InterviewSolution
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InterviewSolution
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suppose the ends of the coil in the previousproblem are connected to a resistane of 100 Omega. Neglecting the resistance of the coil, find the heat produced in the circuit in one minute. |
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Answer» Solution :Since`H=int_(0)^(1 MIN) i^2 R.dt` `=int_(0)^(1min) (B^2A^2omega^2)/(R^2) sin omega t.R dt` `=(B^2 A^2 omega^2)/(2R). Int_(0)^(1 min) (1-cos 2 omega t).dt` `=(B^2A^2omega^2)/(2R)(1-(sin 2 omega t)/(2 omega))_(0)^(1 min)` `(B^2 A^2 omega^2)/(2R)(60-(sin.2xx80xx2 pi//60xx60)/(2xx80xx2 pi//60))` `=(60)/(2R)xxpi^2r^4xxB^2xx(80xx(2 pi)/(60))^2` `` `= 60/200xx10xx64/9xx10xx625xx10^(-8)xx10^(-4)` `=(625xx6xx64)/(9xx2)xx10^(-11)=1.33xx10^(-7)J.` |
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