Phys 012 1st Edition Lecture 16 Outline of Last Lecture II. Problem: Find the force on the rectangular coil as it enters the magnetic field B at constant velocity v.a. Entering field: flux is increasing induced electromagnetic force (εmf) induced currenti. ΦB = ΣBΔAcosθ = Blyii. εind = -N (ΔΦB/Δt) = -NBL(Δy/Δt) = -NBLviii. Iind = εind/R = -NBLv/R1. Causes forces on sides of coil inside B; forces up and down cancel; net force to the lefiv. Fnet = NILBsinθ = N[(NBLv)/R]LB = (N2B2L2v)/R (to the lef)b. Inside field: no change in flux no εmf or current inducedc. Leaving field: flux decreasing induced εmf induced current in opposite direction as enteringi. Net force also to the lef with same magnitudeIII. Lenz’s Law: the polarity of an induced εmf is such that it produces a current whose magnetic field opposes the change in magnetic flux through the loopa. Ex) Conductive bar moving in a circuit in a magnetic fieldb. Ex) Circuit next to a wire carrying a rapidly decreasing currentThese 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.c. Ex) Closed loop of wire inside a closed current where switch has just been closedd. Ex) Magnet next to a wire circuit wrapped around a conductive metal rodOutline of Current Lecture IV. Problem: A conductive metal bar of mass m, making contact with an electric circuit with resistance R, is falling straight downward. Find an expression for the terminal velocity of the bar, assuming no air resistance, resistance within the circuit wires, and no friction between the bar and the wires.a. ΦB = ΣBΔAcosθb. As bar falls, Iind. is counter clockwise.c. Forces acting on bar: FB oriented upward, Fgrav. oriented downward.i. FB = ILBii. Fgrav. = mgd. ΣF = ILB – mg = ma; a = (ILB)/m – ge. εind. = -N (ΔΦB/Δt) = -BLvf. Iind. = |εind.|/R = (BLv)/Rg. a = {[(BLv)/R]LB}/m – g = {B2L2v}/Rm – gh. At terminal velocity, a = 0.i. {B2L2v}/Rm – g = 0ii. v = (gRm)/( B2L2)V. Self-Inductance *switch has just been closeda. The increasing currents in the diagrams above when the switch is first closed will create a magnetic field and a change in flux. This will cause a back (opposing)
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