InterviewSolution
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InterviewSolution
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A coil of inductance L connects the upper ends of two vertical copper bars separated by a distance l . A horizontal conducting connector of mass m starts falling with zero initial velocity along the bars without losing contact with them. The whole system is located in a uniform magnetic field B perpendicular to the plane of the bars. Find the law of motion x(t) of the connector. |
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Answer» Solution :Let at any TIME `t` VELOCITY of rodis `v` and current in circuit `i` . Induced emf in rod `e=Bvl` `Bvl=L(di)/(dt)` `B(dx)/(dt)l=L(di)/(dt)` `Ldi=Bldx` Integrating `Li=Blx implies i=(Bl)/(L)x` `mg-Bil=ma` `a=g-(Bl)/(m)i=g-(B^(2)l^(2))/(mL)x` `(dv)/(dt)=g-omega^(2)x` (where `omega=(Bl)/(sqrt(mL)))` `(d^(2)v)/(dt^(2))=-omega^(2)(dx)/(dt)=-omega^(2)v` `v` oscillates simple harmonically with ANGULAR frequency `omega=(Bl)/(sqrt(mL))`. The solution of this equation `v=Asin(0megat+phi)` `t=0` , `v=0` , `phi=0` `v=Asinomegat` `(dv)/(dt)=Aomegacosomegat` At `t=0` , `((dv)/(dt)=a=g-(Bl)/(m)i=g(. :. i=0)` `g=Aomega` `implies A=(g)/(omega)` `v=Asinomegat` `v=(g)/(omega)sinomegat` `(dx)/(dt)=(g)/(omega)sinomegat` `int_(0)^(x)dx=(g)/(omega)int_(0)^(t)sinomegadt` `x(g)/(omega)(|-cosomegat|_(0)^(t))/(omega)` `=(g)/(omega^(2))(1-cosomegat)` `v=(g)/(omega)sinomegat` Let `v_(0)((=(g)/(omega))` : velocity AMPLITUDE `V=V_(0)sinomegat` `x=(g)/(omega^(2))(1-cosomegat)=(v_(0))/(omega)(1-cosomegat)`  
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