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 an angular frequency ω. 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 (a) ((Bπ rω)²)/(2R) (b) ((Bπ r²ω)²)/(8R) (c) (Bπ r²ω)/(2R) (d) ((Bπ rω²)²)/(8R)

Answer: (b). Area of the semicircle A = (π r²)/(2), so peak emf e₀ = BAω = (Bπ r²ω)/(2). Mean power: P = (e₀²)/(2R) = 1/2R((Bπ r²ω)/(2))² P = ((Bπ r²ω)²)/(8R) So the answer is (b).

In a uniform magnetic field of induction B a wire in the form of a… | JEE Main 2004 (AIEEE)

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Rotating Conductors & AC Generator

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 an 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: (b)

Area of the semicircle $A = \frac{\pi r^2}{2}$ , so peak emf $e_0 = BA\omega = \frac{B\pi r^2\omega}{2}$ .
Mean power: $P = \frac{e_0^2}{2R} = \frac{1}{2R}\left(\frac{B\pi r^2\omega}{2}\right)^2$
$P = \frac{(B\pi r^2\omega)^2}{8R}$
So the answer is (b).

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Rotating Conductors & AC Generator question from JEE Main 2004 (AIEEE)

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