JEE Main 2006Annual, Paper 1PhysicsEasyElectromagnetic Induction and Alternating Currents149× since 2002The flux linked with a coil at any instant ′t′'t'′t′ is given by ϕ=10t2−50t+250\phi = 10{t^2} - 50t + 250ϕ=10t2−50t+250 The induced emfemfemf at t=3st=3st=3s isAOption A. −190-190−190 VVVBOption B. −10-10−10 VVVCOption C. 101010 VVVDOption D. 190190190 VVVCheckSkip · show solution
