored the experimental setup with two coils of the same radius placed besides one another

let us now consider an example in which two coaxial solenoids induce current into one another

you consider to co-axial solenoids s1 and s2 all slend l-let the radius of the inner

solenoid s1 be r1 and the number of turns per unit length be n1 let the radius of the outer solenoid s

to be r2 and the number of turns per unit length by n2 let n1 and n2 be the total number of

turns in quois s1 and s2 dance we have N 1 is equal to n 1l and n 2 is equal to n2 n

when a current i2 is passed through the coil s to the magnetic flux is set up within the coil and this flux is linked

with the COI s one let us denote this magnetic flux by phi1 let this be the flux button of the coin

then the total flux linkage with solenoid s1 is n 1 Phi 1 equal to m1 2 I 2

let this be equation 1 m12 is called the mutual inductance of solenoid s1 with respect to solenoid s2

it is also referred to as the coefficient of mutual induction the magnetic flux is equal to the

product of the magnetic field strength into the area the magnetic field strength be - due to the current i2 in

solenoid s2 is given by B 2 equal to MU knot n - I - let this be equation 2 area of the coil in solenoid s 2a is

equal to PI R 1 square let this be equation three multiplying equations 2 & 3 we get the

value of the magnetic flux in the coil of solenoid s 1 as Phi 1 equal to MU knot n 2 I 2 into PI R 1 square

let this be equation for also we know that n1 is equal to n1 L

let this be equation size on multiplying equations 4 and 5 and simplifying the right-hand side

we get the expression n 1 Phi 1 equal to MU knot n 1 n 2 pi R 1 square L I 2 let this be equation six

equating the right-hand-side expressions of equation 6 and equation 1 and simplifying

we get em 1 2 is equal to mu not n 1 n 2 PI R 1 square L let this be equation 7 note that for these calculations

we neglected the edge effects and consider the magnetic field to be uniform throughout the length and width

of the solenoid s2 also solenoids of equal lengths were chosen

this is a good approximation keeping in mind that the solenoid is long incline L is much greater than R two

[Music] let us now consider the case in which current i1 is passed through the

solenoid s1 this current produces magnetic which is also linked with the solenoid

s2 let the magnetic flux Burton linked with solenoid s to be Phi 2

then the total flux linkage with solenoid s 2 is n 2 Phi 2 is equal to M 2 1 i1

let this be equation eight m21 is called the mutual inductance of solenoid s 2 with respect to solenoid s

1 it is also referred to as the coefficient of mutual induction

the magnetic flux is equal to the product of the magnetic field strength into the area the magnetic field b 1 due

to the current i1 in solenoid s 1 is given by B 1 is equal to MU knot n 1 i1 let this be equation 9

[Music] area of the coil in solenoid s1 e is equal to PI add 1 square

let this be equation 10 here we assume the flux due to current i1 through s1 is completely consigned to

the solenoid s1 only this is because the solenoids are very long

on multiplying equations nine and ten we obtain the magnetic flux per turn through the coil of solenoid s 2 as Phi

2 equal to mu knot n 1 I 1 into PI R 1 square let this be equation 11

we also know that n 2 is equal to n 2 l that is the equation 12 on multiplying equations 11 and 12 and simplifying the

right-hand side we get the equation n 2 Phi 2 is equal to MU naught n 1 n 2 pi R 1 square L I 1

let this be equation 13 equating the right hand side of equation 13 and equation 8 and simplifying we get

the expression M 2 1 is equal to MU naught n 1 n 2 pi R 1 square L let this be equation 40 comparing equation 7 and

14 we get M 1 2 is equal to M 2 1 let M 1 2 is equal to M 2 1 be equal to M which is equal to MU naught and 1 n 2 pi

R 1 square L note that the mutual inductance is

neither dependent on the current nor the magnetics nuts it is dependent on the geometrical

dimensions of the solenoid this expression is true when the medium within the solenoids is air

if the medium within the solenoids has relative permeability mu R then M is equal to mu naught mu R n 1 n 2 pi R 1

square L [Music]

this equality m12 is equal to m2 one moon's good for long coaxial solenoids and is far more general

even though the in a solenoid s1 is much smaller than the outer solenoid s2 calculation of flux linkage within a

solenoid is possible as it is effectively immersed in the uniform magnetic field due to the current

through the outer solenoid does calculation of mutual inductance m12 is easy

in this case it is difficult to find the flux linkage without a solenoid due to the current through the inner solenoid

this is because the magnetic field due to the inner solenoid would vary across the length as well as the cross section

of the outer solenoid does calculation of mutual inductance m21 becomes difficult

let us now take a look at how a change of current effects mutual inductance [Music]

let us consider an experimental setup in which there is a coin see one that is connected to a galvanometer G

and another koi see two connected to a battery EMF is induced in coil c1 wherever there

is any change in current through coil c2 let's find one be the flux through coil see one of N 1 turns when current in

coil C 2 is i2 a 10-1 v 1 equal to M I to the rate of change of current is equal to the rate

of change of flux this means that DN 1 Phi 1 by DT is equal to DM i 2 by DT

the EMF induced in coil C 1 is given by e 1 is equal to minus T n1 Phi 1 by DT this implies that e1 is equal to minus M

di2 by DT remember only flux and current change hence M is a constant

this equation shows letter veering current in a coin can induce EMS in a neighboring coin

let us now solve an example to reinforce our understanding of mutual inductance a current-i passes through the outer

ring of two coplanar concentric loops of radii r1 and r2 our one is much greater than R 2

what is the mutual inductance between the two loops the magnetic field induced by a

current-i at the center of the ring of radius r1 is B 1 equal to MU knot I by 2 r1

the flux through the ring of radius r2 is given by Phi 2 1 equal to B 1 e 2 this implies that Phi 2 1 is equal to MU

knot I by 2 R 1 into PI R 2 square mutual inductance M is given by sight to one by I on substituting the value of

side to one and simplifying we get the value of mutual inductance as mu not PI R 2 square by 2 R 1

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[Music] you