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A conducting bar of mass m length lis pushed with a speed V_(0) on a smooth horizontal conducting rail containing an inductance L. If the applied magnetic field has inward field induction B find the maximum distance covered by the bar before it stops. Assume that the induced current i=0 when the bar is at the position of the inductor . (##FIITJEE_PHY_MB_05_C02_SLV_018_Q01.png" width="80%"> |
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Answer» Solution :If the bar slides a distance dx the flux LINKAGE `dphi = 8l ` dx `implies `The induced emf = - `(dphi)/(DT ) =- (Bldx )/(dt)` Since the induced emf across the inductor = `(Ldi)/(dt)` `implies -(Bldx)/(dt)= -(Ldi)/(dt)` `implies L int_(0) di = B l int_(0)^(x) dx ` `implies Li =B l x implies i = (Bl)/(L)x` This induced current interacts with the applied magnetic field of induction and imparting a restoring ( magnetic ) force F = `-ILB`. `implies F =- ((Bl)/(L) x)l`B `implies (mvdv)/(dx) =-(B^(2)l^(2))/(L) x` `implies int_(V0)^(0) vdv = -(B^(2)l^(2))/(mL) int_(0)^(s) xdx , -(v_(0)^(2))/(2) = (B^(2)l^(2)s^(2))/(2mL)` `implies S = sqrt((my_(0)^(2)L)/(B^(2)l^(2)))=(sqrt(mL)).(v_(0))/(Bl)` 
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