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

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Transcription:

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

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 by U H = where l H = magnetic field intensity [A/m] U = magnetomotive force [A] or ampere turns l = length of the magnetic circuit [m] example find the magnetic field intensity at the circle U 50A H = = = 159A/m ρ = 5 cm 2πρ 2π 0.05m ( ) 50 A 8/28/2003 Electromechanical Dynamics 2

Magnetic Flux Density For a magnetic flux φ, there exists a magnetic flux density, B, given by B= φ where A B = flux density [T] (tesla) φ = flux in a component [Wb ] (weber) A = cross section area [m 2 ] example find the flux density h = 10cm 6 360 10 Wb B = φ d = 20cm = = 1.8T d h ( 0.01m)( 0.02m) φ = 360µWb 8/28/2003 Electromechanical Dynamics 3 I

Relationship between B-H in Free pace 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, µ 0 B = µ 0 H in the I system, µ 0 = 4π 10-7 H/m [henry/meter] B mt 2.0 1.5 1.0 0.5 0 0 200 400 600 800 1000 1200 A/m H 8/28/2003 Electromechanical Dynamics 4

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, µ r = µ / µ 0 relative permeability is unitless B = µ 0 µ r H for many materials, the relative permeability is not constant but varies nonlinear w.r.t. B 2.0 T B 1.5 1.0 0.5 8/28/2003 Electromechanical Dynamics 5 0 silicon iron (1%) cast iron free space 0 1 2 3 4 5 6 ka/m H

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 flux density that would have been produced in free-space B B µr = 800,000 µ0 H H Example Determine the relative permeability of relay steel (1% silicon) at a flux density of 0.6 T and 1.4 T 8/28/2003 Electromechanical Dynamics 6

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 Φ In I units, where E = t E = induced voltage [V] = number of turns in the coil Φ = change of flux inside the coil [Wb] t = time interval of the flux changes [s] 8/28/2003 Electromechanical Dynamics 7

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 8

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 = B l v E = induced voltage [V] B = flux density [T] l = active length of conductor in the magnetic field [m] v = relative speed of the conductor [m/s] 8/28/2003 Electromechanical Dynamics 9

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 [] F = Bl I sinθ θ = angle between the flow directions of current and flux 8/28/2003 Electromechanical Dynamics 10

Lorentz Force on a Conductor Example a conductor of 1.5 m long is carrying a current of 50 A and is placed in a magnetic field with a density of 1.2 T calculate the force on the conductor if it is perpendicular to the lines of flux calculate the force on the conductor if it is parallel to the lines of flux 8/28/2003 Electromechanical Dynamics 11

Direction of Force A current carrying a current is surrounded by a magnetic field The flux lines of two magnetic fields never cross each other the flux lines of two magnet fields are vectorally added the generated mechanical force tends to push the lines of flux back to an even distribution 8/28/2003 Electromechanical Dynamics 12

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 13

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 8/28/2003 Electromechanical Dynamics 14 φ

Residual Flux When the external magnetic field decreases, the magnetic domains tend to retain their original orientation this is called residual induction φ φ R 8/28/2003 Electromechanical Dynamics 15

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 H c 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 16

Hysteresis Losses With an external ac flux, the B/H characteristics of a magnetic material traces out a curve from (+B m +H m) to (+B r 0) to (0 -H c ) to (-B m -H m ) to (-B r 0) to (0 +H c ) φ magnetic material absorbs energy during each cycle and the energy is dissipated as heat the heat released per cycle [J/m 3 ] is equal to the area [T-A/m] of the hysteresis loop φ 8/28/2003 Electromechanical Dynamics 17

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 rectangularshaped coils The induced currents flowing inside the plate are eddy currents, and flow to oppose the change in flux 8/28/2003 Electromechanical Dynamics 18 I 4 I E I 1 E φ φ φ

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 8/28/2003 Electromechanical Dynamics 19 I C φ I E1 φ I > I E1 E2 I C I E2 I E2

Homework Problems: 1-90, 1-91, 1-97 2-4, 2-5, 2-7, 2-8, and 2-9 8/28/2003 Electromechanical Dynamics 20