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  • A small loop of area of cross section  is lying concentrically

A small loop of area of cross section 10^{-4} \mathrm{~m}^2 is lying concentrically and coplanar inside a bigger loop of radius 0.628 m. A current of 10 A is passed in the bigger loop. The smaller loop is rotated about its diameter with an angular velocity \omega. The magnetic flux linked with the small loop will be

Option: 1

10^{-9} \sin \omega t


Option: 2

10^{-9} \cos \omega t


Option: 3

10^{-10} \sin \omega t


Option: 4

10^{-10} \cos \omega t


Answers (1)

best_answer

We know that \phi=\mathrm{BA} \cos \omega \mathrm{t}
\mathrm{\begin{aligned} & \text { Here } B=\frac{\mu_0 i}{2 r} \\ & \begin{aligned} \therefore \quad & \phi=\frac{\mu_0 \mathrm{i}}{2 \mathrm{r}} \times \mathrm{A} \cos \omega \mathrm{t} \\ & =\frac{\left(4 \times 3.14 \times 10^{-7}\right) 10}{2 \times 0.628} \times 10^{-4} \times \cos \omega \mathrm{t} \\ & =10^{-9} \cos \omega \mathrm{t} \end{aligned} \end{aligned} }

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Divya Prakash Singh

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