PHYS 102 1nd Edition Lecture 20 Outline of Last Lecture I. Calculating Magnetic Fields continued…Outline of Current Lecture II. Magnetic Flux III. Induced Voltage IV. Calculating Induced Voltage Current LectureII. Magnetic Flux A. Magnetic flux φ = BA cos θWhere B is like the number of people, between B lines and the line perpendicular, % the plane of the loop (the normal) If you change the flux from φi to φf, you get induced current (voltage): Δ φ = φf- φi In this case, = (BA cos θ)f -(BA cos θ)i Faraday’s Law III. Induced Voltage Equation ε = -N(Δ φ/ Δ t)Where, ε is the induced voltage, N is the number of loops, and Δ t is the time during the change. The negative symbol in front of the N is significant for the opposing reaction of the coil due to the change of flux. IV. Calculating Induced Voltage ε = -N(Δ φ/ Δ t) ; A = πr2A = π(1.6/1000)2 m2 ; Δ t= 120 ms = 120 X 10-3 s ; Bi = 0 ; Bf = 1.5 T ε = -1{[(BA cos θ)f -(BA cos θ)i]/120 X 10-3}ε = -0.1 X 10-3 Volts l ε l = 0.1 X 10-3 = 0.1 mV = 0.0001 voltsThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.compare this to a AA battery (1.5 volts) 21. ε = -N(Δ φ/ Δ t) = -200 [(φf- φi)/1.80]= -200 {[(BA cos θ)f -(BA cos θ)i]/1.80} ; where Bf= 50 X 10-6 T and θ= 90-28 = 62 degrees= -200[(Af-Ai)(50 X 10-6)(cos 62)/180]= [-200 (39 X 10-4)(50 X 10-6)(cos 62)]/ 180l ε l = 10.2 X 10-6
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