lec29_032607_-_Faradays_Law_IV

lec29_032607_-_Faradays_Law_IV - d Ed = - dt ( B dA)...

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Unformatted text preview: d Ed = - dt ( B dA) Faraday's Law IV Faraday' Relates a time-varying B-field at some location in timeBspace to a curly E-field around that area. E B = B dA = B dA = B dA = BA =B x 1 Monday, 26 March 2007 d B emf = - dt Motional emf Lenz's Law Faraday's Law Lenz' Faraday' d B = Binduced dt d B d = (B x ) dt dt dx =B dt =B v d B =B v dt dB Fingers curl in E NC dt Quantitatively, we related integral of the curly E-field around Ea closed path to the rate of change of the magnetic flux on a surface enclosed by that path. path. Direction: Thumb points in - = emf = E d N turns in the coil: Monday, 26 March 2007 B = B dA d E d = -N dt I= R = ( B dA) B v R emf = - N d B dt 2 B A Single Bar qv B E v B A Rotating Bar v r dr FB = FE qvB = qE vB = E = vB d = vBdr qE = vBdr A conducting bar of length rotates with a constant angular speed about a pivot at one end. A uniform B-field is directed Bperpendicular to the plane of rotation as shown. Find the motional emf induced between the ends of the bar. Monday, 26 March 2007 = Ed = E d =E d =E Monday, 26 March 2007 E= vB = Induced emf in an airplane: B = 0.0005 T v = 340 m/s (Mach 1) = 10 m = B vdr = B rdr 1 0 2 = B 2 = vB = 1.7 V! 3 4 1 Question getting something for nothing Suppose you have this brilliant idea: A Ferris wheel has radial metallic spokes between the hub and the circular rim (~ 10 m). These spokes move in the (~ m). magnetic field of the Earth (0.5 e-4 T), so each spoke acts like (0.5 e- T), a bar in the previous example. You plan to use the emf generated by the rotation of the Ferris wheel to power the light-bulbs on the wheel. Suppose the period of rotation for lightthe Ferris wheel is 1 minute. Will this idea work? minute. Question follow up Even if you could find light-bulbs that operate using a lightpotential difference on the order of fractions of millivolts, we millivolts, would still have energy difficulties. Each spoke must be part of a circuit to provide a voltage to the light-bulbs, and consequently, it must carry a current. lightSince the current-carrying spokes are in the magnetic field, a currentmagnetic force is exerted on each spoke in the direction opposite the direction of motion. As a result, the motor of the Ferris wheel must supply more energy to perform work against this magnetic "drag" force. drag" Therefore, the motor must ultimately provide the energy that is operating the light-bulbs, and nothing is gained for free! lightMonday, 26 March 2007 6 2 2 = 0.10 s -1 T 60 s 1 1 B. No = B 2 = 0.5 10- 4 T 0.10 s -1 (10 m )2 2 2 C. I don't know don' -4 = 2.5 10 V A. Yes = ( )( ) Monday, 26 March 2007 = 0.25 mV 5 Generators - AC Coil Loop Magnet Slip Rings Magnet Generators - DC Coil Loop Commutator max t Magnet Brushes Magnet Brushes t Monday, 26 March 2007 7 Monday, 26 March 2007 8 2 Motors Generators in Reverse Eddy Currents & Magnetic Breaks B Induced emf! emf! Induced I! F =0 emf I F =0 F Lenz's law: This current must experience a magnetic force Lenz' that opposes the rotation of the disk F = I B I must point down Monday, 26 March 2007 9 Monday, 26 March 2007 Interaction between Eddy Current & B-field causes a braking action on the disk! 10 Eddy Currents & Magnetic Breaks v1 Eddy Currents & Magnetic Breaks S N N S v2 v3 induced I induced emf induced emf induced Binduced Binduced emf induced I induced emf induced B Binduced I induced B S N S N emf induced I induced induced I induced F = I B Monday, 26 March 2007 F = I B F = I B 11 Monday, 26 March 2007 12 3 ...
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This note was uploaded on 04/08/2008 for the course PHGN 200 taught by Professor Japguy during the Fall '07 term at Mines.

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