Chapter 6 Electromagnetic Induction

Law of Electromagnetic Induction Inductances

Energy and Force

1. Law of Electromagnetic Induction

2. Inductances

3. Energy in Steady Magnetic Fields

4. Magnetic Forces

1. Law of Electromagnetic Induction

From physics we know that when the magnetic flux through a

closed coil is changing, an induced electromotive force e will be

generated in the coil, with the relation

te

d

d

where the positive direction of the electromotive force e and that of

the magnetic flux comply with the left hand rule.

If the magnetic flux is increased with time, the direction of the

induced electromotive force and that of the magnetic flux obey the left

hand rule. if the magnetic flux is decreased with time, they will obey

the right hand rule.

e I

The induced magnetic flux caused by the

induced current in the coil always resists the

change of the original magnetic flux. The

induced magnetic flux is called the reaction

magnetic flux, and the induced electromotive

force is called the back electromotive force.

When the induced electromotive force is generated in the coil,

there is an electric field to push the charges to move in the coil, and

this induced electric field is denoted as E.

te

l d

dd

lE

The line integral of the induced electric field intensity around

the closed coil is equal to the induced electromotive force in the coil

e, i.e.

Considering , we have S

SB d

Sl tSBlE dd

Which is called the law of electromagnetic induction, and it shows that

when the magnetic field through a closed coil is changing, an induced

electric field will be generated in the coil.

The law of electromagnetic induction shows that a time-varying

magnetic field can produce a time-variable electric field.

Based on Stokes’ theorem, from the above equation we have

0d)(

S

BE

S t

Since the equation holds for any area S, the integrand must be

zero, so that

t

BE

which is called the differential form of law of electromagnetic

induction, and it means that the negative time rate of change of

the magnetic flux density at a point is equal to the curl of the time-

variable electric field intensity at that point.

The law of electromagnetic induction is one of basic laws

for time-varying electromagnetic fields, and it is also one of

Maxwell’s equations.

t

BE

2. Inductances

In a linear medium, the magnetic flux through the closed circuit

is also proportional to the current I.

IL

It is called the inductance of the circuit, with the unit henry (H), a

nd the inductance can be also considered as the magnetic flux linkage

per unit current.

In linear media, the inductance of a circuit depends only on the

shape and the sizes, but not on the current.

The magnetic flux linked with the current I is called the magnetic

flux linkage with the current I, and it is denoted as . The ratio of t

o I is denoted by L, hence

The magnetic flux linkage is different from the magnetic flux, and

it is associated with a current.

If the magnetic flux is linked with a current N times, then the

magnetic flux linkage will be increased by N times. If only a part of the

magnetic flux is linked to a current, the magnetic flux linkage must be

reduced proportionately.

I

N

IL

Suppose we have two loop currents,

the magnetic flux linkage linked with

current I1 consists of two parts: one is

generated by the magnetic flux caused by

current I1 itself, and it is denoted as .

dl1

O

z

y

x

dl2

l2l1 I2I1

r2 - r1

r2

r1

A loop coil with N turns the magnetic flux linkage with the current

is = N , and the inductance of the loop coil with N turns is

Another is produced by the magnetic flux at loop l1 by current I2.

The magnetic flux linkage linked with current I2 is

22212

If the surrounding medium is linear, then all the ratios, , ,

and are independent of the currents since all the magnetic flu

x linkages are proportional to the current generating them.

1

11

I

2

12

I

2

22

I

1

21

I

where L11 is called the self-inductance of loop l1, and M12 is called the

mutual inductance from loop l2 to loop l1.

Similarly, we define 2

2222 I

L

1

2121 I

M

where L22 is called the self-inductance of loop l2, and M21 is called the

mutual inductance from loop l1 to loop l2.

Hence, the magnetic flux linkage linked with current I1 is

12111

1

1111 I

L

2

1212 I

MLet

Substitute the above parameters L11 ， L22 ， M12 ， and M21 into

the above equation, we have

2121111 IMIL 2221212 ILIM

For linear homogeneous media, we can prove that

2112 MM

Since we can find the mutual inductances between any two loop

circuits as follows:

2 1

12

2121

dd

π4 l lM

rr

ll

1 2 21

1212

dd

π4 l lM

rr

ll

Considering ， we have21121221 ,dddd rrrrllll

2112 MM

2 1

12

2121

dd

π4 l lM

rr

ll

1 2 21

1212

dd

π4 l lM

rr

ll

In electronic devices, if we need to increase the magnetic coupling

between two coils, the two coils should be placed parallel to each other.

If the magnetic coupling needs to be eliminated, they should be perpen

dicular to each other.

The mutual inductance could be positive or negative, while

the self-inductance is always positive.

If everywhere, the mutual inductances be zero. 21 dd ll

If everywhere, the mutual inductances will be maximum.21 d //d ll

Example 1. Calculate the mutual inductance between an infinitely

long straight line and a rectangular coil. The line and the coil are at the

same plane, and in vacuum.

a

b

drr

D

0I1I2

z

S2

Solution: Select cylindrical coordinate system,

and let the infinitely long straight line to be at

the z-axis. The magnetic flux density produced

by current I1 is then

eBr

I

π210

1

The magnetic flux linkage 21 with

current I2 by current I1 is

2

d121 SSB

If the flowing direction of the current I2 is clockwise, dS and B1 h

ave the same direction. Then

bD

D D

bDaIr

r

aI

101021 ln

π2d

1

π2

We have 0lnπ20

1

2121

D

bDa

IM

If the flowing direction of the current I2

is counter clockwise, then the B1 and dS are

opposite, and M21< 0.

Example 2: Calculate the inductance per unit length of a coaxial

line carrying a direct current.

Solution: Assume the radius of the

inner conductor of the coaxial line is a, the

interior radius of the outer conductor is b,

and the exterior radius is c.

b

c

a

O

a

b

drr

D

0I1I2

z

S2

In the coaxial line, we construct a

rectangular circuit from a longitudinal

section of unit length, left side width a

and right width (c b).

The current in the inner conductor i

s that on the left side, while the current i

n the outer conductor is that on the right

side.

The inductance per unit length of a coaxial line is

IL1

where I is the current in the coaxial line, and is the magnetic flux

linkage per unit length with the current I.

a IO b

c

r cbaO

dr

II

e

The magnetic flux linkage with current I

is formed by three parts: the first one is in

the outer conductor, the second is between

the inner and the outer conductors, and the

third is in the inner conductor.

a IO b

c

r cbaO

dr

II

e

Since the thickness of the outer conductor

is usually very thin, the magnetic flux linkage

in the outer conductor can be neglected.

The magnetic flux density Bo between the inner and the outer

conductors is given by

eB

r

I

π20

o

The magnetic flux produced by this magnetic field is called outside

magnetic flux , then the outside magnetic flux per unit length iso

a

bIrB

b

a

b

aSln

π2ddd 0

o

o oo

reBSB

The magnetic flux density Bi in the

inner conductor is

20

i π2 a

IrB

ra

Ird

π2d

20

i

a

bIrBr

b

a

b

aSln

π2ddd 0

o

o oo

eBSB

The outside magnetic flux is linked

completely with the current I, and it is

equal to the magnetic flux linkage with

the current I. o

The magnetic flux produced by this magnetic field is called inside

magnetic flux , then the magnetic flux through the area of unit le

ngth and width dr isi id

a IO b

c

r cbaO

dr

II

e

This magnetic flux is only linked with

the partial current I between 0 and r in

the inner conductor, instead of the full

current I.

ra

Ir

I

Id

π2dd

4

30

ii

From this we find the magnetic flux linkage i formed by the

magnetic field in the inner conductor that links with the full

current I as a I

0

0ii π8

d

ra

Ird

π2d

20

i

Therefore, to the full current I, the

magnetic flux linkage formed by this

magnetic flux should be scaled down to be

a IO b

c

r cbaO

dr

II

e

π8ln

π200io

1

a

b

IL

where the first term is called external inductance, and the second

term is called internal inductance.

when a coaxial line is operating with time-varying electro-

magnetic field, the magnetic flux in the inner and the outer

conductors can both be neglected.

a

bL ln

π20

1

Then the total magnetic flux linkage with the full current I

as . The inductance per unit length of the coaxial line is

thereforeio

Hence, the inductance per unit length of the coaxial line is equal

to the external inductance, given by

3. Energy in Steady Magnetic Fields

If an impressed source is applied to a circuit, a current will be

generated in the circuit. In the process of establishing the current,

the reaction magnetic flux in the circuit will resist the increment of

the current.

Assume the current is increased very slowly so that radiation

loss can be neglected, all energy provided by the impressed

source will be stored in the magnetic field around the circuit.

Based on the work done by the impressed source, the energy stored

in the magnetic field can be calculated.

In order to overcome the back electromotive force due to the

reaction magnetic flux and to maintain the current, the impressed

source has to do work.

Suppose the current in a loop is gradually increased from zero

to the final value I, and the magnetic flux through the loop is also

increased from zero to the final value .

tU

d

d

If the current in the loop is i(t) at time t, then at this moment

the instantaneous power sent to the loop is

ttiUtitP

d

d)()()(

The work done by the impressed source in dt is

d)(d)(d tittPW

te

d

dThe back electromotive force is

In order to overcome the back electromotive force, the impressed

source has to produce a voltage , given byeU

At any moment the relationship between the magnetic flux linkage

with a current loop and the current is

)()( tLit

)()( tLit

Considering the inductance L of a loop is independent of the loop

current i, we find the work done by the impressed source in dt as

itLiW d)(d

When the loop current has increased to the final value I, the total

work W done by the impressed source is

I

LIitLiW

0

2

2

1d)(

The magnetic flux linkage with a current loop is just the

magnetic flux through the loop, hence

If the energy of the magnetic field is denoted as Wm , it must be

given by 2m 2

1LIW 2

m2

I

WL or

For some loop circuits, it is very convenience to calculate the

inductance by the above equation.

Considering , the energy of the magnetic field produced

by a loop circuit carrying the current I can be rewritten asI

L

IW 2

1m

where is the magnetic flux linkage for the current I.

The work establishes the current I in the loop, and the current

produces the magnetic field in space.

Since the current is increased very slowly, the radiation loss

can be neglected, and all work done by the impressed source will be

converted into the energy stored in the magnetic field.

For N current loops, we can let all currents of the loops be

increased slowly at the same rate from zero to the final value.

Based on energy conservation, the final energy is independent of

the process in reaching the final state.

NjNjjjjjj IMILIMIM 2211

When all currents are increased at the same rate, the magnetic

flux linkage is also increased at the same rate. Assume the current

of the j-th loop at the moment t is , where Ij is the final va

lue, and is the rate coefficient ( ).

jj Itti )()(

10

N

jjj

N

jjj IttiW

11

d)(d)(d

Since the magnetic flux linkage of a loop is proportional to th

e current, the magnetic flux linkage j of the j-th loop is

Then the work done by the impressed source in N loops in dt is

When all currents are increased to the final values, the total

work W done by the impressed source is

WW d

The magnetic energy produced by the N loops carrying the final

currents is

1

0 1

m dN

jjjIW

N

jjjIW

1m 2

1 That is

If the magnetic flux linkages and the currents of all loops are known,

the magnetic energy caused by these loop currents can be determined.

The magnetic flux through a circuit can be expressed in terms of t

he vector magnetic potential A as . Hence the magnetic flu

x linkage with the j-th circuit can also be expressed by the vector magn

etic potential A as

l

dlA

j

d lj lA

N

jl j

j

IW1

m d2

1jlA

Then the magnetic energy caused by N loop currents can be

expressed in terms of the vector magnetic potential A as

where A is the composite vector magnetic potential produced by all

currents at the j-th circuit.

If the currents are distributed continuously in the volume V and

, then the right side of the above equation becomes a

volume integral. In this case, the magnetic energy can be written as

VI dd Jl

V

VW m d

2

1JA

If the currents are distributed on a surface S, the magnetic energy is

given by

S S SW m d

2

1JA

The magnetic energy density

Considering ， we haveJH

V

VW d2

1m HA

Using the vector identity , the above

equation can be rewritten as

AHHAAH )(

VVWVV

d2

1d)(

2

1 m AHAH

where V is the region in which the current is found.

SAHAH d)(2

1d)(

2

1

SV

V

Apparently, if the domain of integration is extended to infinity,

the above equation still holds. Let S be the surface of a sphere with

infinite radius. Applying divergence theorem, the first integral

becomes

When the current is distributed in a region of finite volume, the

magnetic field intensity at large distance is inversely proportional to the

square of the distance, while the vector magnetic potential is inversely

proportional to the distance. Hence the surface integral over the surface

at infinity will be zero, we have

0d)(

SAHS

Considering ， we hav

e

BA VWV

d)(2

1 m BH

where V is the volume occupied by the magnetic field. Obviously, the

integrand stands for the density of the magnetic energy.

If the magnetic energy density is denoted by small letter wm , then

BH 2

1mw

SAHAH d)(2

1d)(

2

1

SV

V

For linear isotropic media, , and the magnetic energy

density can be rewritten as

HB

2m 2

1Hw

Since the magnetic energy is proportional to the square of the

magnetic field intensity, the magnetic energy does not obey the

superposition principle.

Example. Calculate the magnetic energy per unit length in a

current-carrying coaxial line. Suppose the steady current is I, the

radius of the inner conductor is a, the radius of the outer

conductor is b, with thickness neglected, and the region between

the conductor is vacuum.

Solution: The inductance of a coaxial line per unit length

is

a

bL ln

π2π800

Hence, the magnetic energy per unit length in the coaxial line as

a

bIILIW ln

π4π162

1 20

202

1m

We can also calculate the magnetic energy in the coaxial line by

using magnetic energy density.

20

ii π2 a

IrBH

The magnetic field intensity inside the inner conductor is

Hence, the magnetic energy inside the inner conductor per unit

length is

π16dπ2

π22

1d

2

1 20

0

2

202im

Irr

a

IrVHW

a

V

The magnetic field intensity Ho between the inner and the outer

conductors is

r

IBH

π20

oo

Therefore, the magnetic energy per unit length between the inner

and the outer conductors is

a

bIrrHW

b

aln

π4dπ2

2

1 20

2o0mo

The total magnetic energy per unit length in the coaxial line is th

en , which is the same as before.)( momi WW

Since , it is very simple to calculate the inductance of a

circuit by using the magnetic energy.

2m2

I

WL

4. Magnetic Forces

dl1

O

z

y

x

dl2

l2l1

I2I1

r2 - r1

r2

r1

The magnetic force F acting on

the current element Idl is

BlF dI

12221 dd BlF I

where the magnetic flux density B1 produced by the current I1 as

1 3

12

121101

)(d

π4)(

l

I

rr

rrlrB

Hence, the force F21 due to the magnetic field B1 on the loop

current l2 is

1 2 3

12

121122021

)](d[d

π4 l l

II

rr

rrllF

Then the force dF21 of the magnetic

field caused by the current I1 on the

current element I2dl is

In the same way, the force F12 due to the magnetic field B2 caused

by the current I2 on the current loop l1 is

1 2 3

21

212211012

)](d[d

π4 l l

II

rr

rrllF

The above two equations are called Ampere’s law.

Based on Newton’s third law, we have , which can be p

roved directly from the equations. 1221 FF

If the shape of the loop circuit is complex, it is very difficult to

evaluate the integral, and a close form expression cannot even be

found. In order to calculate the magnetic force, the method of virtual work

can be used.

Here we directly apply the generalized forces and coordinates to

derive the general formulas for obtaining the magnetic force.

Under the influence of the generalized force F of the magnetic field

caused by the current I1, assume one generalized coordinate of the loop

circuit l2 has an increment dl , leading to an increment of the magnetic e

nergy dWm.

lFWW ddd m

Two cases are to be considered:

(a) If the currents I1 and I2 are unchanged, the case of a constant cur

rent system results. Then

2211m d2

1d

2

1d IIW

The works done by the impressed source for two circuits, respectively,

are111 dd IW 222 dd IW

Then the total work dW done by the impressed source in the two loop

circuits should be equal to the sum of the work by the generalized magn

etic force and the increment of the magnetic energy so that

m21 d2ddd WWWW

Then the total work dW done by the impressed source in the two lo

op circuits is

lFWW dd2d mm i.e.

We find constant

m

Il

WF

(b) If the magnetic flux linkages with all circuits are unchanged,

we have the case of a constant magnetic flux system.

lFW dd0 m

We findconstant

m

Φl

WF

Since there is no change in the magnetic flux, no new electromotive

force is produced in the two circuits. Hence, the impressed source does

zero work, i.e.

Magnetic forces are used widely, such as in electromagnets, maglev b

earings, and maglev vehicles, all of them being based on the action of ma

gnetic forces.

Example 1. Calculate the magnetic force between an infinitely long

current-carrying wire and a rectangular current-carrying loop. The

dimensions and the position of the loop are shown in the figure.

a

b

D

0I1I2

Solution: Let it be the constant current system.

The magnetic force between the wire and the

loop is therefore

constant

m

Il

WF

2211m 2

1

2

1 IIW where

2221212

2121111

ILIM

IMIL

Since MMM 2112

MIILILIW 21222211

21m 2

1

2

1We obtain

Taking the separation D as the generalized coordinate l, and since the

self-inductance L11 and L22 are independent of the separation D, the

magnetic force is

D

MIIF

21

The mutual inductance M is D

bDaM

ln

π20

DbD

abIIF

)(π2021

We find

where the negative sign shows that it is an attractive force.

If the direction of one of the currents is opposite, the term involving

M becomes negative, leading to a repelling force.

Example 2. Find the attractive force of the electromagnet.

B0

S

I

l

Solution: Since the iron core can be

approximately considered as a perfect

magnetic conductor, the magnetic field

intensity in the iron core is zero, and

the magnetic energy is also zero.

In this way, the magnetic energy exists only in two gaps. Hence,

the total magnetic energy is

0

20

0

20

m 2

12

SlB

SlB

W

In view of this, in order to find the attractive force of the

electromagnetic magnet, it is easy to consider this case as a constant

magnetic flux system. Hence, we have

0

20

0

2

constant

m

SB

Sl

WF

where the negative sign shows that it is an attractive force.

In addition, we can see that the attractive force is proportional

to the cross-sectional area S of the iron core and the square of the

magnetic flux density in the gaps.

Considering the magnetic flux in the gaps is , we obtain SB0

S

lW

0

2

m