Lecture 23: WED 11 MARPowerPoint PresentationForces Between WiresSummaryRemember Gauss Law for E-Fields?New Gauss Law for B-Fields!Ampere’s law: Closed LoopsSample ProblemAmpere’s Law: Example 1Ampere’s Law: Example 2Ampere’s Law: Example 2 (cont)Slide 12SolenoidsLecture 23: WED 11 Lecture 23: WED 11 MARMARAmpere’s law Ampere’s law Physics 2102Jonathan DowlingAndré Marie Ampère (1775 – 1836)QuickTime™ and aTIFF (LZW) decompressorare needed to see this picture.Exam 02 and Midterm GradeQ1&P1, Dr. Schafer, Office hours: MW 1:30-2:30 pm, 222B NicholsonQ2&P2, Dr. Gaarde, Office hours: TTh 2:30-3:30 pm, 215B NicholsonQ3&P3, Dr. Lee, Office hours: WF 2:30-3:30 pm, 451 NicholsonQ4&P4, Dr. Buth, Office hours: MF 2:30-3:30 pm, 222A NicholsonExam 02 Average: 63/100You can calculate your midterm letter grade, as posted on paws, via this handy formula used by all sections:[(MT01+MT02)*500/200+HWT/761]*30]/530*100 = Midterm ScoreHere MT01 and MT02 are your scores on the first and second midterms and HWT is your total homework points for HW01-07 only.Once you have this Midterm Score you then can get your letter grade via:A: 90-100 B: 80-89 C: 60-79 D: 50-59 F: below 50.Details on the grading policy are at: http://www.phys.lsu.edu/classes/spring2009/phys2102/I have posted both midterm exam grades and your midterm class score on WebAssign.aIBπμ2101=Magnetic field due to wire 1 where the wire 2 is,1221BILF =aI2I1LaIILπμ2210=aIILπμ2210=Force on wire 2 due to this field,FForces Between WiresForces Between WiresNeil’s Rule: Same Currents – Wires Attract!Opposite Currents – Wires Repel!Neil’s Rule: Same Currents – Wires Attract!Opposite Currents – Wires Repel!SummarySummary• Magnetic fields exert forces on moving charges:• The force is perpendicular to the field and the velocity.• A current loop is a magnetic dipole moment. • Uniform magnetic fields exert torques on dipole moments. • Electric currents produce magnetic fields:•To compute magnetic fields produced by currents, use Biot-Savart’s law for each element of current, and then integrate.• Straight currents produce circular magnetic field lines, with amplitude B=0i/2r (use right hand rule for direction). • Circular currents produce a magnetic field at the center (given by another right hand rule) equal to B=0i/4r • Wires currying currents produce forces on each other: Neil’s Rule: parallel currents attract, anti-parallel currents repel.Given an arbitrary closed surface, the electric flux through it isproportional to the charge enclosed by the surface.qFluxR=R0!q∫=⋅≡ΦSurface0εqAdErr∫=⋅≡ΦSurface0εqAdErrRemember Gauss Law for E-Remember Gauss Law for E-Fields?Fields?New Gauss Law for B-New Gauss Law for B-Fields!Fields!No isolated magnetic poles! The magnetic flux through any closed “Gaussian surface” will be ZERO. This is one of the four “Maxwell’s equations”.∫=• 0AdB∫=• 0AdBThere are no SINKS or SOURCES of B-Fields!What Goes IN Must Come OUT!There are no SINKS or SOURCES of B-Fields!What Goes IN Must Come OUT!The circulation of B (the integral of B scalar ds) along an imaginary closed loop is proportional to the net amount of current traversing the loop.The circulation of B (the integral of B scalar ds) along an imaginary closed loop is proportional to the net amount of current traversing the loop.i1i2i3dsi4)(3210loopiiisdB −+=⋅∫μrrThumb rule for sign; ignore i4If you have a lot of symmetry, knowing the circulation of B allows you to know B.Ampere’s law: Closed Ampere’s law: Closed Loops Loops rB ⋅drsLOOP—∫=μ0ienclosedSample ProblemSample Problem•Two square conducting loops carry currents of 5.0 and 3.0 A as shown in Fig. 30-60. What’s the value of ∫B∙ds through each of the paths shown?Path 1: ∫B∙ds = 0•(–5.0A+3.0A)Path 2: ∫B∙ds = •(–5.0A–5.0A–3.0A)Ampere’s Law: Example Ampere’s Law: Example 11•Infinitely long straight wire with current i.•Symmetry: magnetic field consists of circular loops centered around wire.•So: choose a circular loop C -- B is tangential to the loop everywhere!•Angle between B and ds = 0. (Go around loop in same direction as B field lines!)∫=⋅CisdB0μrr∫==CiRBBds0)2( μπRiBπμ20=RiBπμ20=RMuch Easier Way to Get B-Field Around A Wire: No Cow-Culus.Ampere’s Law: Example Ampere’s Law: Example 22•Infinitely long cylindrical wire of finite radius R carries a total current i with uniform current density•Compute the magnetic field at a distance r from cylinder axis for:–r < a (inside the wire)–r > a (outside the wire)•Infinitely long cylindrical wire of finite radius R carries a total current i with uniform current density•Compute the magnetic field at a distance r from cylinder axis for:–r < a (inside the wire)–r > a (outside the wire)iCurrent out of page, circular field lines∫=⋅CisdB0μrr∫=⋅CisdB0μrrAmpere’s Law: Example 2 (cont)Ampere’s Law: Example 2 (cont)∫=⋅CisdB0μrrCurrent out of page, field tangent to the closed amperian loop enclosedirB0)2( μπ =22222)(RrirRirJienclosed=== πππriBenclosedπμ20=202 RirBπμ=202 RirBπμ=For r < RFor r>R, ienc=i, soB=0i/2R = LONG WIRE!For r>R, ienc=i, soB=0i/2R = LONG WIRE!Need Current Density J!R02iRμπrBOAmpere’s Law: Example 2 (cont)Ampere’s Law: Example 2 (cont)r <RB ∝ rr <RB ∝ rr >RB ∝ 1/ rr >RB ∝1/ rr <RB =μ0ir2πR2r <RB =μ0ir2πR2r >RB =μ0i2πrOutside LongRWire!r >RB =μ0i2πrOutside LongRWire!SolenoidsSolenoidsinBinhBhisdBinhhLNiiNihBsdBisdBenchencenc0000)/(000μμμμ=⇒=⇒=•===+++=•=•∫∫∫rrrrrrThe n = N/L is turns per unit length.The n = N/L is turns per unit
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