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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 Here, `r_(1) = (1)/(20) m, N_(1) = 2 xx 10^(4), l = 1 m` `r_(2) = (1)/(100) m, N_(2) = 100, (dI)/(dt) = (4 - 0)/(0.05) = 80 A//s` `R = 10 pi^(2)` `e = M (dI)/(dt) = (mu_(0) N_(1) N_(2) A_(2))/(l) xx (dI)/(dt)` `= 4 pi xx 10^(-7) (2 xx 10^(4)) 10^(2) pi ((1)/(100))^(2) xx 80` `= 640 pi^(2) xx 10^(-5)` `Q = I xx dt = (e)/(R ) xx dt = (640 pi^(2) xx 10^(-5))/(10 pi^(2)) xx 0.05` `= 32 xx 10^(-6) C = 32 mu C` |
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