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JEE Main 2004
Exam
JEE Main
Year
2004
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2004 · Physics · Electromagnetic Induction

In a uniform magnetic field of induction B a wire in the form of...

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

Options

  1. ABπr2ω2R\frac{B \pi r^2 \omega}{2 R}
  2. B(Bπr2ω)22R\frac{\left(B \pi r^2 \omega\right)^2}{2 R}
  3. C(Bπrω)22R\frac{(\mathrm{B} \pi \mathrm{r} \omega)^2}{2 \mathrm{R}}
  4. D(Bπrω2)28R\frac{\left(B \pi r \omega^2\right)^2}{8 R}

Correct answer

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

Step-by-step solution

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}$Now,Now, \lt \sin ^2 \omega t>=1 / 2(meanvalue)∴⟨(mean value) \therefore\langle\mathrm{P}\rangle=\frac{\left(\mathrm{B} \pi \mathrm{r}^2 \omega\right)^2}{8 \mathrm{R}} \text{. }$
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