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Unformatted text preview: ECE312 Autumn 2010: Problem Set 2 Solutions Problem 1 : (a) Magnetic flux Ψ = integraltextintegraltext B · ds over the loop area. Because the loop is an “open” surface, we can choose either ± ˆ z to be the direction of ds . This choice will be related to how we interpret any induced EMF later on. Let’s choose ds to be in the ˆ z direction, and here B is a constant function of space, so we simply have Ψ = ( πb 2 ) parenleftbigg μ I ( t ) 2 b parenrightbigg = 1 2 πbμ I ( t ) (b) Faraday’s Law for a nonmoving circuit: V tr emf = ∂ Ψ ∂t = 1 2 πbμ dI ( t ) dt (c) For an inductor, we know V = L dI dt with the voltage polarity shown in the figure below. Therefore we have the inductance L = πbμ / 2 . Note the minus sign can be dropped because of the oppositional definition inherent in the voltage of the inductor circuit element. (d) Using Lenz’s law, the induced EMF will create currents that oppose the change in flux. At t = 0, the current is increasing in the ˆ φ direction, so the flux Ψ is increasing in the +ˆ z direction. Induced currents will createdirection....
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 Fall '08
 Johnson,J
 Electromagnet, Flux, Cos, Trigraph, motional emf, vEMF

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