Lecture 12 Magnetism of Matter: Maxwell’s Equations Chp. 32PowerPoint PresentationWhat is Mutual Inductance? MSlide 6FerromagnetismWhat is the atomic origin of magnetism?Spin Magnetic Dipole Moment of the ElectronSlide 10Slide 11Slide 12(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 15Slide 16Maxwells 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 lawNew Question: Can 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 graphSummary of Maxwell Equations Integral formLecture 12 Magnetism of Matter: Maxwell’s Equations Chp. 32•Cartoon •Warm-up problem•Opening Demo•Topics–Finish up Mutual inductance–Ferromagnetism –Maxwell equations–Displacement current•DemosWhat is the magnetic energy stored in a solenoid or coil€ dUBdt= Lididt€ dUB= Lidi€ dUB0UB∫= Lidi0i∫€ UB= Lidi0i∫=12Li2€ UB=12Li2For an inductor LNow define the energy per unit volume € uB=UBAlArea A l€ uB=12Li2Al=Lli22A€ Ll= μ0n2A€ uB=12Li2Al=12μ0n2i2€ uB=B22μ0€ B = μ0ni€ uE=E22ε0The energy density formula is valid in generalWhat 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 2I1I275. 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 by€ M =NΦi=Nμ0l2πln 1+ba ⎛ ⎝ ⎜ ⎞ ⎠ ⎟M =NΦi(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? The number of iron atoms in the iron bar isN =7.9g cm3( )5.0cm( )1.0cm2( )55.847g mol( )6.022×1023mol( )=4.3×1023. Thus, the dipole moment of the bar isμ = 2.1 × 10−23J T( )4.3 × 1023( )= 9.03A ⋅m2.(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)(b) τ =μBsin90o= 9.03A⋅m2( )1.5T( )=13.5N ⋅m(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.  8 9 m25 cmA = 1 cm2€ B =μ0μ2πz3€ B =4π ×10−7NA28.9A.m2π × 0.05m(.01m)2€ B =4 ×10−78.9N / A.m5 ×10−6€ B = 0.71 TBT=μ0μ2πLdzz3.06.01∫=(μ0μ2πL)(−2z20.060.01) = (μ0μ2πL)(−2(0.01)2) € BT= dB∫=μ02πdμz3∫dμ =μALAdz =μLdzz.PBigBite is a 50 ton electromagnet with a 25 cm by 100 cm gapB = 1 TeslaMaxwells Equations:In 1873 he wrote down 4 equations which govern all classical electromagnetic phenomena.You already know two of them. € 1. ΦE=r E .dA∫= qenc/ε0 € 2. ΦB=r B .dA∫= 0A magnetic field changing with time can produce an electric field: Faraday’s law € 3. r E .dr s ∫= −dΦBdtElectric lines curl around changing magnetic field linesLine integral of the electric fieldaround the wire equals the change of Magnetic flux through the areaBounded by the loopExampleNew Question: Can a changing electric field with time produce an magnetic field?.Yes it can and it is calledMaxwell’s law of induction € r B .dr s ∫= μ0ε0dΦEdtMaxwell’s law of induction € r B .dr s ∫= μ0ε0dΦEdtConsider the charging of our