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A stsnding wave `y = 2A sin kx cos omegat` is set up in the wire `AB` fixed at both ends by two vertical walls (see Fig. 3.197). The region between the walls contains a constant magnetic field `B`. Now answer the following question: In the above qestion, the time when the emf becomes maximum for the time isA. (a) `(2pi)/(omega)`B. (b) `(pi)/(omega)`C. ( c) `(pi)/(2omega)`D. (d) `(pi)/(4omega)` |
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Answer» Correct Answer - C ( c) `y = 2A sin kx cos omegat` `v = (dy)/(dt) = -2A sin kx omega sin omegat` `v_(max) = -2A omega sin kx, k = (3pi)/(AB)` `e = int_(0)^(l = AB) bv_(max)dx = -2aomegaBint_(0)^(AB) sin kx dx = + (2omegaAB)/(k)[cos(3pi)/(AB)AB - cos theta] = (-4(AB)omega)/(k)` `omegat = (pi)/(2)` `t = (pi)/(2omega)` For second harmonic `k = (2pi)/(AB)` |
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