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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 is decreased at a constant rate form + 2 A to - 2 A in 0.05 s. Find the e.m.f. induced in the coil. Also, find the total charge flowing through the coil during this time, when the resistance of the coil is `10 pi^(2) ohm`. |
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Answer» Here, `r_(1) = 0.1//2 = 0.5 m, N_(1) = 2 xx 10^(4)`, `N_(2) = 100, r_(2) = 0.01 m, dI = - 2 - 2 = - 4 A` `dt = 0.05 s, R = 10 pi^(2) ohm, e = ? Q = ?` Mutual inductance of the two solenoids is `M = mu_(0) N_(1) N_(2) A_(2) = mu_(0) N_(1) N_(2) (pi r_(2)^(2))` `4 pi xx 10^(-7) xx 2 xx 10^(4) xx 100 xx pi (0.01)^(2)` `M = 8 pi^(2) xx 10^(-5) H` `e = (M dI)/(dt) = 8 pi^(2) xx 10^(-5) xx ((-4))/(0.05) = 6.31 xx 10^(-2) V` Changing induced, `q = I dt = (e)/(R ) dt` `= (6.31 xx 10^(-2) xx 0.05)/(10 pi^(2))` `q = 3.2 xx 10^(-5) C` |
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