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Magnetic field creation (example of a problem) Three long, straight wires are parallel to each other and perpendicular to the plane of the paper. Their mutual location is shown in Figure below. The currents in wires 1 and 2 are the same, I 1 = I 2 =I. What is the direction of current in wire 3, and what is the ratio I 3 /I 1 such that the net magnetic field at empty corner is zero? 2
Solution a a a a 3
Ampere s law ΣB l = µ I, // l 0 vacuum Andre Ampere, French mathematician and physicist 1775-1836 4
Magnetic field of a long wire with current ΣB l = µ I, // l 0 vacuum 1) Select the path, dashed line 2) Apply Ampere s law B 2 π r = µ I 0 B µ I 0 = 5 2π r
B field inside a conducting rod A long, cylindrical conductor is solid throughout and has a radius R. Electric charges flow parallel to the axis of the cylinder and pass uniformly through the entire cross section. The arrangement is, in effect, a solid tube of current. The 2 current per unit cross-sectional area (i.e., the current density) is I 0 /( πr ). Use Ampere s law to show that the magnetic field inside the conductor at a 2 distance r from the axis is µ 0I0r / (2πR ). (Hint: For closed path, use a circle of radius r perpendicular to and centered on the axis. Note that the current through any surface is the area of the surface times the current density.) 6
Magnetic field inside a solenoid 1) Select the path, dashed line 2) Apply Ampere s law Along circular path, very little current crosses the plane of the loop; magnetic field outside the solenoid is very small B = µ 0nI ΣB l l = I, // µ 0 Assume, the number of the coil turns per unit of length is n. Then, along the rectangular path, the current through the loop is n a and the sum in the Ampere s law is so that I B a Current I vacuum B w a 7
Magnetic field inside a toroid Consider a plastic doughnut wound by N turns of wire that carries the current I. What would be B field inside the core if one assumes that the field is almost uniform there? I N Β Β R Air inside 8
Magnetic field inside a toroid Β Amperian loop I Air inside R N Β ΣB l =µ I // l 0 Currents crossing the loop B 2π R = µ 0NI NI B = R µ 0 2 π 9
Magnetic properties of materials Para-, Dia- and Ferro- magnetic materials 10
Magnetic field in a material Magnetism of atoms is originated from: 1) circular motion of electrons around Boh r orbits and 2) spinning around their axes 11
Paramagnetic material All motions give non-zero B field, individual atoms act as small permanent magnets, all B fields are aligned with the external magnetic field Paramagnetic material can be magnetized and magnetic field inside material is larger than the one that would have been in vacuum Example: Aluminum 12
Diamagnetic material, no external B field 13
Diamagnetic material, external B field is applied r = mv qb Diamagnetic material (gold) is difficult to magnetize. Resultant magnetic field is in opposite direction with respect to the external B field 14
Strong paramagnetic: Ferro-magnetism Iron, nickel, cobalt, chromium dioxide In some materials a group of atoms (>10 16 ) can have common properties forming so called magnetic domains Domains can be enhanced through application of external magnetic field and they can stay in new state for quite a time One can make a permanent magnet 15
Effect of ferromagnetic material 16
Fringing effect for an air gap N Some flux lines travel outside the gap Φ lines S 17
Magnetic circuits: Why to bother? To make a record on a tape, voice signal is amplified electrically and electric current sets up varying magnetic field in the doughnut core and the air gap. Because of fringing, some flux lines are spread outside of the gap and they magnetize the tape thus leaving the signal signatures on the magnetic tape (they are stored as permanent magnet and can be recovered by reversing the process) 18
Induced currents and voltages B field changes as we move permanent magnet Area decreases as we stretch! Induced current and voltages are generated when 1)B field through the loop or 2) area of the loop in B field is changing. Changes in magnetic flux Φ=ΒΑ is what controls the effect.
Application: Cruise control device
Voltage created by a conductor moving in magnetic field and crossing magnetic field lines (motional emf) 21
Motional emf E m = L V B F q = qe = qe / L = q vel B E = L vel B Charge separation Magnetic force DC generator
Problem: Bar on rails as a DC motor A metal bar is moved along the frictionless rails with constant speed V. The rails are immersed in magnetic field B directed upwards as shown. Find 1) Polarity of the motional emf 2) Direction and magnitude of the current induced in the rails. 3) Magnitude of the force F. Given are B, L, R and speed V. 4) Power delivered to the resistor R and how much is it as % of the total mechanical power (efficiency of the generator ) 23
Summary of equations ΣB l l = I, // µ 0 vacuum E m = L V B 24