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Chapter 27 Electromagnetic Induction

- Faradys Law

- Discovery of Faradys law of induction

- Faradys Law

- Faradys 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

- The flux through the circuit may be changed in

several different ways - B may be made more intense.
- The coil may be enlarged.
- The coil may be moved into a region of stronger

field. - The angle between the plane of the coil and B may

change.

- Faradys Law

- Faradys law of induction (contd)

- Faradys Law

- Induced electric field

Consider work done in moving a test charge

around the loop in one revolution of induced emf.

Work done by emf

Work done by electric field

emf 2prE for a circular loop

For a circular current loop

In general

Faradays law rewritten

- Lenzs Law

- Direction of induced emf and Lenzs law

Extend Faradys law to solenoids with N turns

Why the minus sign and what does it mean?

Number of turns

Lenzs Law

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.

or the induced emf and induced current are in

such a direction as to oppose the change that

produces them!

- Lenzs Law

- Example 1

- Lenzs 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

FE

FB

- Motional Electromotive Force

- Origin of motional electromotive force I (contd)

- Motional Electromotive Force

- Origin of motional electromotive force II

v constant

B

.

Bind

- Motional Electromotive Force

- Origin of motional electromotive force II

(contd)

- Motional Electromotive Force

- Origin of motional electromotive force II

(contd)

- Motional Electromotive Force

- Origin of motional electromotive force II

(contd)

- Motional Electromotive Force

- Origin of motional electromotive force II

(contd)

- Motional Electromotive Force

- Origin of motional electromotive force III

- Motional Electromotive Force

- Origin of motional electromotive force III

(contd)

- Motional Electromotive Force

- Origin of motional electromotive force III

(contd)

- 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

Also we have

- Motional Electromotive Force

- Example A bar magnet and a loop (contd)

loop

For the example just considered, let us see what

happens in a small interval dt. The relative

displacement ?loopdt causes a small area of B

field to enter the loop. For a length dL of the

loop the ddFB passing inside is d(dA)B dL

?loopdt sin? B. We can see this as

Integrating this expression right round the

circuit ( i.e. over dL) shows that this ?B

interpretation recovers Faradays law. You will

also see that the sign of E is consistent with

Lenzs Law.

- Motional Electromotive Force

v

v

- Example A generator (alternator)

top

The armature of the generator is rotating in a

uniform B field with angular velocity ? this can

be treated as a simple case of the E ?B field.

On the ends of the loop ?B is perpendicular to

the conductor so does not contribute to the emf.

On the top ?B is parallel to the conductor and

has the value E ?B cos ? ?RB cos ?t. The

bottom conductor has the same value of E in the

opposite direction but the same sense of

circulation.

B

bottom

- Eddy Current

- Eddy current Examples

- Eddy Current

- Eddy current Examples (contd)

- 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 ring In the field as shown below. What

does Faradays Law say will happen?

Current will flow in the ring. What will happen

If there is no ring present?

Now consider a hypothetical path Without any

copper ring.There will be an induced Emf

with electric field lines as shown above.

In fact there will be many concentric circles

everywhere in space.

The red circuits have equal areas. Emf is the

same in 1 and 2, less in 3 and 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?

E is parallel to dl

Notice that there is no wire or loop of wire. To

find E use Faradays Law.

Example with numbers

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

Find E at r 5.2 cm

Find E at 12.5 cm

- 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 FB). 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 FB when I 1 amp.

The unit of inductance is the henry

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,

So L is proportional to n2 and the volume of the

solenoid

Self Inductance

- Calculation of self inductance A solenoid

(contd)

Example the L of a solenoid of length 10 cm,

area 5 cm2, with a total of 100 turns is

L 6.2810-5 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

Self Inductance

- Calculation of self inductance A toroidal

solenoid

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

Then when the current increases uniformly from

0.0 to 6.0 A in 3.0 ms, the self-induced emf E

will be

- 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.

When I is first established, we have a finite

(self-induced emf)

The source of I does work against the

self-induced emf in order to raise I to its final

value.

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

- Self Inductance

- Inductor

A circuit device that is designed to have a

particular inductance is called an inductor or a

choke. The usual symbol is

a

I

variable source of emf

L

b

- Mutual Inductance

- Transformer and mutual inductance

The 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 F2 is

given by

M2 1Mutual Inductance of the coils

Generally, M 1 2 M 2 1

Also

- Mutual Inductance

- Changing current and induced emf

Consider two fixed coils with a varying current

I1 in coil 1 producing magnetic field B1. The

induced emf in coil 2 due to B1 is

proportional to the magnetic flux through coil 2

f2 is the flux through a single loop in coil 2

and N2 is the number of loops in coil 2. But we

know that B1 is proportional to I1 which means

that F2 is proportional to I1. The mutual

inductance M is defined to be the constant

of proportionality between F2 and I1 and depends

on the geometry of the situation.

The induced emf is proportional to M and to the

rate of change of the current .

- Mutual Inductance

- Example

Now consider a tightly wound concentric solenoids.

Assume that the inner solenoid carries current

I1 and the magnetic flux on the outer solenoid

FB2 is created due to this current. Now the flux

produced by the inner solenoid is

The flux through the outer solenoid due to this

magnetic field is

- Mutual Inductance

- Example of inductor Car ignition coil

Two ignition coils, N116,000 turns, N2400 turns

wound over each other. l10 cm, r3 cm. A current

through the primary coil I13 A is broken in 10-4

sec. What is the induced emf ?

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 lt 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.

Kirchhoffs Loop Rule

Power balance

Multiply by I

- The R-L Circuit

- Current growth in an R-L circuit (contd)

Rate at which energy is stored up in the inductor.

Power supplied by the battery

Power dissipated as heat in the resistor

If energy in inductor is

then

or

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

So, the energy stored in an inductor carrying

current I is

- The R-L Circuit

- Current growth in an R-L circuit (contd)

Kirchhoffs Loop Rule

I then increases until finally dI/dt 0

At t 0, I 0

Compare with

Current in an LR circuit as function of time

- The R-L Circuit

- Current growth in an R-L circuit (contd)

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

t)

- The R-L Circuit

- Current growth in an R-L circuit (contd)

Now we raise e to the power of each side

- The R-L Circuit

- Discharging an R-L circuit

Add switch S2 to be able to remove the battery.

And add R1 to protect the battery so that it is

protected when both switches are closed.

First S1 has been closed for a long enough time

so that the current is steady at its final value

I0.

At t0, close S2 and open S1 to effectively

remove the battery. Now the circuit abcd carries

the current I0.

Kirchhoffs loop rule

- The R-L Circuit

- Discharging an R-L circuit (contd)

Now lets calculate the total heat produced in

resistance R when the current decreases from I0

to 0.

Rate of heat production

Energy dissipated as heat in the resistor

The current as a function of time

The total energy

The total heat produced equals the energy

originally stored in the inductor

- The L-C Circuit

- Complex number and plane

real part Re(z)x, imaginary part Im(z)y

Complex number z x iy

- The L-C Circuit

- Simple harmonic oscillation

- The L-C Circuit

- Simple harmonic oscillation (contd)

- The L-C Circuit

- Simple harmonic oscillation (contd)

- The L-C Circuit

- An L-C circuit and electrical oscillation

S

Consider a circuit with an inductor and

a capacitor as shown in Fig. Initially

the capacitor C carries charge Q0

At t0 the switch closes and charge flows through

inductor producing self-induced emf.

The current I is by definition

Acceleration equation for a mass on a spring

Kirchhoffs loop rule

- The L-C Circuit

- An L-C circuit and electrical oscillation

(contd)

The solution of this equation is simple harmonic

motion.

Now lets figure out what A and f are. For that

choose initial condition as I(0)0 and Q(0)Q0.

Then AQ0 and f0.

The charge and current are 90o out of phase with

the same angular frequency w I is at maximum when

Q0, and Q is at maximum when I0.

- The L-C Circuit

- An L-C circuit and electrical oscillation

(contd)

The charge and current are 90o out of phase with

the same angular frequency w I is at maximum when

Q0, and Q is at maximum when I0.

-I(t)

- The L-C Circuit

- An L-C circuit and electrical oscillation

(contd)

The electric energy in the capacitor

The electric energy oscillates between its

maximum Q02 and 0.

The magnetic energy in the inductor

The magnetic energy oscillates between its

maximum Q02 /(2C) and 0.

UtotUeUm constant

Ue(t)

Um(t)

- The L-R-C Circuit

- Another differential equation

- The L-R-C Circuit

- Another differential equation (contd)

- The L-R-C Circuit

- Another differential equation (contd)

- The L-R-C Circuit

- An L-R-C circuit and electrical damped

oscillation

At t0 the switch is closed and a capacitor with

initial charge Q0 is connected in series across

an inductor.

Initial condition

A loop around the circuit in the direction of

the current flow yields

Since the current is flowing out of the capacitor,

- The L-R-C Circuit

- An L-R-C circuit and electrical damped

oscillation (contd)

If R2lt 4LC, the solution is

Note that if R0,no damping occurs.