On the left is a region of uniform magnetic field B1 into thepage, and adjacent
ID: 1758187 • Letter: O
Question
On the left is a region of uniform magnetic field B1 into thepage, and adjacent on the right is a region of uniform magneticfield B2 also into the page. The magnetic field B2 is smaller thanB1 (B2 < B1). You pull a rectangular loop of wire of length w,height h, and resistance R from the frist region into the secondregion, on a constant force F to the right, and you notice that theloop doesn't accelerate but moves with a constant speed. A loop of wire is dragged through twomagnetics. Calcuate this constant speed v in terms of the knownquantities B1,B2,w,h,R,and F, and explain your calculationcarefully. Also show the approximate surface-charge distribution onthe loop.Explanation / Answer
Given : Magnetic field Induction left region is : B1 into the page. Magnetic field in hte second region is : B2 into the page. Length of the rectangular loop is : w Height is : h Resistance is : R This loop is pulled by applying force " F " from first region to secong region. Workdone in moving this loop is : W = F * distancemoved ---------------(1) But we know that : W = Uf - Ui = B2 - B1 = ( B2 - B1) -----------------(2) From (1) & (2) ( B2 - B1) = F * distancemoved -----------(3) But = N i A = iA ( since it is a single loop ) = i ( w * h) and Velocity(v) = distance / time ==> distance = v * t Hence equation (3) becomes : i ( w * h) * (B2 - B1 ) = F * v * t Hence velocity is : v = [ i ( w *h) * ( B2 - B1 ) ] / F * t If we consider this velocity per unit time then : v = [ i ( w *h) * ( B2 - B1 ) ] / F From Ohm's law : V = i R i = V /R Where V = potential. ==> velocity is : v = [ V ( w * h) * (B2 - B1 ) ] / F *RRelated Questions
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