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Unformatted text preview: ECE312 Autumn 2010: Problem Set 2 Solutions Problem 1 : (a) Magnetic flux = integraltextintegraltext B ds over the loop area. Because the loop is an open surface, we can choose either z to be the direction of ds . This choice will be related to how we interpret any induced EMF later on. Lets choose ds to be in the z direction, and here B is a constant function of space, so we simply have = ( b 2 ) parenleftbigg I ( t ) 2 b parenrightbigg = 1 2 b I ( t ) (b) Faradays Law for a non-moving circuit: V tr emf =- t =- 1 2 b 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 = b / 2 . Note the minus sign can be dropped because of the oppositional definition inherent in the voltage of the inductor circuit element. (d) Using Lenzs law, the induced EMF will create currents that oppose the change in flux. At t = 0, the current is increasing in the direction, so the flux is increasing in the + z direction. Induced currents will createdirection....
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