40 60 80 1000246Q.1 - 8.0Q.2 - 5.9Q.3 - 7.9Q.4 - 6.9Mean = 71.9Median = 74Sigma = 20Number of studentsScore on test (%)Exam 2 statisticsExam 2 statisticsLecture 17 Lecture 17 --The Magnetic Field of a CurrentThe Magnetic Field of a CurrentChapter 33 Chapter 33 --Tuesday March 20thTuesday March 20th•A brief review of the exam•Reminder about Lorentz contraction/time dilation•Interaction force between parallel currents•The magnetic field of a current•The magnetic field of a moving charge•The Biot-SavartlawReading: pages 749 thru 760 (up to section 33Reading: pages 749 thru 760 (up to section 33--5) in HRK5) in HRKRead and understand the sample problemsRead and understand the sample problemsNext WebAssign deadline will be on Monday at 11:59pmNext WebAssign deadline will be on Monday at 11:59pmHomework set (Ch. 33): I will post these later todayHomework set (Ch. 33): I will post these later todayPractice problems:Ch. 33 Practice problems:Ch. 33 --Ex. 5, 15, 21; Prob. 1Ex. 5, 15, 21; Prob. 1Time dilationTime dilationSSDDSS’’DD2'DtcΔ=2alvt=Δllhh()() ()2222222222222222224221'1',where and1hahvtlDlDlvttDccvDttccvttcγγ ββΔ⎛⎞=+=+⎜⎟⎝⎠⎡⎤Δ⎛⎞ ⎛ ⎞Δ= = +⎢⎥⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠⎢⎥⎣⎦⎛⎞⎛⎞−Δ= =Δ⎜⎟⎜⎟⎝⎠⎝⎠⇒Δ = Δ = =−Lorentz contractionLorentz contractionTurn the time dilation problem on its sideTurn the time dilation problem on its sideLoSVelocity v relative to SS’L + vt1L − vt22'oLtcΔ=11 2 21222222Fromthe timedilationresult12 2'111oooL vt ct L vt ctLLLctttcv cv c vLLcttccvLLLγββγβ+= −=Δ= + = + =−+−Δ=Δ = =−−−⇒= =−i.e. length L measured in S is shorter than the proper length Lo, as measured in S’, by a factor of γ.A major stumbling block......A major stumbling block......Can you see the problem?Can you see the problem?The laws of physics are the same in all inertial reference frameThe laws of physics are the same in all inertial reference frames.s.v, iv, iμμ μππ π⎛⎞⎟⎜== = =⎟⎜⎟⎜⎝⎠2;22 2oo oii iFL iLBiL Brr rThe resulting magnetic forceThe resulting magnetic forcerL121 2 1 22oiFiLBiLdμπ==The magnetic force between two wiresThe magnetic force between two wires21 2 1i=×FLBGGG7410 Tm/Aoμπ−=× ⋅μochosen so that when i1= i2= 1 A, and L = d = 1 m, F21= 2×10-7NAccording to rules developed earlier, B must be down.2oiBrμπ=The magnetic field due to a wire in 3DThe magnetic field due to a wire in 3DLike Coulomb’s Law:(For an infinite line charge)2oErλπε=12212oLi iFdμπ=21 2 1i=×FLBGGG⊗⊗i1i2FF2121BB11LetLet’’s look ends look end--on at the wires (2D problem)on at the wires (2D problem)12122oLi iFdμπ=12 1 1i=×FLBGGG⊗⊗i1i2FF1212BB22LetLet’’s look ends look end--on at the wires (2D problem)on at the wires (2D problem)⊗⊗i1i2BB22⊗⊗i1BB11FF1212FF2121LetLet’’s look ends look end--on at the wires (2D problem)on at the wires (2D problem)2oiBrμπ=The magnetic field due to a wire in 3DThe magnetic field due to a wire in 3DLike Coulomb’s Law:(For an infinite line charge)2oErλπε=μλελ→ = →⇒→=1; oodqidsdq ds vdqMagnetic field of a moving point chargeThe magnetic field due to a moving chargeThe magnetic field due to a moving charge• The field strength is directly proportional to the magnitude of the velocity vand the charge q.• If vreverses direction or qchanges sign, then so does the direction of the magnetic field B.• The field is zero at points along the direction of v(forward as well as backward).• The field Bis tangent to circles drawn about the velocity vin planes perpendicular to the velocity. The direction of vis determined by the right-hand-rule.• The field decreases like 1/r2along lines perpendicular to the motion of q.()μπ=×2geometrical factor4ovdqdBrThe magnetic field due to a moving chargeThe magnetic field due to a moving chargeLike Coulomb’s Law:(For a point charge)214oqErπε=μπ=×2..4ovqBGFr22ˆ4sin4ooqrqvBrμπφμπ×==vrBGGμε→→1;ooqvqThe BiotThe Biot--Savart LawSavart Law22sin sin44oovdq i dsdBrrμφμφππ==23ˆ44ooid iddrrμμππ××==sr srBGGGGThe BiotThe Biot--Savart LawSavart
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