Chapter 27: Electromagnetic Induction Farady’s Law Discovery of Farady’s law of induction.

Chapter 27: Electromagnetic Induction

Farady’s Law

Discovery of Farady’s law of induction

Farady’s Law

Farady’s law of induction

An emf in volts is induced in a circuit that is equal to the time rate of change of the total magnetic flux in webers threading (linking) the circuit:

Bd

dt

The flux through the circuit may be changed in several different ways

1) B may be made more intense.

2) The coil may be enlarged.

3) The coil may be moved into a region of stronger field.

4) The angle between the plane of the coil and B may change.

Farady’s Law

Farady’s law of induction (cont’d)

sd

Farady’s Law

Induced electric field

Work done by emf : 0qemfW Work done by electric field: )2(00 rEqsdEqsdF

Consider work done in moving a test charge around the loop in one revolution of induced emf.

sdEemf

dt

dsdE B

Faraday’s law rewritten

emf = 2rE for a circular loop

In general :For a circular currentloop

Lenz’s Law

Direction of induced emf and Lenz’s law

BdN

dt Why the minus sign and

what does it mean?

The sign of the induced emf is such that it tries to produce a current that would create a magnetic flux to cancel (oppose) the original flux change.

Lenz’s Law

or – the induced emf and induced current are in such a direction as to oppose the change that produces them!

Extend Farady’s law to solenoids with N turns:

Number of turns

Lenz’s Law

Example 1

B due to induced current

B due to induced current

Lenz’s Law

Example 2

• The bar magnet moves towards loop.

• The flux through loop increases, and an emf induced in the loop produces current in the direction shown.

• B field due to induced current in the loop (indicated by the dashed lines) produces a flux opposing the increasing flux through the loop due to the motion of the magnet.

Motional Electromotive Force Origin of motional electromotive force I

FB

FE

Motional Electromotive Force Origin of motional electromotive force I (cont’d)

Motional Electromotive Force Origin of motional electromotive force II

B.Bind

v: constant

Motional Electromotive Force Origin of motional electromotive force II (cont’d)


generalin )( sdBd

loop conducting closed afor emf motional : )( sdB

E

E

Motional Electromotive Force Origin of motional electromotive force III

Motional Electromotive Force Origin of motional electromotive force III (cont’d)

Motional Electromotive Force

A bar magnet and a loop (again)

In this example, a magnet is being pushed towards a closed loop.

The number of field lines linking the loop is evidently increasing.

There is relative motion between the loop and the field lines and an observer at any point in the metal of the loop, or the charges in the loop, will see an E field

E υ Bobs ��

Also we have

E ds the Faraday emf. ��


Example: A bar magnet and a loop (cont’d)

For the example just considered, let us see what happens in a small interval dt. The relative displacement υloopdt causes a small area of B field to enter the loop. For a length dL of the loop the ddΦB passing inside is d[(dA)B] = dL υloopdt sinθ B. We can see this as

loop loop( ) ddA B (dL υ ) B dL (υ B)

dL E

B

B

d d dt dt

dd d

dt

��

��

Integrating this expression right round the circuit ( i.e. over dL) shows that this υ×B interpretation recovers Faraday’s law. You will also see that the sign of E is consistent with Lenz’s Law.

loop


Example: A generator (alternator)

The armature of the generator is rotating in a uniform B field with angular velocity ω this can be treated as a simple case of the E = υ×B field.

On the ends of the loop υ×B is perpendicular to the conductor so does not contribute to the emf. On the top υ×B is parallel to the conductor and has the value E = υB cos θ = ωRB cos ωt. The bottom conductor has the same value of E in the opposite direction but the same sense of circulation.

AB E ds 2 cos cos the Faraday emf.LRB t AB t ��

top

bottom

vv

B

Eddy Current

Eddy current: Examples

Eddy Current

Eddy current : Examples (cont’d)

Eddy Current

Eddy current prevention

The orange represents a magnetic field pointing into the screen and let say it is increasing at a steady rate like 100 gauss per sec. Then we put a copper ringIn the field as shown below. What does Faradays Law say will happen?

Current will flow in the ring. What willhappen If there is no ring present?

Now consider ahypothetical pathWithout any copper ring.There will be an induced Emf withelectric field lines asshown above.

In fact therewill be many concentriccircles everywhere in space.

The red circuitshave equal areas.Emf is the same in 1 and 2, less in 3and 0 in 4. Note no current flows. Therefore, no thermal energy is dissipated

Example:A magnetic field is to the board (screen) and uniform inside a radius R. The magnetic field is increasing at a steady rate. What is the magnitude of the induced field at a distance r from the center?

Notice that there is no wire or loop of wire. To find E use Faraday’s Law.

dt

drEEdl m 2

2rBBAm

dt

dBrrB

dt

d

dt

d m 22 )(

dt

dBrrE 22

Rrdt

dB

r

RE

2

2

Rrdt

dBrE

2

E is parallel to dl

R

x xx

x

x

xxx

x

x r

ld

Field circulates around B field

B

dt

dBRrE 22

Example with numbers

Suppose dB/dt = - 1300 Gauss per sec and R= 8.5 cm

Rrdt

dBrE

2

Find E at r = 5.2 cm

Find E at 12.5 cm

Rrdt

dB

r

RE

2

2

E (0.085m)2

2(0.125m)0.13T 0.0038 V /m 3.8mV /m

E (0.052m)

20.13T 0.0034 V /m3.4mv /m

Self Inductance Self induction

When a current flows in a circuit, it creates a magnetic flux which links its own circuit. This is called self-induction. (‘Induction’ was the old word for the flux linkage ΦB).

The strength of B is everywhere proportional to the I in the circuit so we can write

L is called the self-inductance of the circuit

L depends on shape and size of the circuit. It may also be thought as being equal to the flux linkage ΦB when I = 1 amp.

The unit of inductance is the henry

2Wb T m1 H 1 1

A A

LIB

Self Inductance Calculation of self inductance : A solenoid Accurate calculations of L are generally difficult. Often the answer depends even on the thickness of the wire, since B becomes strong close to a wire.

In the important case of the solenoid, the first approximation result for L is quite easy to obtain: earlier we had

Hence

Then,

IN

B0 I

ANNABB

2

0

AnAN

IL B 2

0

2

0

lengthunit per

turnsofnumber the: n

So L is proportional to n2 and the volume of the solenoid

Self Inductance Calculation of self inductance: A solenoid (cont’d)

Example: the L of a solenoid of length 10 cm, area 5 cm2, with a total of 100 turns is

L = 6.28×10−5 H

0.5 mm diameter wire would achieve 100 turns in a single layer.

Going to 10 layers would increase L by a factor of 100. Adding an iron or ferrite core would also increase L by about a factor of 100.

The expression for L shows that μ0 has units H/m, c.f, Tm/A obtained earlier

lengthunit per

turnsofnumber the: n

An

ANL 2

0

2

0

Self Inductance Calculation of self inductance: A toroidal solenoid

)2(2

20

2

0 rAnr

AN

I

NL B

r

NIABAB

20

The magnetic flux inside the solenoid:

Then the self-inductance of the solenoid:

If N = 200 turns, A = 5.0 cm2 , and r = 0.10 m:

H. 40H 1040

m) 10.0(2

)m 100.5(m)](200) Wb/(A104[

6

2427

L

Then when the current increases uniformly from 0.0 to 6.0 A in 3.0 s,the self-induced emf will be:

V. 80A/s) 10H)(2.0 1040( 66

dt

dIL )(

dt

d B

Self Inductance Stored energy in magnetic field

Why is L an interesting and very important quantity?

This stems from its relationship to the total energy stored in the B field of the circuit which we shall prove below.

The source of I does work against the self-induced emf in order to raise I to its final value.

When I is first established, we have a finite

(self-induced emf)

I

m

U

m ULIIdILdUm

0

2

0 2

1

2

2

1LIUm

dt

dIL

dt

d B

dt

dILII

dt

dUm power = work done per unit time

Self Inductance Stored energy in magnetic field: Example

Returning to our expression for the energy stored in an inductance we can use it for the case of a solenoid. Using formulae we have already obtained for the solenoid

and

Hence:

Energy per unit volume in the field

nIB 0 AnI

L B 20

AB

n

BAnLIUm

0

22

0

20

2

22

1

2

1

0

2

2B

A

Uu m

m

Self Inductance Inductor

A circuit device that is designed to have a particular inductance is calledan inductor or a choke. The usual symbol is:

L

I

a

b

variablesource of emf

b toa from increases potential 0

,0/ If

b toa from drops potential0

,0/ If

ab

ab

baab

V

dtdI

V

dtdIdt

dILVVV

Mutual Inductance

Transformer and mutual inductanceThe classic examples of mutual inductance are transformers for power conversion and for making high voltages as in gasoline engine ignition.

A current I1 is flowing in the primary coil 1 of N1 turns and this creates flux B which then links coil 2 of N2 turns.

The mutual inductance M2 1 is defined such that the induction Φ2 is given by

M2 1—Mutual Inductance of the coils

Also Generally, M 1 2 = M 2 1

121222 IMIL

212111 IMIL

Mutual Inductance Changing current and induced emf

1

2

11

1

222 ;

dI

dM

dt

dIM

dt

dI

dI

d

dt

d

The induced emf is proportional to M and to the rate of change of thecurrent .

Consider two fixed coils with a varying current I1 in coil 1 producingmagnetic field B1. The induced emf in coil 2 due to B1 is proportionalto the magnetic flux through coil 2: 22212 NAdB

is the flux through a single loop in coil 2 and N2 is the number of loopsin coil 2. But we know that B1 is proportional to I1 which means that 2 isproportional to I1. The mutual inductance M is defined to be the constant ofproportionality between 2 and I1 and depends on the geometry of the situation.

1

22

1

2

I

N

IM

Mutual Inductance

ExampleNow consider a tightly wound concentricsolenoids. Assume that the inner solenoidcarries current I1 and the magnetic fluxon the outer solenoid is created dueto this current. Now the flux producedby the inner solenoid is:

/ where 111101 NnInB The flux through the outer solenoid due to this magnetic field is:

12

11202

112112 )()(2

IrnnrBNABNB

. generalin ; )( 12212

11201

212 MMMrnn

IM B

Mutual Inductance

Example of inductor: Car ignition coil

Two ignition coils, N1=16,000 turns, N2=400 turns wound over each other.=10 cm, r=3 cm. A current through the primary coil I1=3 A is broken in10-4 sec. What is the induced emf ?

1-41

21120

121

1122

s 103

)(; 2

Adt

dI

rnnI

Mdt

dIM B

V 000,62 Spark jumps across gap in a spark plug and ignites a gasoline-air mixture

The R-L Circuit Current growth in an R-L circuit

Consider the circuit shown. At t < 0 the switch is open and I = 0.

The resistance R can include the resistance of the inductor coil.

The switch closes at t = 0 and I begins to increase, Without the inductor the full current would be established in nanoseconds. Not so with the inductor.

Kirchhoff’s Loop Rule: 0 0dI

IR Ldt

Multiply by I:2

0dI

I I R LIdt

Power balance

The R-L Circuit Current growth in an R-L circuit (cont’d)

20

dII I R LI

dt Power supplied

by the battery

Power dissipated as heat in the resistor

If energy in inductor is mU

mdU dILI

dt dt

then:

or mdU L I dI

Integrate from t = 0 (I = 0) to t = (I = If)

2

0 0

1

2

mf fU I

mf m fU dU L I dI LI

So, the energy stored in an inductor carrying current I is :

21

2mU LI

Rate at which energy is stored up in the inductor.


Kirchhoff’s Loop Rule:

0 0dI

IR Ldt

0

0

dI

dt L

I then increases until

finally dI/dt = 0

0fI R

Current in an LR circuit as function of time

At t = 0+, I = 0

Compare with:


0L dII

R dt R

0

dI Rdt

LI

R

Integrating between (I = 0, t = 0) and (I = I, t = t)

0

0

/ln

/

I R Rt

R L


Now we raise e to the power of each side

0

0

/

/

Rt

LI Re

R

0 0Rt

LI eR R

0 1Rt

LI eR

The R-L Circuit Discharging an R-L circuit

Add switch S2 to be able to removethe battery. And add R1 to protectthe battery so that it is protectedwhen both switches are closed.

First S1 has been closed for a longenough time so that the current issteady at its final value I0.

At t=0, close S2 and open S1 to effectively remove the battery. Now the circuit abcd carries the current I0.

Kirchhoff’s loop rule:

0dt

dILIR

LRteII /0

The R-L Circuit Discharging an R-L circuit (cont’d)

Now let’s calculate the total heat produced in resistance R when the currentdecreases from I0 to 0.

Rate of heat production: RIdt

dWP 2

Energy dissipated as heat in the resistor:

0

2RdtIdWW

The current as a function of time:LRteII /

0

The total energy: 20

/220 2

1LIRdteIW LRt

The total heat produced equals the energy originally stored in the inductor

The L-C Circuit Complex number and plane

Complex number : z = x + iy real part Re(z)=x, imaginary part Im(z)=y

The L-C Circuit Simple harmonic oscillation

The L-C Circuit Simple harmonic oscillation (cont’d)

Acceleration equation for a mass on a spring

The L-C Circuit An L-C circuit and electrical oscillation

SConsider a circuit with an inductor and acapacitor as shown in Fig. Initially thecapacitor C carries charge Q0

At t=0 the switch closes and charge flowsthrough inductor producing self-induced emf.

dt

dIL

The current I is by definition: dt

dQI

Kirchhoff’s loop rule: 0C

Q

dt

dIL

0d

c.f. 02

2

2

2

xm

k

dt

x

C

Q

dt

QdL

The L-C Circuit An L-C circuit and electrical oscillation (cont’d)

The solution of this equation is simple harmonic motion.

xxm

k

dt

xdQQ

LCtd

Qd 22

2

2

2

c.f. 1

)cos( c.f. )cos( tAxtAQ

Now let’s figure out what A and are. For that choose initial conditionas: I(0)=0 and Q(0)=Q0. Then A=Q0 and

)2/cos()sin()(,)cos()( 000 tQtQtItQtQ

The charge and current are 90o out of phase with the same angular frequency is at maximum when Q=0, and Q is at maximum when I=0.


)2/cos()sin()(,)cos()( 000 tQtQtItQtQ

The charge and current are 90o out of phase with the same angular frequency is at maximum when Q=0, and Q is at maximum when I=0.

-I(t)


The electric energy in the capacitor:

)(cos2

1

2

1

2

1 220

2

tC

Q

C

QQVU ce

The electric energy oscillates between its maximum Q02 and 0.

The magnetic energy in the inductor:

LCt

C

QtQLLIUm

1)(sin

2

1)(sin

2

1

2

1 22022

022

The magnetic energy oscillates between its maximum Q02 /(2C) and 0.

Utot=Ue+Um constant

Ue(t) Um(t)

The L-R-C Circuit Another differential equation

The L-R-C Circuit Another differential equation (cont’d)

The L-R-C Circuit An L-R-C circuit and electrical damped oscillation

At t=0 the switch is closed and a capacitorwith initial charge Q0 is connected in seriesacross an inductor.

Initial condition: 0)0(;0 0 IQQ

A loop around the circuit in the direction of thecurrent flow yields:

0 IRdt

dIL

C

Q

Since the current is flowing out of the capacitor,dt

dQI

01

2

2

QLCdt

dQ

L

R

dt

Qd

The L-R-C Circuit An L-R-C circuit and electrical damped oscillation (cont’d)

If R2< 4LC, the solution is:

LCLRteQtQ LRt /1 and )]2/([' where'cos)( 22)2/(0

Note that if R=0,no damping occurs.

Chapter 27: Electromagnetic Induction Farady’s Law Discovery of Farady’s law of induction.

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Transcript of Chapter 27: Electromagnetic Induction Farady’s Law Discovery of Farady’s law of induction.