Chapter 2: Fundamentals of Magnetismfrick/EE4220-EM_Dynamics/lecture2.pdf · Chapter 2:...

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8/28/2003 Electromechanical Dynamics 1 Chapter 2: Fundamentals of Magnetism

Transcript of Chapter 2: Fundamentals of Magnetismfrick/EE4220-EM_Dynamics/lecture2.pdf · Chapter 2:...

8/28/2003 Electromechanical Dynamics 1

Chapter 2: Fundamentals of Magnetism

8/28/2003 Electromechanical Dynamics 2

Magnetic Field Intensity

• Whenever a magnetic flux, φ, exist in a conductor or component, it is due to the presence of a magnetic field intensity, H, given bywhere– H = magnetic field intensity [A/m]

– U = magnetomotive force [A] or ampere turns

– l = length of the magnetic circuit [m]

– examplefind the magnetic field intensity at the circle

l

UH =

ρ = 5 cm( ) A/m159

m05.02

A50

2===

ππρU

H

50 A

8/28/2003 Electromechanical Dynamics 3

Magnetic Flux Density

• For a magnetic flux φ, there exists a magnetic flux density, B, given bywhere– B = flux density [T] (tesla)

– φ = flux in a component [Wb ] (weber)

– A = cross section area [m2]

– examplefind the flux density

AB

φ=

( )( ) T8.1m02.0m01.0

Wb10360 6

=×==−

hdB

φd = 20cm

h = 10cm

φ = 360µWb

I

8/28/2003 Electromechanical Dynamics 4

Relationship between B-H in Free Space

• In free space, the magnetic flux density B is directly proportional to the magnetic field intensity H

• The constant of proportionality for free space is called the permeability constant, µ0

– in the SI system, µ0 = 4π×10-7 H/m[henry/meter]

HB 0µ=

0 200 400 600 800 1000 1200 A/m

mT2.0

1.5

1.0

0.5

0

H

B

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B-H Characteristic of Magnetic Material

• The flux density is influenced by the magnetic property of the material in which the flux passes– instead of specifying a permeability for every material, a

relative permeability is defined, µr = µ / µ0

– relative permeability is unitless

• for many materials, the relative permeability isnot constant but variesnonlinear w.r.t. B

HB rµµ0= 2.0 T

1.5

1.0

0.5

0

H

B

0 1 2 3 4 5 6 kA/m

cast iron

silicon iron (1%)

free space

8/28/2003 Electromechanical Dynamics 6

Determining Relative Permeability

• One can find the relative permeability in a material by taking the ratio of the flux density in the material to the fluxdensity that would have been produced in free-space

– ExampleDetermine the relativepermeability of relaysteel (1% silicon) at aflux density of 0.6 Tand 1.4 T

H

B

H

Br 000,800

0

≈=µ

µ

8/28/2003 Electromechanical Dynamics 7

Faraday’s Law

• Electromagnetic induction– If the flux linking a loop (or turn) varies as a function of

time, a voltage is induced between its terminals

– The value of the induced voltage is proportional to the rate of change of flux

• In SI units,where– E = induced voltage [V]

– N = number of turns in the coil

– ∆Φ = change of flux inside the coil [Wb]

– ∆t = time interval of the flux changes [s]

tNE

∆∆Φ=

8/28/2003 Electromechanical Dynamics 8

Faraday’s Law

– Example• a coil of 2000 turns surrounds a flux of 5 mWb produced

by a permanent magnet

• the magnet is suddenly withdrawn causing the flux inside the coil to drop uniformly to 2 mWb in 0.100 s

• what is the induced voltage?

8/28/2003 Electromechanical Dynamics 9

Voltage Induced in a Conductor

• It is often easier to calculate the induced voltage on a segment of conductor instead of the voltage on a coil

where– E = induced voltage [V]

– B = flux density [T]

– l = active length of conductor in themagnetic field [m]

– v = relative speed of the conductor [m/s]

vlBE =

8/28/2003 Electromechanical Dynamics 10

Lorentz Force

• A current-carrying conductor sees a force when placed in a magnetic field– fundamental principle for the operation of motors

– the magnitude of the force depends upon orientation of the conductor with respect to the direction of the field

– force is greatest when the conductor is perpendicular to the field

• The Lorentz or electromagnetic force:where– F = force acting on the conductor [N]

– θ = angle between the flow directions of current and flux

θsinIlBF =

8/28/2003 Electromechanical Dynamics 11

Lorentz Force on a Conductor

– Example• a conductor of 1.5 m long is

carrying a current of 50 A andis placed in a magnetic fieldwith a density of 1.2 T

• calculate the force on the conductor if it is perpendicularto the lines of flux

• calculate the force on theconductor if it is parallelto the lines of flux

8/28/2003 Electromechanical Dynamics 12

Direction of Force

• A current carrying a current issurrounded by a magnetic field

• The flux lines of two magneticfields never cross each other– the flux lines of two magnet

fields are vectorally added

– the generated mechanical forcetends to push the lines of fluxback to an even distribution

8/28/2003 Electromechanical Dynamics 13

Direction of Force

• Right hand rule– Fingers point in the direction of current flow

(positive to negative)

– Bend fingers into the direction of the magnetic field (north to south)

– Thumb points in the direction of force

8/28/2003 Electromechanical Dynamics 14

Residual Flux

• Metals that have a strong magnetic attraction can be modeled as being composed of many molecular size magnets– orientation of the magnets are

normally random

– by applying an external magnetic field (e.g. using a coil with a current flow), the molecular size magnets will align themselves with the external field

NS

NS

NS

N S

NS

NS

NS

N S

φ

8/28/2003 Electromechanical Dynamics 15

Residual Flux

• When the external magnetic fielddecreases, the magnetic domainstend to retain their original orientation– this is called residual induction

NS

NS

NS

NS

φR

NS

NS

NS

NS

φ

8/28/2003 Electromechanical Dynamics 16

Hysteresis Loop• To eliminate the residual flux, a

reverse coil current is required to generate a field H in the opposite direction– the magnetic domains gradually

change their previous orientation until the flux density becomes zero

– Hc is the coercive force

– energy is required to overcome the molecular friction of the domains to any changes in direction

• AC magnets have ac flux changing continuously and will map out a closed curve call a hysteresis loop

8/28/2003 Electromechanical Dynamics 17

Hysteresis Losses

• With an external ac flux, the B/Hcharacteristics of a magnetic material traces out a curve from– (+Bm +Hm) to (+Br 0) to (0 -Hc) to

(-Bm -Hm) to (-Br 0) to (0 +Hc)

• magnetic material absorbs energy during each cycle and the energy is dissipated as heat– the heat released per cycle [J/m3]

is equal to the area [T-A/m] of the hysteresis loop

NS

NS

NS

NS

N S

NS

NS

NS

φ

φ

8/28/2003 Electromechanical Dynamics 18

Eddy Currents• An ac flux φ linking a rectangular-shaped

conductor induces an ac voltage E across its terminals

• If the conductor terminals are shorted, a substantial current flows

• The same flux linking smaller coils induce lesser voltages and lower currents

• A solid metal plate is basically equivalent to a densely packed set of rectangular-shaped coils

• The induced currents flowing inside the plate are eddy currents, and flow to oppose the change in flux

φ

φ

φ

E

I1I4

IE

8/28/2003 Electromechanical Dynamics 19

Eddy Current Losses• Eddy currents become a problem

when iron must carry an ac flux– eddy currents flow throughout the

entire length of the iron core

– resistance in the iron causes energy to be dissipated as heat

• Losses can be reduced by splitting the core into sections (lamination)– subdividing causes the losses to

decrease progressively

– varnish coatings insulate the laminates from current flows

– silicon in the iron increases the resistance

φ

φ

ICIE1

ICIE2

21 EE II >

IE2

8/28/2003 Electromechanical Dynamics 20

Homework

• Problems:– 1-90, 1-91, 1-97

– 2-4, 2-5, 2-7, 2-8, and 2-9