Lect16 - Physics 212 Lecture 16 Motional EMF Conductors...

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Unformatted text preview: Physics 212 Lecture 16 Motional EMF Conductors moving in B field Induced emf !! Induced emf 50 40 30 20 10 0 Confused Avg = 2.7 Confident Physics 212 Lecture 16, Slide 1 Music Who is the Artist? A) B) C) D) E) Susan Tedeschi Susan Tedeschi Marcia Ball Shawn Colvin Bonnie Raitt Bonnie Raitt Melissa Etheridge I guess I’m iin a bluesy n mood this week mood Wonderful albums Physics 212 Lecture 16, Slide 2 Physics 212 Lecture 16 Motional EMF Conductors moving in B field Induced emf !! Induced emf 50 40 30 20 10 0 Confused Avg = 2.7 Confident Physics 212 Lecture 16, Slide 3 The Big Idea B When a conductor moves through a region containg a magnetic field: Magnetic forces may be exerted on the charge carriers in the conductor XXXXXXXXX F L +- F = qv × B v F XXXXXXXXX + These forces produce a charge separation in the conductor - + This charge distribution creates an electric field in the conductor The equilibrium distribution is reached when the forces from the electric and magnetic fields cancel The equilibrium electric field produces a potential difference (emf) in the conductor 05 E - qvB = qE FB + E = vB V = EL FE + E - V = vBL Physics 212 Lecture 16, Slide 4 Preflight 2 BB A B C D Rotate picture by 90o Bar a XXXXXXXXX No force on charges No charge separation No E field No emf a b F vX + - B +F XXXXXXXXX Fa = 0 :22 Fb = qvB qvB v Bar b Opposite forces on charges Charge separation E = vB emf = EL = vBL 80 60 40 20 0 Physics 212 Lecture 16, Slide 5 BB Equivalent circuit A B C D Changing Area Rotate picture by 90o A B V Preflight 4 70 60 XXXXXXXXX 50 Bar F B +F XXXXXXXXX :22 I R Fb = qv0B v0 40 Opposite forces on charges Charge separation E = v0B emf = EL = v0BL 30 20 10 0 Physics 212 Lecture 16, Slide 6 BB F Energy External agent must External exert force F to the right to maintain constant v constant I A B C D Changing Area A B C D E Preflight 50 5 Preflight 5 40 Counterclockwise Current F = IL × B vBL F = LB R :22 This energy is This dissipated in the resistor! resistor! F points to left vBL 2 P = Fv = LBv = I R R 30 20 10 0 Physics 212 Lecture 16, Slide 7 BB Let’s step step through this one through 60 50 40 30 20 10 Preflight 11 20 0 Physics 212 Lecture 16, Slide 8 L E + v t Only leading side has charge separation emf = BLv (cw current) BLv - LE + E + t v Leading and trailing sides have charge separation emf = BLv – BLv = 0 (no current) BLv - LE + 20 Only trailing side has charge separation emf = BLv (ccw current) BLv v t Physics 212 Lecture 16, Slide 9 Preflight 11 20 Physics 212 Lecture 16, Slide 10 10 Changing B field BB Preflight 6 50 -∏ ∏+ B ∏ B ∏∏∏ + v - ∏B ∏∏∏ B 40 30 20 10 I 20 0 Physics 212 Lecture 16, Slide 11 11 Generator: Changing Orientation BB On which legs of the loop is charge separated? A) B) B) C) D) 22 Top and Bottom legs only Top Front and Back legs only All legs None of the legs v×B parallel to top and bottom legs Perpendicular to front and back legs Physics 212 Lecture 16, Slide 12 12 Generator: Changing Orientation BB L w At what angle θ is emf the largest? At is emf (A) θ = 0 (B) θ = 45o (C) θ = 90o (C) (D) emf is same at all angles (D) emf v×B Largest for θ = 0 (v perp to B) Largest (v perp F w ε = 2 EL = 2 L = 2 Lv × B = 2 L( )ωB cosθ = ωAB cos(ωt ) q 2 22 Physics 212 Lecture 16, Slide 13 13 Changing Orientation BB Preflight 8 50 v v×B ∏ + v I X v×B 40 B 30 20 10 0 20 Physics 212 Lecture 16, Slide 14 14 Putting it together Change Area of loop Change magnetic field through loop Change orientation of loop relative to B Faraday’s Law Φ ≡ B⋅ A dΦ ε =− dt Physics 212 Lecture 16, Slide 15 15 Example Problem A rectangular loop (h=0.3m L=1.2 m) with total resistance of 5Ω is moving away from a long straight wire carrying total current 8 amps. What is the induced current in the loop when it is a distance x=0.7 m from the wire? I v h x L •Conceptual Analysis: •Long straight current creates magnetic field in region of the loop. •Vertical sides develop emf due to motion through B field •Net emf produces current •Strategic Analysis: •Calculate B field due to wire. •Calculate motional emf for each segment •Use net emf and Ohm’s law to get current Physics 212 Lecture 16, Slide 16 16 Example Problem A rectangular loop (h=0.3m L=1.2 m) with total resistance of 5Ω is moving away from a long straight wire carrying total current 8 amps. What is the induced current in the loop when it is a distance x=0.7 m from the wire? BB I v h x L What is the direction of the B field produced by the wire in the region of the loop? A) Into the page B) Out of the page C) Left D) Right E) Up Physics 212 Lecture 16, Slide 17 17 Example Problem A rectangular loop (h=0.3m L=1.2 m) with total resistance of 5Ω is moving away from a long straight wire carrying total current 8 amps. What is the induced current in the loop when it is a distance x=0.7 m from the wire? BB B into page I v h x L What is the emf induced on the left segment? A) Top is positive B) Top is negative C) v×B Zero Physics 212 Lecture 16, Slide 18 18 Example Problem A rectangular loop (h=0.3m L=1.2 m) with total resistance of 5Ω is moving away from a long straight wire carrying total current 8 amps. What is the induced current in the loop when it is a distance x=0.7 m from the wire? BB B into page I v h x L What is the emf induced on the top segment? A) left is positive v×B B) left is negative perpendicular to wire C) Zero Physics 212 Lecture 16, Slide 19 19 Example Problem A rectangular loop (h=0.3m L=1.2 m) with total resistance of 5Ω is moving away from a long straight wire carrying total current 8 amps. What is the induced current in the loop when it is a distance x=0.7 m from the wire? BB B into page I v h x L What is the emf induced on the right segment? A) top is positive B) top is negative C) v×B Zero Physics 212 Lecture 16, Slide 20 20 Example Problem A rectangular loop (h=0.3m L=1.2 m) with total resistance of 5Ω is moving away from a long straight wire carrying total current 8 amps. What is the induced current in the loop when it is a distance x=0.7 m from the wire? BB B into page I v h x L Which expression represents the emf induced in the left wire (A) ε left = µ0 I Lv 2π x (B) ε left = µ0 I hv 2π x (C) ε left = µ0 I 2π ( L + x ) ε = Bhv Lv µ0 I B= 2πx µ0 I ε= hv 2πx Physics 212 Lecture 16, Slide 21 21 Example Problem A rectangular loop (h=0.3m L=1.2 m) with total resistance of 5Ω is moving away from a long straight wire carrying total current 8 amps. What is the induced current in the loop when it is a distance x=0.7 m from the wire? BB B into page I v h x L Which expression represents the emf induced in the right wire (A) ε right = 2π ( L + x ) hv µ0 I hv 2π x µ0 I = Lv 2π (h + x ) (B) ε right = (C) ε right µ0 I ε = Bhv B= µ0 I 2π ( L + x) ε= µ0 I 2π ( L + x) hv Physics 212 Lecture 16, Slide 22 22 Example Problem A rectangular loop (h=0.3m L=1.2 m) with total resistance of 5Ω is moving away from a long straight wire carrying total current 8 amps. What is the induced current in the loop when it is a distance x=0.7 m from the wire? BB B into page I v h x L Which expression represents the total emf in the loop? (A) ε loop = µ0 I µ0 I hv + hv 2π x 2π ( L + x ) Iloop = (B) µI µ0 I ε loop = 0 hv − hv 2π x 2π ( L + x ) Iloop = (C) ε loop = 0 ε loop R µ0 I 1 1 hv − 2π R x x + L Physics 212 Lecture 16, Slide 23 23 ...
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