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Magnetic Forces, Materials and Devices. INEL 4151 ch 8 Dr. Sandra Cruz-Pol Electrical and Computer Engineering Dept. UPRM. http://www.treehugger.com/files/2008/10/spintronics-discover-could-lead-to-magnetic-batteries.php. Applications. Motors Transformers MRI More….
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Magnetic Forces, Materials and Devices INEL 4151 ch 8 Dr. Sandra Cruz-Pol Electrical and Computer Engineering Dept. UPRM http://www.treehugger.com/files/2008/10/spintronics-discover-could-lead-to-magnetic-batteries.php
Applications Motors Transformers MRI More… http://www.youtube.com/watch?v=0ajvcdfC65w
Forces due to Magnetic fieldsB=mH= magnetic field densityH=magnetic field intensity Force can be due to: • B on moving charge, Q • B on current • 2 currents B is defined as force per unit current element
Forces on a Charge Analogous to the electric force: We have magnetic force: The total force is given by: Note that Fm is perpendicular to both u and B If the charge moving has a mass m, then: Usually Fm < Fe
Forces on a current element The current element can be expressed as: So we can write: For closed path Line I=[A] current element Surface K=[A/m] current element Volume J=[A/m2] current element
Force between two current elements • Each element produces a field B, which exerts a force on the other element. I1 R21 I2
P.E. 8.4 Find the force experienced by the loop if I1=10A, I2=5A, ro=20cm, a=1cm, b=30cm Divide loop into 4 segments. z z I1 F2 F1 b I2 F3 r ro r a F4
For segment #1, Force #1 Since I1 is infinite long wire: z I1 I2 b F1 r a ro
For segment #2, The B field at segment #2 due to current 1. z F2 I1 I2 b r a ro
For segment #3, Force #3 The field at segment 3: z I1 I2 b F3 r a ro
For segment #4, The B field at segment #4 due to current 1. z I1 I2 b r a ro F4
The total force en the loop is I1=10A, I2=5A, ro=20cm, a=1cm, b=30cm • The sum of all four: • Note that 2 terms cancel out: z F2 F1 F3 r F4 Attracts!
Magnetic Torque and Moment Inside a motor/generator we have many loops with currents, and the Magnetic fields from a magnet exert a torque on them. The torque in [N m]is: • Where m is the magnetic dipole moment: • Where S is the area of the loop and an is its unit normal. This applies if B is uniform z B SIDE VIEW B
Magnetic Torque and Moment The Magnetic torque can also be expressed as: The torque in [N m]is:
Magnetization (similar to Polarization for E) • Atoms have e- orbiting and spinning • Each have a magnetic dipole associated to it • Most materials have random orientation of their magnetic dipolesif NO external B-field is applied. • When a B field is applied, they try to align in the same direction. • The total magnetization [A/m]
Magnetization • The magnetization current density [A/m2] • The total magnetic density is: • Magnetic susceptibility is: • The relative permeability is: • Permeability is in [H/m].
Classification of Materials according to magnetism Non-magnetic mr=1 Ex. air, free space, most materials in their natural state. Magnetic mr≠1 Diamagnetic mr≤1 Electronic motions of spin and orbit cancel out. lead, copper, Si, diamonds, superconductors (mr=B=0). Are weakly affected by B Fields. Paramagnetic mr≥1 (air, platinum, tungsten, platinum ) Temperature dependent. Not many uses except in masers Ferromagnetic mr>>1 Iron, Ni,Co, steel, alloys Loose properties if heated above Curie T (770C) Nonlinear: mr varies
B-H or Magnetization curve • When an H-field is applied to ferromagnetic material, it’s B increases until saturation. • But when H is decreased, B doesn’t follow the same curve.
Hysteresis Loop • Some ferrites, have almost rectangular B-H curves, ideal for digital computers for storing information. • The area of the loop gives the energy loss per volume during one cycle in the form of heat. • Tall-narrow loops are desirable for electric generators, motors, transformers to minimize the hysteresis losses.
Magnetic B.C. • We’ll use Gauss Law & • Ampere’s Circuit law
B.C.: Two magnetic media • Consider the figure below: B1n B1 D S Dh m1 B1t B2n m2 B2 B2t
B.C.: Two magnetic media • Consider the figure below: H1n H1 q1 a b K m1 H1t Dh H2n m2 H2 c d H2t D w
P.E. 8.8 Find B2 • Region 1 described by 3x+4y≥10, is free space • Region 2 described by 3x+4y≤10 is magnetic material with mr=10 Assume boundary is current free. OJO: Error en Solucionario!
P.E. 8.7 B-field in a magnetic material. • In a region with mr=4.6 • Find H and susceptability mA/m
How to make traffic light go Green when driving a bike or motorcycle • Stop directly on top of induction loop on the street • Attach neodymiummagnets to the vehicle • Move on top of loop • Push crossing button • Video detectors • http://www.wikihow.com/Trigger-Green-Traffic-Lights • http://www.labreform.org/education/loops.html
Inductors • If flux passes thru N turns, the total Flux Linkage is • This is proportional to the current I • So we can define the inductance as: • then:
When more than 1 inductor *Don’t confuse the Magnetization vector, M, with the mutual inductance!
Self -inductance • The total energy in the magnetic field is the sum of the energies: • The positive is taken if currents I1 and I2 flow such that the magnetic fields of the two circuits strengthen each other. See table 8.3 in textbook with formulas for inductance of common elements like coaxial cable, two-wire line, etc.
P.E. 8.10 solenoid A long solenoid with 2 x 2 cm cross section has iron core (permeability is 1000x ) and 4000 turns per meter. If carries current of 0.5A, find: • Self inductance per meter Note L is Independent of current
Magnetic Circuits • Magnetomotive force • Reluctance • Like V=IR • Ex: magnetic relays, motors, generators, transformers, toroids • Table 8.4 presents analogy between magnetic and electric circuits In units of ampere-turns
Magnetic Circuits Electric circuit rules apply!
Ex. Find current in coil needed to produced flux density of 1.5T in air gap.Assume mr=50 and all cross sectional area is 10 cm2
8.42 A cobalt ring (mr=600) has mean radius of 30cm. • If a coil wound on the ring carries 12A, calculate the N required to establish an average magnetic flux density of 1.5 Teslas in the ring. radius of 30cm
Force on Magnetic materials Relay • N turns, current I • B.C. B1n=B2n(ignore fringing) • Total energy change is used to displace bar a distance dl. • S=cross sectional area of core
P.E. 8.16 U-shape electromagnet • Will lift 400kg of mass(including keeper + weight) • Iron yoke has mr=3,000 • Cross section =40cm2 • Air gap are 0.1mm long • Length of iron=50cm • Current is 1A Find number of turns, N Find force across one air gap Esto nos da el B en el air gap.
MagLevmagnetic Levitation • Use diamagnetic materials, which repel and are repelled by strong H fields. • Superconductors are diamagnetic. • Floating one magnet over another • Maglev train 312 mph • Regular train • 187 mph