Lect15 - Physics 212 Lecture 15 Ampere’s Law rr ∫ B •...

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Unformatted text preview: Physics 212 Lecture 15 Ampere’s Law rr ∫ B • d l = µo I enclosed Physics 212 Lecture 15, Slide 1 Music Who is the Artist? A) B) C) D) E) BB Arturo Sandoval Tiempo Libre Chucho Valdes & Afro-Cuban Messengers Freddie Omar con su banda Freddie su Los Hombres Calientes Los Calientes Why?? He did a spectacular show at Krannert Saturday night !! He Krannert Legendary Cuban piano player Legendary Hasn’t been in the U.S. in seven years Hasn been A real treat !! real Physics 212 Lecture 15, Slide 2 Your Comments “For infinite sheets when you break up the For We will use a simulation to illustrate We integral to its segments, I don't know how the the contribution to integral B dot dl integral was worked out to have gotten BL for from different segments. from each segment. When are you going to put up the review material for the 2nd exam on the website? V ery soon.. review ery soon.. “Can we discuss Ampere's Law and actually Can solving problems using it? Thank you! solving This will be our calculation today “B fields inside wires” “Let's talk about why our lecture needs Let's to stop having side conversations during lecture. It's getting extremely irritating and distracting.” and “People should really keep it down in lecture. People It's getting really annoying.” It's “Bob Dylan” 05 Oct 22 at Assembly Hall “Oustside is not a word.” “I saw you listening to your ipod at saw ipod at Champaign Surplus. (Probably picking out the next artist for this slide.)” next AGREED !! AGREED Talking is encouraged during clicker questions, but NOT OTHERWISE questions, 40 30 20 10 0 Confused Confident Physics 212 Lecture 15, Slide 3 Infinite current-carrying wire rr LHS: ∫ B • d l = ∫ Bd l = B ∫ dl = B ⋅ 2πR RHS: I enclosed = I µo I B= 2πR General Case :05 Physics 212 Lecture 15, Slide 4 Practice on Enclosed Currents Ienclosed = I 80 Ienclosed = I Preflight 6 Preflight 4 Preflight 2 Ienclosed = I 80 Ienclosed = 0 Ienclosed = 0 Ienclosed = 0 70 60 60 60 50 40 40 40 20 20 :08 30 20 0 0 10 0 Physics 212 Lecture 15, Slide 5 Cylindrical Symmetry Preflight 8 X BB X X X Enclosed Current = 0 Check cancellations 50 40 30 20 10 0 :22 Physics 212 Lecture 15, Slide 6 Ampere’s Law I into screen :12 rr ∫ B ⋅ dl I enc Physics 212 Lecture 15, Slide 7 rr ∫ B ⋅ d l = µo I enc Ampere’s Law dl B dl B dl :14 B Physics 212 Lecture 15, Slide 8 rr ∫ B ⋅ d l = µo I enc Ampere’s Law dl B B dl B dl :16 Physics 212 Lecture 15, Slide 9 rr ∫ B ⋅ d l = µo I enc Ampere’s Law dl B B dl B dl :16 Physics 212 Lecture 15, Slide 10 10 Which of the following current Which distributions would give rise to the B.dL distribution at the right? A :18 B BB C Physics 212 Lecture 15, Slide 11 11 :19 Physics 212 Lecture 15, Slide 12 12 :19 Physics 212 Lecture 15, Slide 13 13 :19 Physics 212 Lecture 15, Slide 14 14 Match the other two: A :21 B BB Physics 212 Lecture 15, Slide 15 15 Preflight 10 80 60 Use the right hand rule and curl your Use fingers along the direction of the current. current. :22 40 20 0 Physics 212 Lecture 15, Slide 16 16 Simulation :23 Physics 212 Lecture 15, Slide 17 17 Solenoid Several loops packed tightly together form a uniform magnetic field inside, and nearly zero magnetic field outside. 1 2 4 3 From this simulation, we can assume a constant field inside the solenoid and zero field outside the solenoid, and apply Ampere’s law to find the magnitude of the constant field inside the solenoid !! rr ∫ B • d l = µo I enc BL + 0 + 0 + 0 = µo I enc :28 r r 3r r 4r r 1r r ∫ B • d l + ∫ B • d l + ∫ B • d l + ∫ B • d l = µo I enc 2 1 2 3 BL = µo nLI n = # turns/length 4 B = µo nI Physics 212 Lecture 15, Slide 18 18 y Example Problem An infinitely long cylindrical shell with inner radius a and outer radius b carries a uniformly distributed current I out of the screen. a Sketch |B| as a function of r. • Conceptual Analysis – – I x b Complete cylindrical symmetry (can only depend on r) ⇒ can use Ampere’s law to calculate B B field can only be clockwise, counterclockwise or zero! rr ∫ B • d l = µo I enc B ∫ dl = µo I enc For circular path concentric w/ shell • Strategic Analysis Calculate B for the three regions separately: 1) r < a 2) a < r < b 3) r > b :31 Physics 212 Lecture 15, Slide 19 19 y Example Problem I BB r a What does |B| look like for r < a ? rr ∫ B • d l = µo I enc b x r so B = 0 0 (A) (A) :33 (B) (B) (C) Physics 212 Lecture 15, Slide 20 20 y Example Problem I r a What does |B| look like for r > b ? BB b x rr ∫ B • d l = µo I enc I (A) (A) :35 (B) (B) (C) Physics 212 Lecture 15, Slide 21 21 y Example Problem What does |B| look like for r > b ? rr LHS: ∫ B • d l = ∫ Bd l = B ∫ dl = B ⋅ 2πr RHS: dl I r B a I enclosed = I b x µo I B= 2πr (A) (A) :36 (B) (B) (C) Physics 212 Lecture 15, Slide 22 22 y Example Problem I What is the current density j (Amp/m2) in the conductor? (A) (A) I j= 2 πb (B) (B) j = I / area :40 I j= 2 2 πb +πa a (C) BB b x I j= 2 2 πb −πa area = π b2 − π a 2 I j= 2 2 πb −πa Physics 212 Lecture 15, Slide 23 23 y Example Problem I r a What does |B| look like for a < r < b ? (A) (A) :43 (B) (B) BB b x (C) Physics 212 Lecture 15, Slide 24 24 y Example Problem What does |B| look like for a < r < b ? rr ∫ B • d l = µ oI enc B ⋅ 2πr = µ o⋅ :45 I π (b 2 − a 2 ) (A) (A) B ⋅ 2πr = µ o⋅ jAenc ⋅ π (r − a ) 2 2 (B) (B) I r a µo I (r 2 − a 2 ) B= ⋅ 2πr (b 2 − a 2 ) (C) b x Starts at 0 and increases almost linearly Physics 212 Lecture 15, Slide 25 25 y Example Problem An infinitely long cylindrical shell with inner radius a and outer radius b carries a uniformly distributed current I out of the screen. a Sketch |B| as a function of r. I x b :48 Physics 212 Lecture 15, Slide 26 26 y Follow-Up Add an infinite wire along the z axis carrying current I0. What must be true about I0 such that there is some value of r, a < r < b, such that B(r) = 0 ? A) |I0| > |I| AND I0 into screen A) |I a I I0 X BB x b B) |I0| > |I| AND I0 out of screen B) |I C) |I0| < |I| AND I0 into screen C) |I D) |I0| < |I| AND I0 out of screen D) |I E) There is no current I0 that can produce B = 0 there E) There B will be zero if total current enclosed = 0 :48 Physics 212 Lecture 15, Slide 27 27 ...
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