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MIT 8 02T - Ampere’s Law

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1 1 W09D1: Sources of Magnetic Fields: Ampere’s Law Today’s Reading Assignment Course Notes: Sections 9.3-9.4, 9.7, 9.10.2 Announcements No Math Review W09 PS 7 due Tuesday W10 at 9 pm in boxes outside 32-082 or 26-152 Next reading Assignment W09D2 Faraday’s Law Course Notes: Sections 10.1-10.4 2 3 Review: Creating Magnetic Fields: Biot-Savart2 4 The Biot-Savart Law Current element of length carrying current I produces a magnetic field at the point P: dB =µ04πI ds ׈rr2 ds B(r) =µ04πId′s × (r −′r )r −′r3wire∫5 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: 6 Group Problem: Line Integral of Magnetic Field B ⋅ dscircular path∫Calculate line integral of magnetic field about circular path for very long current carrying wire3 7 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. 8 3rd Maxwell Equation: Ampere’s Law B ⋅ dsclosed path∫=µ0J ⋅ˆn daopen surface∫∫Open surface is bounded by closed path. 9 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 boundary4 10 Concept Question: Line Integral The integral expression B ⋅ dsclosed path∫1. is equal to the magnetic work done around a closed path. 2. is an infinite sum of the product of the tangent component of the magnetic field along a small element of the closed path with a small element of the path up to a choice of plus or minus sign. 3. is always zero. 4. is equal to the magnetic potential energy between two points. 5. None of the above. 11 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 ⋅ dsclosed path∫=µ0J ⋅ˆn daopen surface∫∫12 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 ⋅ dsclosed path∫=µ0J ⋅ˆn daopen surface∫∫5 13 Zero Current Enclosed B ⋅dsclosed path∫=µ0I2πr2r2dθCD∫+µ0I2πr1r1dθAB∫= 0Same argument holds For (b) more complicated path consisting of radial and tangential legs, and (c) continuous path (a) (b) (c) For path shown in (a) 14 Line Integral: Arbitrary Path B ⋅ dscircular path∫= −µ0I+= B ⋅ dsclosed path∫= 0 ⇒B ⋅ dsarbitrary path∫=µ0ICirculate clockwise Circulate counterclockwise Zero enclosed current Circular path Arbitrary path Superposition of two cases must add to zero 15 Concept Questions: Ampere’s Law6 16 Concept Question: Ampere’s Law Integrating B around the loop shown gives us: 1. a positive number 2. a negative number 3. zero 17 Concept Question: Ampere’s Law Integrating B around the loop in the clockwise direction shown gives us: 1. a positive number 2. a negative number 3. zero 18 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 ⋅dsoriented closed path∫="B times length"or "zero"⎧⎨⎩ B ⋅dsclosed path∫=µ0J ⋅ˆn daopen surface∫∫ Ienc=µ0J ⋅ˆn daopen surface∫∫7 19 Infinite Wire I!A cylindrical conductor has radius R and a uniform current density with total current I. we shall find the direction and magnitude of the magnetic field for the two regions: (1) outside wire (r ≥ R) (2) inside wire (r < R) 20 Worked Example: Ampere’s Law Infinite Wire I I B Amperian Closed Path: B is Constant & Parallel Current penetrates surface 21 Example: Infinite Wire Region 1: Outside wire (r ≥ R) B ⋅ ds∫ B =µ0I2πrˆθ = B ds∫ = B 2πr( ) =µ0Ienc =µ0ICylindrical symmetry à Amperian Circle B-field counterclockwise8 22 Group Problem: Magnetic Field Inside Wire I!We just found B(r>R) Now you find B(r<R) 23 Infinite Wire: Plot of B vs. r 202 RIrBinπµ=rIBoutπµ20=24 Group Problem: Non-Uniform Cylindrical Wire A cylindrical conductor has radius R and a non-uniform current density with total current: Find B everywhere J = J0Rrˆn9 25 Other Geometries 26 Two Loops 27 Two Loops Moved Closer Together10 28 Multiple Wire Loops 29 Multiple Wire Loops – Solenoid http://youtu.be/GI2Prj4CGZI 30 Demonstration: Long Solenoid http://tsgphysics.mit.edu/front/?page=demo.php&letnum=G%2018&show=011 31 Magnetic Field of Solenoid loosely wound tightly wound For ideal solenoid, B is uniform inside & zero outside Horiz. comp. cancel 32 Magnetic Field of Ideal Solenoid B∫⋅ ds =B ⋅ ds1∫+B ⋅ ds2∫+B ⋅ ds3∫+B ⋅ ds4∫Using Ampere’s law: Think! B ⊥ ds along sides 2 and 4B = 0 along side 3⎧⎨⎩ Ienc= nlI n: # of turns per unit length B ⋅ ds = Bl =µ0nlI∫ B =µ0nlIl=µ0nI n = N / L : # turns/unit length = Bl + 0 + 0 + 033 Ampere’s Law: Infinite Current Sheet I Amperian Loops: B is Constant & Parallel OR Perpendicular OR Zero I Penetrates B B12 34 Group Problem: Current Sheet A sheet of current (infinite in the y & z directions, of thickness d in the x direction) carries a uniform current density: Find the direction and magnitude of B as a function of x. J = Jˆk35 Surface Current Density A very thin sheet of current of width w carrying a current I in the positive z-direction has a surface current density For sheet of thickness d , width w, and current I K = Kˆk K = I / w I = Jdw = Kw ⇒ J = K / d36 Solenoid is Two Current


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