Lecture 17: THU 18 MAR 10Remember Gauss Law for E-Fields?Gauss’s Law for B-Fields!Ampere’s law: Closed LoopsSlide 5PowerPoint PresentationSample ProblemAmpere’s Law: Example 1Ampere’s Law: Example 2Ampere’s Law: Example 2 (cont)Slide 11SolenoidsSlide 13Slide 14Magnetic Field of a Magnetic DipoleSlide 16Slide 17Lecture 17: THU 18 Lecture 17: THU 18 MAR 10MAR 10Ampere’s law Ampere’s law Physics 2102Jonathan DowlingAndré Marie Ampère (1775 – 1836)QuickTime™ and aTIFF (LZW) decompressorare needed to see this picture.Given an arbitrary closed surface, the electric flux through it isproportional to the charge enclosed by the surface.qFlux7=70!q∫=⋅≡ΦSurface0εqAdErr∫=⋅≡ΦSurface0εqAdErrRemember Gauss Law for E-Remember Gauss Law for E-Fields?Fields?Gauss’s Law for B-Fields!Gauss’s Law for B-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—∫=μ0ienclosedThe 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. € r B ⋅dr s loop∫= μ0(i1− i2)Thumb rule for sign; ignore i3If 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—∫=μ0ienclosed rB ⋅drsLOOP—∫=μ0ienclosedenc Calculation of . We curl the fingersof the right hand in the direction in which the Amperian loop was traversed. We notethei direction of the thumb. All currents inside the loop to the thumb are counted as .All currents inside the loop to the thumb are counted as .All currents outside the loop are not countpaantiparalrallel positivelel negativeenc 1 2ed. In this example : .i i i= −Sample 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 so B is tangential to the loop everywhere!•Angle between B and ds is 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 Calculus.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 Long7Wire!r >RB =μ0i2πrOutside Long7Wire!SolenoidsSolenoidsinBinhBhisdBinhhLNiiNihBsdBisdBenchencenc0000)/(000μμμμ=⇒=⇒=•===+++=•=•∫∫∫rrrrrrThe n = N/L is turns per unit length.The n = N/L is turns per unit length.QuickTime™ and aYUV420 codec decompressorare needed to see this picture.Magnetic Field in a Toroid “Doughnut” Solenoid QuickTime™ and a decompressorare needed to see this picture.Toriod Fusion Reactor: Power NYC For a Day on a Glass of H2O€ B =μ0Ni2πr€ B =μ0Ni2πrMagnetic Field of a Magnetic Magnetic Field of a Magnetic DipoleDipole302/32202)(2)(zzRzBμπμμπμrrr≈+=302/32202)(2)(zzRzBμπμμπμrrr≈+=All loops in the figure have radius r or 2r. Which of these arrangements produce the largest magnetic field at the point indicated?A circular loop or a coil currying electrical current is a magnetic dipole, with magnetic dipole moment of magnitude =NiA. Since the coil curries a current, it produces a magnetic field, that can be calculated using Biot-Savart’s law:Magnetic Dipole in a B-Field rF =0rτ =rμ ×rBU =−rμgrBForce is zero in homogenous fieldTorque is maximum when 7=790°Torque is minimum when 7=70° or 180°Energy is maximum when 7=7180°Energy is minimium when 7=70°QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and aAnimation decompressorare needed to see this picture.Neutron Star a Large Magnetic
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