Exam 2 covers Ch. 28-33, Lecture, Discussion, HW, LabElectric current produces magnetic fieldLaw of Biot-SavartMagnetic field from long straight wire: DirectionCurrent dependenceDistance dependenceWhy?Long straight wireField from a circular loopMagnetic field from loopMagnetic field from a current loopSolenoid electromagnetComparing Electric, MagneticA shortcut: Ampere’s lawAmpere’s law‘Testing’ Ampere’s lawUsing Ampere’s lawQuick QuizTue. Oct. 28, 2008 Physics 208, Lecture 17 1Exam 2 covers Ch. 28-33,Lecture, Discussion, HW, LabChapter 28: Electric flux & Gauss’ lawChapter 29: Electric potential & workChapter 30: Electric potential & field (exclude 30.7)Chapter 31: Current & ConductivityChapter 32: Circuits(exclude 32.8)Chapter 33: Magnetic fields & forces(exclude 33.3, 33.6, 33.10, Hall effect)Exam 2 is Tue. Oct. 28, 5:30-7 pm, 2103 ChTue. Oct. 28, 2008 Physics 208, Lecture 17 2Electric current produces magnetic fieldCurrent (flow of electric charges )in wire produces magnetic field.That magnetic field aligns compass needleCurrentMagnetic fieldTue. Oct. 28, 2008 Physics 208, Lecture 17 3Law of Biot-SavartEach element of current produces a contribution to the magnetic field.rIds€ dB =μo4πIds ׈ r r2B out of pagedIdBrTue. Oct. 28, 2008 Physics 208, Lecture 17 4Magnetic field from long straight wire:DirectionWhat direction is the magnetic field from an infinitely-long straight wire?I € dr B =μo4πIdr s ׈ r r2xyTue. Oct. 28, 2008 Physics 208, Lecture 17 5Current dependenceHow does the magnitude of the B-field change if the current is doubled?I € dr B =μo4πIdr s ׈ r r2xyA) Is halved B) QuadruplesC) Stays sameD) DoublesE) Is quarteredTue. Oct. 28, 2008 Physics 208, Lecture 17 6Distance dependenceHow does the magnitude of the B-field at 2 compare to that at 1?I € dr B =μo4πIdr s ׈ r r2xy12A) B2=B1B) B2=2B1C) B2=B1/2D) B2=4B1E) B2=B1/4Tue. Oct. 28, 2008 Physics 208, Lecture 17 7Why?Biot-Savart saysWhy B(r) 1/r instead of 1/r2 ? € dr B =μo4πIdr s ׈ r r2 € r r Large contribution from this current element.Decreases as 1/r2ISmall contribution from this current element.~ independent of rTue. Oct. 28, 2008 Physics 208, Lecture 17 8Long straight wireAll current elements produce B out of pagexa€ r = x2+ a2r€ dB =μo4πIds ׈ r r2=μo4πIr2sinθ=μo4πIr2ar=μo4πIax2+ a2( )3 / 2€ B =μoIa4πdxx2+ a2( )3 / 2=−∞∞∫μoI2πaAdd them all up:Tue. Oct. 28, 2008 Physics 208, Lecture 17 9Field from a circular loopEach current element produce dBAll contributions add as vectorsAlong axis, allcomponents cancelexcept for x-compTue. Oct. 28, 2008 Physics 208, Lecture 17 10Magnetic field from loopWhich of these graphs best represents the magnetic field on the axis of the loop?BzzxyBzBzBzA.B.C.D.zzzzTue. Oct. 28, 2008 Physics 208, Lecture 17 11Magnetic field from a current loopOne loop: field still loops around the wire.Many loops: same effectTue. Oct. 28, 2008 Physics 208, Lecture 17 12Solenoid electromagnetSequence of current loops can produce strong magnetic fields.This is an electromagnetTue. Oct. 28, 2008 Physics 208, Lecture 17 13Comparing Electric, MagneticBiot-Savart: calculate B-field from current distribution.Resulting B-field is a vector, and…complication: current (source) is a vector!Coulomb: calculate E-field from charge distributionResulting E is a vector but charge (source) is not a vectorTue. Oct. 28, 2008 Physics 208, Lecture 17 14A shortcut: Ampere’s lawIntegral around closed path proportional to current passing through any surface bounded by path.Ampere’s law€ B • ds∫= μoIclosed pathsurface bounded by pathIRight-hand ‘rule’: Thumb in direction of positive currentCurled fingers show direction integrationTue. Oct. 28, 2008 Physics 208, Lecture 17 15Ampere’s lawSum up component of B around pathEquals current through surface. Ampere’s law€ B • ds∫= μoIclosed pathsurface bounded by pathI € r B Component of B along pathTue. Oct. 28, 2008 Physics 208, Lecture 17 16‘Testing’ Ampere’s lawLong straight wire , Br € B r( )=μoI2πrr€ B • ds∫IB(r)B||dspath has constant rpath length = 2πr€ B • ds∫=μoI2πrds∫€ B • ds∫=μoI2πrds∫=μoI2πrds∫€ B • ds∫=μoI2πrds∫=μoI2πrds∫=μoI2πr2πr = μoICircular pathSurface bounded by pathTue. Oct. 28, 2008 Physics 208, Lecture 17 17Using Ampere’s lawCould have used Ampere’s law to calculate BrIB(r)€ B • ds∫ = Bds =∫ B ds∫ = B2πr = μoI ⇒ B =μoI2πrCircular pathSurface bounded by pathB||dsB constant on pathpath length = 2πrTue. Oct. 28, 2008 Physics 208, Lecture 17 18Quick QuizSuppose the wire has uniform current density. How does the magnetic field change inside the wire?A. Increases with rB. Decreases with rC. Independent of rD. None of the aboverB€ B∫• ds = 2πrB= μoIcut= μoIπr2πR2⇒ B r(
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