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A rectangular coil of area 'A', having number of turns N, is rotated at 'f' revolutions per second in a uniform magnetic field B, the field being perpendicular to the coil. Prove that the maximum emf induced in the coil is 2 pi fNBA. |
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Answer» Solution :Consider a rectangular coil PQRS of N turns, each of area A, held in a uniform magnetic field `vecB`. Let the coil be rotated at a steady angular velocity w about its own axis. Let at any instant t, normal to the area (i.e., the area vector `vecA`) subtends an ANGLE `THETA = omega t` from direction of magnetic field `vecB`. Then, at that MOMENT, the magnetic flux linked with the coil is ` phi_(B) = NvecB.vecA = NB Acos theta = NB Acos omegat` `therefore` Induced emf `varepsilon = - (dphi_(B))DT = - d/dt (N B A cos omegat)` ` =-N B A d/dt(cos omegat) = N B A omegat` where `varepsilon_(0) = N B A omega` = maximum value of induced emf. If the coil be ROTATING at .f. revolutions per second, then `omega = 2pif` and the maximum induced emf `varepsilon_(0) = NBAomega = 2pifNBA` 
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