This preview shows page 1-2-3-4-5-6-7-8-9-10-67-68-69-70-71-72-73-74-75-136-137-138-139-140-141-142-143-144-145 out of 145 pages.
1 W12D2 Exam 3 Review: Summary of Presentations and Concept Questions2 W06D2 Magnetic Forces and Sources of Magnetic Fields W06D2 Magnetic Force on Current Carrying Wire, Sources of Magnetic Fields: Biot-Savart Law Reading Course Notes: Sections 8.3, 9.1-9.23 Outline Magnetic Force on Current Carrying Wire Sources of Magnetic Fields Biot-Savart Law4 Review: Lorentz Force Law F = qE +v ×B( ) Fmag= qv ×B Felec= qEForce on charged particles in electric and magnetic fields Electric Force Magnetic Force5 dFmag= dqv ×BMagnetic Force on Current-Carrying Wire dqv = dqdsdt=dqdtds = Ids dFmag= Ids ×B Fmag= I ds ×Bwire∫6 Magnetic Force on Current-Carrying Wire Fmag= I dswire∫⎛⎝⎜⎞⎠⎟×BIf the wire is a uniform magnetic field then If the wire is also straight then Fmag= I (L ×B)7 Magnetic Field of Moving Charge Moving charge with velocity v produces magnetic field: B =µo4πqv xˆrr2:ˆrunit vector directed from charged object to P rˆpermeability of free space µ0= 4π× 10−7T ⋅ m ⋅ A−1P8 The Biot-Savart Law Current element of length carrying current I produces a magnetic field at the point P: dB =µ04πI ds ׈rr2 ds B(r) =µ04πId′s × (r −′r )r −′r3wire∫9 The Right-Hand Rule #2 dirB(P) =ˆz ׈r =ˆϕ dir ds =ˆz10 The Biot-Savart Law: Infinite Wire Magnetic Field of an Infinite Wire Carrying Current I from Biot-Savart: See W06D3 Problem Solving http://web.mit.edu/8.02t/www/materials/ProblemSolving/solution05.pdf B =µ0I2πyˆk B =µ0I2πrˆθMore generally:11 Magnetic Field Generated by a Current Loop http://web.mit.edu/viz/EM/visualizations/magnetostatics/calculatingMagneticFields/RingMagInt/RingMagIntegration.htm12 Example: Coil of Radius R In the circular part of the coil… ds ⊥ˆrrˆsdIII dB =µ0I4πds ׈rr2=µ0I4πdsr2=µ0I4πR dθR2=µ0I4πdθRBiot-Savart: → | ds ׈r |= ds B =µ0I4πRdθ02π∫=µ0I2R13 W07D1 Magnetic Dipoles, Force and Torque on a Dipole, Experiment 2 W07D1 Magnetic Dipoles, Torque and Force on a Dipole, Experiment 2: Magnetic Dipole in a Helmholtz Coil http://web.mit.edu/8.02t/www/materials/Experiments/expMagForcesDipoleHelmholtz.pdf Reading Course Notes: Sections 8.4, 8.6.4, 8.10.4, 8.13, 9.5, 9.914 Magnetic Field of Bar Magnet (1) A magnet has two poles, North (N) and South (S) (2) Magnetic field lines leave from N, end at S15 Magnetism – Bar Magnet Like poles repel, opposite poles attract16 Conservation of Magnetic Flux: E ⋅ dAS∫∫=qinε0 B ⋅ dAS∫∫= 0Magnetic Dipole Moment µ≡ IAˆn ≡ IAhttp://web.mit.edu/viz/EM/visualizations/magnetostatics/calculatingMagneticFields/RingMagField/RingMagField.htmIf the current loop is a uniform magnetic field then becasue 18 Magnetic Force on Current Loop in Uniform Field Fmag= I dsloop∫⎛⎝⎜⎞⎠⎟×B =0 I ds =0loop∫19 Torque on Current Loop Place rectangular current loop in uniform B field ˆiˆjˆk Magnetic moment points out of the page torque tries to align the magnetic moment vector in the direction of the magnetic field µ τ =µ ×B τ =µ ×B =µBˆj = IABˆj20 Dipoles don’t move??? This dipole rotates but doesn’t feel a net force in a uniform magnetic field But dipoles can feel magnetic force.N S 21 Force on Magnetic Dipole in Non-Uniform Field What makes the field pictured? Bar magnet below dipole, with N pole on top. It is aligned with the dipole pictured, they attract! N S µ22 Force on Magnetic Dipole N S N S UDipole= -µ ⋅B F = −∇(− µ ⋅B) ⇒F =∇( µ ⋅B)Fz(0,0,z) =µz∂Bz∂z µ ↑Magnetic Field Profiles for Experiment 224 W09D1: Sources of Magnetic Fields: Ampere’s Law Tod ay’s Reading Assignment Course Notes: Sections 9.3-9.4, 9.7, 9.10.225 3rd Maxwell Equation: Ampere’s Law B ⋅ dsclosed path∫=µ0J ⋅ˆn daopen surface∫∫Open surface is bounded by closed path.26 Current Enclosed JCurrent density Ienc=J ⋅ˆn dAopensurfacce S∫∫ Current enclosed is the flux of the current density through an open surface S bounded by the closed path. Because the unit normal to an open surface is not uniquely defined this expression is unique up to a plus or minus sign.27 Ampere’s Law: The Idea In order to have a non-zero line integral of magnetic field around a closed path, there must be current punching through any area with path as boundary28 Sign Conventions: Right Hand Rule Integration direction clockwise for line integral requires that unit normal points into page for open surface integral Current positive into page, negative out of page : B ⋅ dsclosed path∫=µ0J ⋅ˆn daopen surface∫∫29 Sign Conventions: Right Hand Rule Integration direction counterclockwise for line integral requires that unit normal points out of page for open surface integral Current positive out of page, negative into page : B ⋅ dsclosed path∫=µ0J ⋅ˆn daopen surface∫∫30 Applying Ampere’s Law 1. Identify regions in which to calculate B field. 2. Choose Amperian closed path such that by symmetry B is zero or constant magnitude on the closed path! 3. Calculate 4. Calculate current enclosed: 5. Apply Ampere’s Law to solve for B: check signs B ⋅ dsoriented closed path∫="B times length"or "zero"⎧⎨⎩ B ⋅ dsclosed path∫=µ0J ⋅ˆn daopen surface∫∫ Ienc=µ0J ⋅ˆn daopen surface∫∫31 Example: Infinite Wire Region 1: Outside wire (r ≥ R) B ⋅ ds∫ B =µ0I2πrˆθ = B ds∫ = B 2πr( ) =µ0Ienc =µ0ICylindrical symmetry à Amperian Circle B-field counterclockwise32 Magnetic Field of Solenoid loosely wound tightly wound For ideal solenoid, B is uniform inside & zero outside Horiz. comp. cancel33 Magnetic Field of Ideal Solenoid B∫⋅ ds =B ⋅ ds1∫+B ⋅ ds2∫+B ⋅ ds3∫+B ⋅ ds4∫Using Ampere’s law: Think! B ⊥ ds along sides 2 and 4B = 0 along side 3⎧⎨⎩ Ienc= nlI n: # of turns per unit length B ⋅ ds = Bl =µ0nlI∫ B =µ0nlIl=µ0nI n = N / L : #
View Full Document
Unlocking...