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A horizontal rod AB of length 1 is rotated with constant angular velocity omega about a vertical axis passing through end A. A non-uniform charge of the linear charge density lambda=lambda_(0)x is distributed on the rod where x is the distance from the endAand lambda_(0) is a constant.Find the ratio of the energy desity of magnetic field and electric field on the vertical line passing through A, at a point whose distance is 1 from the end A. |
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Answer» where `B_(x)=int dB sin theta` `B_(y)=int dB cos theta` where `dB=(mu_(0))/(4pi).((lambda_(0)X)(Xomega)dx)/r^(2)` similarly `dE=(K(lambda_(0)X))/r^(2)dx` `E=E_(x)hati+E_(y)hatj` where `E_(x)=int dB cos theta` `E_(y)=int dB sin theta` Energy density ratio `=(B^(2)//2mu_(0))/(1/2 epsilon_(0)E^(2))=B^(2)/E^(2).c^(2)` put the vaue of `B` &`E` and solve. [Do not solve completely, it is VEY lengthy] 
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