chapter5_45_52

chapter5_45_52 - AAE 590E 5.2 MAGNETOSTATICS Ch5 45 AAE...

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Unformatted text preview: AAE 590E 5.2 MAGNETOSTATICS Ch5 45 AAE 590E QUIZ Experiment 1: Experiment 2: conducted by Oersted in 1820 Wire on top of Compass N Compass S Wire under Compass NO Current Current I Current 2I Ch5 46 AAE 590E Observations Oersted conducted these experiments in 1820. Conclusions: A wire with no current produces no magnetic field. Magnitude of magnetic field produced by current of moving charges depends on amount of current. Magnetic field due to current appears to be to direction of current. Direction of magnetic field due to current depends on location with respect to wire (on top/under compass). Wire on top of Compass N Compass S Wire under Compass ! B wire ! B wire ! B wire ! B wire ! B Earth Current I ! B Earth ! B Earth Current 2I ! B Earth NO Current Ch5 47 AAE 590E Magnetic Field Magnetic Field Lines of Moving Charges Magnetic Field Lines of Magnets Ch5 48 AAE 590E So far, we have established that "Stationary Charges" produce electric fields (constant with respect to time): ELECTROSTATICS. Following, we will examine and develop relationships associated with moving charges/steady currents (constant with respect to time): MAGNETOSTATICS. Ch5 49 AAE 590E Biot-Savart Law ! B BiotSavart Law: ! 0 q ! # ^ v r B= " 2 4! r Permeability of Free Space: 0 = 4! " 10 0 #7 #7 $T & % ' ! v ! rscript +q q ! B N A 2 = 1.26 "10 #6 H m ! v Tm 2 = 10 4! m C s Sign of Moving Charge: Product of velocity and charge (+/) will provide the direction of RightHandRule. Point Charge: Coulomb's Law describes the relationship between charge and electric field. Biot-Savart Law describes the relationship between a moving charge and magnetic field it induces. Ch5 50 AAE 590E Magnetic Field of Straight Wire Step I: Steady Current distribution through the wire: !C $ Q I = = const. # = A & t "s % Divide distribution into small pieces. y I y ! R ! r! ! r Step II: ! !B Choose origin, convenient location at half point. Derive expression for magnetic field due to one y. Separation Vector x Magnitude Unit Vector Velocity Magnetic Field ! ! ! R = r ! r " = (x,0,0) ! (0, y,0) = (x, !y,0) z ! R = x 2 + y2 = r ! R 1 ^ = ! = (x, !y,0) r R r ! ! I ! l I !y ^ v= = y q q ! I !y I x!y !B = 0 # 2 ^ $ ^ = % 0 # 3 ^ y r z 4" r 4" r Ch5 51 AAE 590E Magnetic Field of Straight Wire Step III: Every piece of the straight wire contributes some magnetic field in the z direction. Summation L B= 0 + 4! Ix" # $ 2 1 L 2 (x 2 +y 0 3 2 2 ) dy = 0 4! x " x 2 + (L 2)2 " LI Bwire = 4! r " r 2 + (L 2)2 " LI Step IV: Case Study Case I: L >> r Bwire = 0 2I 4! " r Ch5 52 ...
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This document was uploaded on 01/10/2012.

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