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In the figure shown PQRS is a fixed resistanceless conducting frame in a uniform and constant magnetic field of strength B. A rod EF of mass m length l and resistance R can smoothly move on this frame.A capacitor charged to a potential difference V_(0)initially is connected as shown in the figure.Find the velocity of the rod as function of time t if it is released at t=0 from rest. |
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Answer» Applying loop law we get `q/C-iR=Blv=0` ..(1) The force on rod is `F=m(dv)/(dt)=Bil`..(2) differentiating equation (1) and (2) we get`-i/C-R(di)/(dt)-Bl(dv)/(dt)=0`...(3) From equation (2) and (3) we get `(di)/(dt)=-[(B^(2)l^(2))/(mR)+1/(RC)]i` `rArr (di)/(dt)=-Kdt`..(4) where `K=(B^(2)l^(2)C+m)/(MRC)` at `t=0 sec, q=CV_(0)` and `v=0` `therefore` from equation (1) the current at `t=0` is `i_(0)=V_(0)/R` integrating equation `underset(V_(0)//R)overset(i)int(di)/l=-K underset(0)overset(t)int dt` we get `i=V_(0)/R e^(-Kt)` from eqation (2) `(dv)=(Bl)/mi dt` or `dv=(Blv_(0))/(mR) e^(-Kt)dt` integarating the equation `underset(0)overset(v)int dv=underset(0)overset(t)int (Bl)/m V_(0)/R e^(-Kt)dt` `rArr v=(BlV_(0))/(mR)(-1/K)[e^(-Kt-e^(0))]=(BlV_(0))/(mRK)[1-e^(-Kt)]` By substituting `K` we get `v=(BlCB_(0))/(m+B^(2)l^(2)C)(1-e^(((B^(2)l^(2))/(mR) 1/(RC))t))` 
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