Presentation on theme: "Lecture 11 Magnetism of Matter: Maxwell’s Equations Chp"— Presentation transcript:
1Lecture 11 Magnetism of Matter: Maxwell’s Equations Chp Lecture 11 Magnetism of Matter: Maxwell’s Equations Chp. 32 Wednesday MorningCartoon -. Opening Demo -Warm-up problemPhysletTopicsFinish up Mutual inductanceFerromagnetismMaxwell equationsDisplacement currentExamDemos
2What is Mutual Inductance? M When two circuits are near one another and both have currents changing, they can induce emfs in each other.12I1I2On 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.
471. 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 byBWe 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.
5B e Thus, the induced emf’s are Therefore, the total emf across both coils iswhich is exactly the emf that would be produced if the coils were replaced by a single coil with inductance
6(b) How could the coils in this figure be reconnected to yield an equivalent inductance of 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.eThe 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.
7B The emf across coil 1 is Similarly, the emf across coil 2 is The total emf across both coils isThis is the same as the emf that would be produced by a single coil with inductance
875. 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.(a) The flux over the loop cross section due to the current i in the wire is given byThus,
9(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,
10FerromagnetismIron, 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 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 bringingup an magnet and hear the induced currents in the coil. Barkhausen EffectTwo other types of magnetic behavior are paramagnetism ordiamagnetism.
11What is the atomic origin of magnetism? Electron spinning on its axisElectron orbiting around the nucleus
12Spin Magnetic Dipole Moment of the Electron S is the angular momentum due to the electron’s spin. Ithas units kg.m2/s. m 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.
13Therefore, only the z component of m can be measured Therefore, only the z component of m can be measured.Its two possible values are:Corresponding to the two values ofthe electron spin quantum number +1/2and -1/2The above quantity is called the Bohr magneton and is equal to:The magnetic moment of the electron is the prime origin of ferromagnetism in materials.
1422. 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 isThus, the dipole moment of the bar is
15(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.5 cm.PzA = 1 cm2m= 8.9 A.m2
16BigBite is a 50 ton electromagnet with a 25 cm by 100 cm gap B = 1 Tesla
18Maxwells Equations: In 1873 he wrote down 4 equations which govern all classical electromagnetic phenomena.You already know two of them.
19A magnetic field changing with time can produce an electric field: Faraday’s law Line integral of the electric fieldaround the wire equals the change ofMagnetic flux through the areaBounded by the loopElectric lines curl aroundchanging magnetic field linesExample
20Faraday’s Law B is increasing in magnitude Note that induced E field is in such adirection that the B fieldit produces opposes theoriginal B field.Note there is no electric potentialassociated with the electric fieldinduced by Faraday’s Law
21Can a changing electric field with time produce an magnetic field: Yes it can and it is calledMaxwell’s law of induction.
22Maxwell’s law of induction Consider the charging of our circularplate capacitorB field alsoinduced atpoint 2.When capacitor stops chargingB field disappears.
23Find the expression for the induced magnetic field B that circulates around the electric field lines of a charging circular parallel plate capacitorr < RFlux withinthe loop ofradius rEBrRr< Rr > R
24Ampere-Maxwell’s Law Maxwell combined the above two equations to form oneequationThis term has units of currentHow do we interpret this equation?
25What is the displacement current? This is called thedisplacement current idThe term is really is a transfer of electric and magnetic energy from oneplate to the other while the plates are being charged or discharged.When charging stops, this term goes to zero. Note it is time dependent.
26Show that the displacement current in the gap of the two capacitor plates is equal to the real current outside the gapCan I detect the magnetic field associated with displacement current?
27Calculation of id First find the real current i For the field inside a parallel plate capacitorSolving for qThis is the real current i charging the capacitor.Next find the displacement currentdisplacement current =real current. No chargeactually moves acrossthe gap.
28Calculate Magnetic field due to displacement current Current is uniformly spread over the circular plates of the capacitor.Imagine it to be just a large wire of diameter R. Then use theformula for the magnetic field inside a wire.Inside the capacitorOutside the capacitor
29Question 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 graphWhere is the radius R0 and R on the graph?
3037. A parallel-plate capacitor has square plates 1 37. A parallel-plate capacitor has square plates 1.0 m on a side as shown in the figure. A current of 2.0 A charges the capacitor, producing a uniform electric field between the plates, with perpendicular to the plates.(a) What is the displacement current id through the region between the plates?At any instant the displacement current id in the gap between the plates equals the conduction current i in the wires. Thus, id = i = 3.0 A.(b) What is dE/dt in this region?The rate of change of the electric field is
31(c) What is the displacement current through the square, dashed path between the plates? The displacement current through the indicated path is
32(d) What is around this square, dashed path? The integral field around the indicated path is(e) What is the value of B on this path?????
34Warm up set 10 Due 8:00 am Tuesday HRW6 31.TB.02.  Suppose this page is perpendicular to a uniform magnetic field and themagnetic flux through it is 5 Wb. If the page is turned by 30° around an edge the fluxthrough it will be:4.3 Wb10 Wb5.8 Wb2.5 Wb5 Wb2. HRW6 31.TB.08.  Faraday's law states that an induced emf is proportional to:the rate of change of the electric fieldthe rate of change of the magnetic fieldzerothe rate of change of the magnetic fluxthe rate of change of the electric flux3. HRW6 31.TB.09.  The emf that appears in Faraday's law is:around a conducting circuitperpendicular to the surface used to compute the magnetic fluxthroughout the surface used to compute the magnetic fluxnone of thesearound the boundary of the surface used to compute the magnetic flux