InterviewSolution
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InterviewSolution
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In the given arrangement the square loop of area `10 cm^(2)` rotates with an angular velocitys `omega` about its diagonal. The loop is connected to a inductance of `L = 100 mH` and a capacitance of `10 mF` in series. The lead wires have a net resistance of `10 Omega`. Given that `B = 0.1 T` and `omega = 63 "rad"//s` Find the rms currentA. `6 xx 10^(5) A`B. `5 xx 10^(-5) A`C. `4 xx 10^(-5) A`D. `7 xx 10^(-5) A` |
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Answer» Correct Answer - C `phi = BA cos omega t` `e = - (d phi)/(dt) = BA omega sin omega t` `i_("rms") = (V_("rms"))/(Z) = (BA omega // sqrt(2))/(sqrt((omega L - (1)/(omega C))^(2) + R^(2)))` `= (0.1 xx 10^(-3) xx 63)/(sqrt(2 { (63 x 0.1 - (1)/(63 xx 0.01))^(2) + (10)^(2)}))` `= 4.0 xx 10^(-5) A` |
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