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When a rod of length l is rotated with angular velocity of `omega` in a perpendicular field of induction B , about one end , the emf across its ends isA. `Bl^(2)omega`B. `0.5 Bl^(2)omega`C. `Bl omega`D. `0.5 Blomega` |
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Answer» Correct Answer - B Induced e.m.f. `(e) = (d phi)/(dt) = B cdot (dA)/(dt)` But `dA = pil^(2) and dt=T and T = (2 pi)/(omega)` `:. e = (B cdot pi l^(2))/((2pi//omega))=1/2 Bl^(2)omega`. |
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