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An external magnetic field is decreased to zero, due to which a current is induced in a circular wire loop of radius r and resistance R placed in the field. This current will not become zero at the instant when B stops changingA. At the instant when external magnetic field stops changing (t=0), the current in the loop is `i_(0)`. The current in the loop as a function of time for `t gt 0` is given by `i_(0)e^(-2Rt//mu_(0)pi)`.B. For the same as in option (a), the current in the loop as a function of time `t=0` is given by `(mu_(0)iR)/(2 r)`.C. The time in which current in loop decreases to `10^(-3) i_(0)` (from t=0) for `R=100 Omega` and `r=5 cm` is given by `(3 pi^(2)1n 10)/(10^(10))s`.D. Fro the same as in option (c), the time in which current in loop decreases to `10^(-3) i_(0)` (from t=0) for `R=100 Omega` and `r-5 cm` is given by `(3pi^(2))/(10^(6))s`. |
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Answer» Correct Answer - A::C Flux linked with loop due to its own magnetic field. `phi=(mu_(0)i)/(2r) (pi r^(2))=(mu_(0)piri)/(2)` emf induced = `-(dphi)/(dt) = e = -(mu_(0)pir)/(2R)(di)/(dt)` `int_(i_0)^(i) (di)/(dt)=-int_(0)^(t) (2R)/(mu_(0)pir)* dt` `i=i_(0)e^(-2RT//mu_(0)pit)` Now, `10^(-3)i_(0)=i_(0)e^(-(2Rt)/(mu_(0)pir))` which gives `t=(3pi^(2)1n10)/(10^(10))s`. |
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