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The figure shows a rectangular conducting frame MNOP of resistance R placed partly in a perpendicular magnetic field vecB and moved with velocity vecv as shown in the fig. 6.52. Obtain the expressions for the (a) force acting on the arm 'ON' and its direction, and (b) power required to move the frame to get a steady emf induced between the arms MN and PO. |
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Answer» Solution :As the rectangular frame MNOP moves into the magnetic field, magnetic flux linked with the frame increases at a rate `Deltaphi = BDeltaA = BvlDeltat` and hence magnitude of induced EMF `|varepsilon|= (Deltaphi)/(Deltat)=(BvlDeltat)/(Deltat)= Bvl` As resistance of frame is R, hence induced current `I = varepsilon/R = (BLV)/R` The induced current flows in an anticlockwise direction i.e., along ONMPO. (a) The force acting on the arm ON is given by `F = | I vecl xx vecB| = I lB = (Blv)/R. lB= (B^(2)l^(2)v)/R` The force is directed opposite to the direction of v (i.e., towards RIGHT). (b) Power required to move the frame to get a steady emf induced between the arms MN and PO, Powre `P = Fv = (B^(2)l^(2)v^(2))/R` |
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