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A coil with 1500 turns, a radius of `5.0` cm and a resistance of `12 Omega` surrounds a solenoid with `240 turns // cm` and a radius of 4 cm , see figure. The current in the solenoid changes at a constant rate from 0 to 20 A in `0.10` s. Calculate the magnitude of the induced current (in mA) in the 1500 turn coil (`pi^(2)=10` Neglect self inductance of the coil). |
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Answer» Correct Answer - `3840` `varepsilon=150xxd/dt(mu_(0)ni)xxpxx((4.5)/100)^(2)` `=150xx4pixx10^(-7)xx(230xx10^(2))/100xx2/0.1xxpixx(4.5xx4.5)/10^(4)` `i=varepsilon/12=(600pi^(2)xx460xx10^(3)xx10^(-7))/12xx(4.5)^(2)/10^(4)` `=0.04657500xx10^(-6)` `approx 5A` |
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