Consider a closed triangular path that runs from

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Unformatted text preview: C. 5 m/s Faraday’s Law: D. 6 m/s E. 9 m/s dΦ dΦ E d = − ⇒ ε = IR = ∫ dt dt As the moving conductor slides downward the area of the loop, and therefore the flux, increases. v = 6 m/s IR = dΦ dA dH =B = BL = BLv dt dt dt v= IR 3.75 A ⋅ 20 Ω = = 6 m/s BL 5T ⋅ 2.5 m Physics 212 Lecture 29, Slide 13 Phasor diagram at t = 0 Phasers What is VC at t = π/(2ω) (A) α + VC max sin α (C) + VC max cos α (B) − VC max sin α (D) − VC max cos α Phasor diagram at t = π/(2ω) VR α VC VL Voltage is equal to projection of phasor along vertical axis Physics 212 Lecture 29, Slide 14 Ampere’s Law Integrals . Two infinitely long wires carrying current run into the page as indicated. Consider a closed triangular path that runs from point 1 to point 2 to point 3 and back to point 1 as shown. Which of the following plots best shows B·dl as a function of position along the closed path? B B The magnetic field points in the azimuthal direction and is oriented clockwise. No current is contained within the loop 1­2­3­1, therefore NOT (A) The magnetic field falls off like 1/r (r is the distance from the wire). From 3 to 1, B∙ dl is < 0 , la rg e s t m a g...
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This note was uploaded on 02/18/2013 for the course PHYS 212 taught by Professor Kim during the Fall '08 term at University of Illinois, Urbana Champaign.

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