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A coducting rod of mass m=0.3kg and length l=4m can slide without friction on two parallel conducting rails. The conducting rails are connected via an inductancce L=3Mh. This system is placed in a region containing uniform magnetic field B=1T pointing into the plane. if the rod is given an initial velocity v_(0)=2m//s, it oscillates with an amplituude Acm. Find the value of 2A |
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Answer» `m(dv)/(dt)=-IBL` `implies(d^(2)v)/(dt^(2))=-((B^(2)l^(2))/(mL))v` `impliesv=v_(0)cos(omegat)` with `omega=(Bl)/(SQRT(mL))` Thus, `(DX)/(dt)=v_(0)cos(omegat)` `impliesx=(v_(0))/(omega) sin omegat` `2A=(2v_(0))/(omega)=3CM` |
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