Electromagnetic Induction

JEE Main level, single correct, medium

46 of 153

Question 46 of 153

In a uniform magnetic field of induction BB a wire in the form of semicircle of radius rr rotates about the diameter of the circle with angular frequency ω\omega. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is RR the mean power generated per period of rotation is

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Answer: (B) (Bπr2ω)22R\frac{\left(B \pi r^2 \omega\right)^2}{2 R}

Magnetic flux =BAcos⁡θ=B⋅πr22cos⁡ωt=B A \cos \theta=B \cdot \frac{\pi r^2}{2} \cos \omega t
∴εind =−dϕdt=12 Bπr2ωsin⁡ωt∴P=εind 2R=B2π2r4ω2sin⁡2ωt4R\begin{aligned} & \therefore \varepsilon_{\text {ind }}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{1}{2} \mathrm{~B} \pi \mathrm{r}^2 \omega \sin \omega \mathrm{t} \\ & \therefore \mathrm{P}=\frac{\varepsilon_{\text {ind }}^2}{\mathrm{R}}=\frac{\mathrm{B}^2 \pi^2 \mathrm{r}^4 \omega^2 \sin ^2 \omega \mathrm{t}}{4 \mathrm{R}} \end{aligned}
Now, <sin⁡2ωt>=1/2 \lt \sin ^2 \omega t\gt =1 / 2 (mean value)
∴⟨P⟩=(Bπr2ω)28R. \therefore\langle\mathrm{P}\rangle=\frac{\left(\mathrm{B} \pi \mathrm{r}^2 \omega\right)^2}{8 \mathrm{R}} \text {. }

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