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In a uniform magnetic field of induction B a wire in the form of a semicircle of radius r rotates about the diameter of the circle with angular frequency omega. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is R, the mean power generated per period of rotation is ...... |
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Answer» `(Bpir^2omega)/(2R)` `=AB cos omegat` `=(pir^2 B cos omegat)/2` ( `because` Area of SEMICIRCLE `A=(pir^2)/2`) INDUCED emf `epsilon=-(dphi)/(dt)` `=-d/(dt)[(pir^2 B cos omegat)/2]` `=+1/2 pi Br^2omega sin omegat` Power `P=epsilon^2/R=(pi^2B^2r^4 omega^2 sin^2 omegat)/(4R)` `therefore lt sin^2 omegat gt =1/2` `therefore` Average power = `(piBr^2omega)^2/(4R)xx1/2` `=(Bpir^2omega)^2/(8R)` |
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