InterviewSolution
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InterviewSolution
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A long solenoid of diameter 0.1 m has `2 xx 10^(4)` turns per meter. At centre of the solenoid is 100 turns coil of radius 0.01 m placed with its axis coinciding with solenoid axis. The current in the solenoid reduce at a constant rate to 0A from 4 a in 0.05 s . If the resistance of the coil is `10 pi^(2) Omega`, the total charge flowing through the coil during this time isA. `32 pi mu C `B. ` 16 mu C `C. `32 mu C `D. `16 pi mu C ` |
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Answer» Correct Answer - C `epsi=-N(d phi)/(dt)` ` |(epsi)/(R)|=(N)/(R)(dphi)/(dt)` `dq=(N)/(R) d phi` `DeltaQ=(N(Deltaphi))/(R)` `Delta Q= (Delta phi_("total"))/(R)` =`((NBA))/(R)` `=(mu_(0) ni pi r^(2))/(R)` Putting values `(4 pi xx 10^(-7) xx 100 xx 4xx piu xx (0.01)^(2))/(10 pi^(2))` `DeltaQ=32 mu C` |
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