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 freq - Physics MCQ AIEEE 2004 2026 | ExamTest.live | ExamTest.live

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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 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 $\frac{B π r ^{2} ω}{2 R}$
B $\frac{( B π r ^{2} ω ) ^{2}}{8 R}$
C $\frac{( B π r ^{2} ω ) ^{2}}{2 R}$
D $\frac{( B π r ^{2} ω ) ^{2}}{4 R}$

Solution & Step-by-step Explanation

The area of the semicircle is A = \frac{π r ^{2}}{2} .The flux through the area at time t is ϕ = B A cos ω t = B \frac{π r ^{2}}{2} cos ω t .The induced emf ε = - \frac{d ϕ}{d t} = B \frac{π r ^{2}}{2} ω sin ω t .This is an AC emf with peak value ε_{0} = \frac{B π r ^{2} ω}{2} .The instantaneous power is P = \frac{ε ^{2}}{R} .Mean power ⟨ P ⟩ = \frac{ε _{0}^{2}}{2 R} : ⟨ P ⟩ = \frac{1}{2 R} (\frac{B π r ^{2} ω}{2})^{2} = \frac{( B π r ^{2} ω ) ^{2}}{8 R}

Practice this question

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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 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

\frac{B π r ^{2} ω}{2 R}

B

\frac{( B π r ^{2} ω ) ^{2}}{8 R}

C

\frac{( B π r ^{2} ω ) ^{2}}{2 R}

D

\frac{( B π r ^{2} ω ) ^{2}}{4 R}

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