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A particle of charge q, mass m is moving in an electric field vecE= EhatiNC^(-1) andmagnetic field vecB =BhatkT . It follows a trajectory from P to Q as shown Velocity at P is Vhati & that at Q is - 2Vhatj, find a) electric field strength b) power developed at P by electric field c) rate of work done at by both fields at Q |
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Answer» <P> Solution :(a) Work done by magnetic FIELD is zero `becausevecFbotvecV` only `vecE` does work on charged particleFrom work energy theorem `DeltaKE=WD` by `vecE` `1/2m(v^(2)-u^(2))=vecEq.vecS` `1/2m(4v^(2)-v^(2))=q(Ehati.2ahati)` `1/2m3v^(2)=2Eq.a`  b) power developed by `vecE` at P is `vecE.vecV=q(Ehati.vhati)=q(vE)=QV((3MV^(2))/(4aq))rArrP=(3mv^(3))/(4a)` c) Power developed by both fields is equal to power developed by `vecE` alone (`because` Work done by `vecB` is 0) `vecP_(0)=vecF.vecV_(0)=q[(vecEhati.(-2Vhatj)]=0` |
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