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Fig. 6.50 shows a rectangular conducting loop PQRS in which the arm PQ is free to move. A uniform magnetic field acts in the direction perpendicular to the plane of the loop. Arm PQ is moved with a velocity v towards the arm RS. Assuming that the arms QR, RS and SP have negligible resistances and the moving arm PQ has the resistance r, obtain the expression for the current in the loop (ii) the force and (iii) the power required move the arm PQ. |
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Answer» SOLUTION :LET in a rectangular loop PQRS placed in a uniform magnetic field B, the arm PQ be moving, as shown in Fig. 6.51 with a constant velocity .v.. Then in time `Deltat`, the arm PQ will move through a small distance `Deltar = v.Deltat` and consequently the area enclosed by the loop decreases by `DeltAA = area PQRS - area P.Q. RS` `= Deltax.l = (vDeltat)`, where l is the length of arm PQ. `therefore` decrease in magnetic flux of loop `Deltaphi_(B) = BDeltaA = B(vDeltat)l` `therefore` Induced emf in the rectangular loop `|varepsilon|= (Deltaphi_(B))/(Deltat) = (B(v Deltat)t)/(Deltat) = Blv`  As arms OR, RS and SP have no resistance and moving arm PQ has the resistance .R., hence (i) induced CURRENT in the loop `I = varepsilon/r = (Blv)/r` (II) Force required to move the arm `PQ, F = BIl = B ((Blv)/r)l = (B^(2)l^(2)v)/r` (iii) Power required to move the arm `PQ, P = Fv = (B^(2)l^(2)v^(2))/r` |
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