BLock 6 Comprehensive Notes
BLock 6 Comprehensive Notes PHYS 242
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This 3 page Study Guide was uploaded by DB on Tuesday September 27, 2016. The Study Guide belongs to PHYS 242 at North Carolina A&T State University taught by Prof. Sandin in Fall 2016. Since its upload, it has received 5 views. For similar materials see General Physics II in Physics 2 at North Carolina A&T State University.
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Date Created: 09/27/16
PHYS 242 BLOCK 6 NOTES Sections 29.1 to 29.7, 30.1 to 30.3 dΦΒ Experimentally, we fεn = –N dt , calledFaraday’s lawof induction. ε is the induced emf(in V)—a potential difference thatmay give aninduced current. N is the number of turns (no unit). Φ Βs the average magnetic flux (in Wb) through each turn at (in s).me dΦΒ Wb Wb dt is therate at whiΒhΦ is changing with tise ). Thus 1s = 1 V. Lenz’s law: The direction of any magnetic induction effect is such as to oppose the cause of the effect. The derivation on page 967 gives a relation that we can write s = υ⊥ ⊥ ⊥l. a ε is the (motional) emf (in V). → m υ⊥, Β⊥, and⊥ are themutually perpendicular componentsof the velocstyυ (in ) , the uniform magnetic fieldB (in T) , and the length vectorl (in m) of a straight segment. Cover up the solutions and carefully work Examples 29.1 to 29.5 and 29.7 to 29.10. In TEST YOUR UNDERSTANDING on page 969, it should say, “The earth’s magnetic field points toward the earth’s south magnetic pole, which is near the earth’s north geographic pole (see Fig. 27.3).” → → Since the emf is a potential difference, and a potential diffntegral of E ·dl , for a single turn (N = 1) Faraday’s law of induction b E ·d l = – B . dt → V N This E is theinduced electric fmeorCin). Cover up the solution and carefully work Example 29.11. Eddy currentsare induced currents that circulate within thconductingmaterial. → → Trying to apply Ampere’s la∫ B ·d l = µ 0 enclin Fig. 29.22 givencl= C through the plane surface, bencl= 0 through the bulging surface. Maxwell resolved this contradicta quantity in the dΦE dielectric called thedisplacement cDrreDti : dt ≡. The displacementcurrent is not an actual motion of charge, but has the SI unit of ampere = amp D A. Notei = 0only if theEelectric fluxΦ is constant. C 2 F ε is the permittivN·m 2iorm ). 2 Φ is the electric flux (inor V·m). E C dΦE N·m2 V·m dt is therate at which the electric flux changes wC·s tores(i). → → Thus, whenno magnetic materials are present, Ampere’s la∫ B ·d l = 0 Ci +D encl iCis the enclosed conduction current (in A) (caused by actual motion of charge) iDis the enclosed displacement current (in A) (caused by changing enclosed electric flux). 1 The four Maxwell’s equations summarize classical electromagnetism. The first four questions on your finalexam will involve matching verbal descriptions to these four equations. For no dielecicaterials,netm Maxwell’s equations reduce to: o→ → Q encl ∫ E ·d A = ε0 , telling us electric field lines can start on positive charges and end on negative charges. ∫ B ·d A = 0, telling us there are evidently no magnetic monopoles on which to start and to end magnetic field lines. → → dΦ E ∫ B ·d l = 0 Ci + 0 dt ), telling us closed magnetic fieldlines are produced by the motion of charge and/or by changing electric flux. ∫ E ·d l = –dΦ B , telling us closed electric field lines are evidently produced only by changing magnetic flux. dt Suppose we have two coils, as in Fig. 30.1. A changingcurrent in coil 1 will produce a changing magnetic field, giving changing magnetic flux througand, by Faraday’s law of induction,an induced emf in coil 2 : dΦ Β2 dΦΒ1 ε2 = –N2 dt , whereN2 B2 = M 21 1Also ε1 = –N1 dt , whereN1 B1 = M 12 2Advancedelectromagnetism d1 di2 courses show M21= M 12= M. Therefore,ε2 = –M dt and ε 1 = –M dt . d1 A ε 2is the emf (in V) induced in coil2 duesolelyto the time rate of changdto(ins) in coil1. di ε 1is the emf (in V) induced in coil1 duesolelyto the time rate of change o(in ) in coil2. dt s M is the mutual inductance (in henry = H). N Φ N Φ Solving bothN2 B2 = Mi1and N 1 B1 = Mi 2or M gives us M = 2 B2 = 1 B1 .In this equation,all 1 2 quantities are never negative. N2and N 1re the numbers of turns in coils 2 and 1 (no units). Φ is the average magnetic flux (in Wb) through each turl due solelyto i , the current (in A) in coil1. B2 1 Φ B1is the average magnetic flux (in Wb) through each turl due solelyt2 i , the current (in A) in coil2. Wb T·m 2 (N/A·m)·m2 J J V·s The inductance unit is one henry, 1 A = = 1 A = 1 A = 1 A2 = 1 (C/s)·A= 1 A = 1 Ω·s. Cover up the solutions and carefully work Examples 30.1 and 30.2. Now suppose we consider a single coil, as in Fig. 30.4. A changing current in that coil will produce a changing magnetic field, giving changing magnetic flux through the coil and,by Faraday’s law of induction, an dΦ NΦ self-induced emε (in V)ε = –N Β , whereNΦB= Li, which givesε = –L i . Thus L = B is the self- dt dt i NΦ B inductanceL (in H) of that coil (often just called theinductanceof that coii)., all four quantities are never negative. Cover up the solutions and carefully work Examples 30.3 and 30.4. 2 Suppose we have two ordinary coils. Each coil will have a self-inductance1(L and 2 ) and the two coils d1 N 1Φ B 1 may also have a mutual inductanceM( ). Thus, for coil1, we can wrε1,self –L 1dt , with 1 = 1 , where (Φ B 1s the average magnetic flux through each of the N1turns of coil 1 due solely to coil 1’s curre1t i . Thus (Φ B 1from coil1’s current) isnot the same average magneticflux as ΦB1 (from coil 2’s current). For coil 2, we di N (Φ ) similarly writε2,self –L 2 2, withL2= 2 B 2 [where (Φ B 2self) isnot the same as ΦB2 (mutual)]. That is, dt 2 we have three possible inductances (L1 2 , and M) arising from four possible average magnetic fluxes [(ΦB 1, (Φ B 2 ΦB1 andΦ ]B2Then, changes in those average magnetic fluxes with time give four emfs ε1,selfε1,mutual, ε 2,selfandε2,mutual). An inductor(sometimes called achoke) is a circuit element used mainly for its inductance. On page 998, the text derives an expression for the magnetic potential energy U (in J) stored in the magnetic field of an inductor of (self-)inductanceL (in H) when carrying a current I (in A): U = . The unit 2 J equivalence 1 H = 1A 2 is helpful in this equation. Cover up the solution and carefully work Example 30.5. 1 We used U = 2 CV and a parallel-plate capacitor to find the energy deu of an electric field.Similarly, 1 we now use U = LI2and a toroid to find the energy density u of a magnetic field. Consider a toroid of small cross section and two layers of wire wound so that its magnetic field is zero the “dough” of its doughnut- shaped core. NΦ µNI µN A First we find its self-inductanceL = B = NBA . We substitute B = and i = I to find L = . By i i 2πr 2πr 1 1 µN A U 2LI2 2 2πr I2 1 µNI 2 1 definition,u =volume = volume = (2πr)A = 2µ 2πr ) = 2µ B . This final expression contains none of the B2 dimensions of the toroid and is true for all linear materials (materials with aµt): u =2µ . J → u is the energy density (mn3) of the magnetic fieldB (in T) . T·m µ (mu) is thepermeabilityof the material (in A ). T·m µ 0 the magnetic constant, is also the permeability of vacµ0 ≡ 4π × 10–7 A ). Thus µ ≡ µ 0y definition for vacuum and also for nonmagnetic materials. Because of therrdinarily weak magnetizations,µ is slightly greater than µ for paramagneticmaterials(if not at very low temperatures)and µ is 0 slightly less than0 for ordinary (not superconducting) diamagnetic materials. We must distinguish betweenU (magneticpotentialenergy), u (energy density), and µ (permeability).To minimize confusion, in this block we willnot also useµ to stand for themagnitude of the magneticdipole moment –6 (as inµ = NIA). However, we may use the metric prefixµ = 10 , as in µ H. 3
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