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(a) Draw a labelled diagram of a.c. generator. Derive the expression for the instantaneous value of the emf induced in the coil. (b) A circular coil of cross-sectional area 200 cm^(2) and 20 turns is rotated about the vertical diameter with angular speed of 50 rad s^(-1) in a uniform magnetic field of magnitude 3.0 xx 10^(-2)T. Calculate the maximum value of the current in the coil. |
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Answer» Solution :(a) See SHORT Answer Question Number 53. (b) Here cross-sectional area `A = 200 cm^(2) = 200 xx 10^(-4)m^(2)`, number of turns N = 20, ANGULAR speed `omega = 50 "rad s"^(-1)` and magnetic field `B = 3.0 xx 10^(-2)T`. `therefore` Maximum VALUE of the induced emf `varepsilon_(m) = NABomega = 20 xx (200 xx 10^(-4))xx(3.0 xx10^(-2))xx50 = 0.6 V`. If R be the resistance of the coil then maximum value of the current `(I_(m))` flowing in the coil is given as : `I_(m) = (varepsilon_(m))/R=(0.6)RA` |
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