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nikhil sahota Grade: 12
        

2 parallel conducting wires are joined by a similar conducting wire . the entire system is placed in a perpendicular magnetic field B . the entire wire is having resistance per unit length of xohms per meter.a conducting rod PQ of negligible resistance is moving with constant velocity v' under action of variable force at t=0 to x=0. determine magnitude of variable force acting on rod as a funtion of time.

6 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

   after time t , distance covered = x = vt


change in flux from initial stage to stage at t = t is d@


d@ = (dA)cos0 = B(dA)


 dA = L(vt)       (change in area)


d@ = BLvt


e = d@/dt = BLv          ..............1


now , I = e/R            .............2              (R is resistance of loop)


since rod is at x distance from initial so net length of parallel wires = 2x                  ( x on both sides)


let Ro is resistance per unit length then resistance of this wire is


R = 2x(Ro)                    


x = vt


R = 2Rovt         ............3


magnetic force on rod = ILB = eLB/R


                               =B2L2v/R                            


                           F  =B2L2/2Rot       ...................4


this is the required force for which rod moves with constant velocity .......


          F inversly proportional to time & F vs time is rectangular hyperbola


 


approve if u like my ans

6 years ago
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