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In the cylindrical region of radius R=2m, there exists a time varying magnetic field B such that (dB)/(dt)=2"Tesla"//sec. A charged particle having charge q=2C is placed at the point P at a distance d from its center O. Now, the particle is moved in the direction perpendicular to line OP by an external agent up to infinity so that there is no gain in kinetic energy of charged particle. Then |
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Answer» WORK DONE by external agent is `4pi` Joule when `d=4m` `E=2pisqrt(x^(2)+d^(2))=piR^(2)k`, where `((dB)/(dt)=k)` `E=(R^(2)k)/(2sqrt(x^(2)+d^(2)))` `W=int_(0)^(oo) dvecE.dvecx` `=int_(0)^(oo) qEcostheta.dx=int_(0)^(oo) (qR^(2)k)/(2sqrt(x^(2)+x^(2)))d/(sqrt(x^(2)+x^(2)))dx=(qR^(2)kpi)/4` Alternate solution By applying Faraday's Law for an IMAGINARY triangular loop having vertex `O,P` and point at infinity, `qintvecE.vec(dl)` can be found out. 
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