UVA PHYS 632 - Lecture 11 Magnetism of Matter: Maxwell’s Equations Chp. 32

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Lecture 11 Magnetism of Matter: Maxwell’s Equations Chp. 32 Wednesday MorningWhat is Mutual Inductance? MPowerPoint PresentationSlide 6Slide 7Slide 8Slide 9FerromagnetismWhat is the atomic origin of magnetism?Spin Magnetic Dipole Moment of the ElectronSlide 13Slide 14(C) Use the dipole formula to find the magnitude and direction of the magnetic field 1cm from the end of the bar magnet on its central axis at P.BigBite is a 50 ton electromagnet with a 25 cm by 100 cm gapSlide 17Maxwells Equations: In 1873 he wrote down 4 equations which govern all classical electromagnetic phenomena.A magnetic field changing with time can produce an electric field: Faraday’s lawSlide 20Can a changing electric field with time produce an magnetic field:Maxwell’s law of inductionFind the expression for the induced magnetic field B that circulates around the electric field lines of a charging circular parallel plate capacitorAmpere-Maxwell’s LawWhat is the displacement current?Show that the displacement current in the gap of the two capacitor plates is equal to the real current outside the gapCalculation of idCalculate Magnetic field due to displacement currentQuestion 11: A circular capacitor of radius R is being charged through a wire of radius R0. Which of the points a, b, c, and d correspond to points 1, 2, and 3 on the graphSlide 30Slide 31Slide 32Summary of Maxwell Equations Integral formWarm up set 10 Due 8:00 am TuesdayLecture 11 Magnetism of Matter: Maxwell’s Equations Chp. 32Wednesday Morning•Cartoon -. Opening Demo - •Warm-up problem•Physlet •Topics–Finish up Mutual inductance–Ferromagnetism –Maxwell equations–Displacement current–Exam•DemosWhat is Mutual Inductance? MWhen two circuits are near one another and both have currents changing, they can induce emfs in each other.On circuit boards you have to be careful you do not put circuits near each other that have large mutual inductance.They have to be oriented carefully and even shielded.221111IMILm+=φ112222IMILm+=φMMM ==21121 2I1I271. Two coils, connected as shown, separately have inductances L1 and L2. Their mutual inductance is M.(a) Show that this combination can be replaced by a single coil of equivalent inductance given by.221MLLLeq++=We assume that the current is changing at (nonzero) rate di/dt and calculate the total emf across both coils. First consider coil 1. The magnetic field due to the current in that coil points to the left. The magnetic field due to current in coil 2 also points to the left. When the current increases, both fields increase and both changes in flux contribute emf’s in the same direction. .)()(2211dtdiMLanddtdiML +−=+−= εεTherefore, the total emf across both coils is Thus, the induced emf’s aredtdiMLL )2(2121++−=+= εεεwhich is exactly the emf that would be produced if the coils were replaced by a single coil with inductance.221MLLLeq++=(b) How could the coils in this figure be reconnected to yield an equivalent inductance of?221MLLLeq−+=We imagine reversing the leads of coil 2 so the current enter at the back of the coil rather than front (as pictured in the diagram). Then the field produced by coil 2 at the site of coil 1 is opposite to the field produced by coil 1 itself. The fluxes have opposite signs. An increasing current in coil 1 tends to increase the flux in that coil, but an increasing current in coil 2 tends to decrease it. The emf across coil 1 isdtdiML )(11−−=εSimilarly, the emf across coil 2 isdtdiML )(22−−=εThe total emf across both coils isdtdiMLL )2(21−+−=εThis is the same as the emf that would be produced by a single coil with inductanceMLLLeq221−+=75. A rectangular loop of N closely packed turns is positioned near a long, straight wire as shown in the figure.(a) What is the mutual inductance M for the loop-wire combination?(b) Evaluate M for N = 100, a = 1.0 cm, b = 8.0 cm, and l = 30 cm. € Φ = Bwireldr =μ0il2πrdr =aa +b∫aa +b∫μ0il2πln raa +b=μ0il2πln(1+ba)(a) The flux over the loop cross section due to the current i in the wire is given byThus,€ M =NΦi=Nμ0l2πln 1+ba ⎛ ⎝ ⎜ ⎞ ⎠ ⎟(b) Evaluate M for N = 100, a = 1.0 cm, b = 8.0 cm, and l = 30 cm. (b) From the formula for M obtained,( )( )⎟⎠⎞⎜⎝⎛+⋅×=−0.10.81ln230.01041007ππ mmHMH5103.1−×=€ M =Nμ0l2πln 1+ba ⎛ ⎝ ⎜ ⎞ ⎠ ⎟FerromagnetismIron, cobalt, nickel, and rare earth alloys exhibit ferromagnetism.The so called exchange coupling causes electron magnetic momentsof one atom to align with electrons of other atoms. This alignment producesmagnetism. Whole groups of atoms align and form domains. (See Figure 32-12 on page 756)A material becomes a magnet when the domains line up adding all themagnetic moments.You can actually hear the domains shifting by bringing up an magnet and hear the induced currents in the coil. Barkhausen EffectTwo other types of magnetic behavior are paramagnetism or diamagnetism.What is the atomic origin of magnetism?Electron spinning on its axisElectron orbiting around the nucleusSpin Magnetic Dipole Moment of the Electron € r μ =−emr S e =1.6 ×10−9Coulombsm = 9.11×10−31kgS is the angular momentum due to the electron’s spin. It has units kg.m2/s. has units of A.m2 - current times areaRecall for a current loop, the magnetic dipole moment = current times area of loopIn the quantum field theory of the electron, S can not be measured. Only it’s component along the z axis can be measured. In quantumphysics, there are only two values of the z component of the electron spin.Therefore, only the z component of can be measured.Its two possible values are:€ μz= ±eh4πmCorresponding to the two values of the electron spin quantum number +1/2and -1/2The above quantity is called the Bohr magneton and is equal to:€ μB= μz=eh4πm= 9.27 ×10−24A.m2The magnetic moment of the electron is the prime origin of ferromagnetism in materials.22. The dipole moment associated with an atom of iron in an iron bar is 2.1x10-23 J/T. Assume that all the atoms in the bar, which is 5.0 cm long and has a cross-sectional area of 1.0 cm2, have their dipole moments aligned.(a) What is the dipole moment of the bar?(b) What torque must be exerted to hold this magnet perpendicular to an external field of 1.5 T? (The density of iron is 7.9 g/cm3)(a) The number of iron atoms in the iron bar is( )( )( )( )(


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UVA PHYS 632 - Lecture 11 Magnetism of Matter: Maxwell’s Equations Chp. 32

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