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A coil of 800 turns and 50cm^(2) are makes 10 rps about an axis in its own plane in a magnetic field of 10gauss perpendicular to this axis. What is the instantaneous induced emf in the coil? |
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Answer» Solution :`A= 50cm^(2) = 50 xx 10^(-4) m^(2)` n= 10 rps, N= 800 B= 100 gauss `= 100 xx 10^(-4) T= 10^(-2)T` Now, `E= E_(0 ) sin omega t= NBA omega sin omega t` `=800 xx 10^(-2) xx 50 xx 10^(-4) xx 2pi xx 10sin (20 pi t)` or `E= 2.5 sin (20 pi t)` VOLT |
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