# Solutions - Homework 11 - Rana, Rishiraj Homework 11 Due:...

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Rana, Rishiraj – Homework 11 – Due: Apr 15 2008, 9:00 am – Inst: Vitaly1Thisprint-outshouldhave11questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.The due time is Centraltime.001(part 1 of 1) 15 pointsA plane loop of wire of area A is placed in aregion where the magnetic field is perpendicu-lar to the plane. The magnitude of B varies intime according to the expressionB=B0e-at.That is, att= 0 the field isB0, and fort >0,the field decreases exponentially in time.Find the induced emf,E, in the loop as afunction of time.1.E=a A B0e-2at2.E=A B0e-at3.E=a B0tB=B0e-atB000~tThe plot ofEversustis similar to theBversustcurve shown in the figure above.keywords:002(part 1 of 1) 15 pointsThe plane of a rectangular coil, 6.5 cm by4 cm, is perpendicular to the direction of auniform magnetic fieldB.If the coil has 68 turns and a total resistanceof 7.6 Ω, at what rate must the magnitude ofBchange to induce a current of 0.11 A in thewindings of the coil?4.E=a A B05.E=a B0e-at6.E=a A B0e-atcorrectCorrect answer: 4.72851 T/s.Explanation:Basic Concepts:Faraday’s Law:E ≡IE·ds=-dΦB=-A B0ddte-a t=a A B0e-a tThat is, the induced emf decays exponentiallyin time.Note:The maximum emf occurs att= 0,whereE=a A B0.Explanation:
dtSolution:Since B is perpendicular to theplane of the loop, the magnetic flux throughthe loop at timet >0 isΦB=B A=A B0e-a tAlso, since the coefficientAB0and the pa-rameter a are constants, and Faraday’s LawsaysE=-dΦBdtthe induced emf can be calculated the fromEquation above:E=-dΦBdtGiven :x= 6.5 cm = 0.065 m,y= 4 cm = 0.04 m,N= 68 turns,r= 7.6 Ω,andI= 0.11 A.The induced emf isE=I R=NdΦdt=Nd(B A)dt=N Ad Bdt,sod Bdt=I RN(x y)=(0.11 A) (7.6 Ω)68 (0.065 m) (0.04 m)= 4.72851 T/s.
Rana, Rishiraj – Homework 11 – Due: Apr 15 2008, 9:00 am – Inst: Vitaly2keywords:003(part 1 of 1) 15 points

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Term
Spring
Professor
Opyrchal
Tags
Work, Magnetic Field, dt, Rana
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