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  • A circular coil of radius 6 cm and 20 turns rotates about its vertical diameter with an angular speed of <img alt="\mathrm{40 \; rad\; s^{-1}}" src="https://entrancecorner.oncodecogs.com/gif.latex?

A circular coil of radius 6 cm and 20 turns rotates about its vertical diameter with an angular speed of \mathrm{40 \; rad\; s^{-1}} in a uniform horizontal magnetic field of magnitude \mathrm{2\times 10^{-2}T} . If the coil form a closed loop of resistance\mathrm{8\Omega }, then the average power loss due to joule heating is:

Option: 1

\mathrm{2.07 \times 10^{-3} \mathrm{~W}}


Option: 2

\mathrm{1.23 \times 10^{-3} \mathrm{~W}}


Option: 3

\mathrm{3.14 \times 10^{-3} \mathrm{~W}}


Option: 4

\mathrm{1.80 \times 10^{-3} \mathrm{~W}}


Answers (1)

best_answer

\mathrm{\mathrm{r}=6 \mathrm{~cm}=6 \times 10^{-2} \mathrm{~m}, \mathrm{n}=20, \omega=40 \mathrm{rads}^{-1}, \mathrm{~B}=2 \times 10^{-2} \mathrm{~T}, \mathrm{R}=8 \Omega}

Maximum emf induced, \mathrm{\varepsilon=\mathrm{NAB} \omega=\mathrm{N}\left(\pi^2\right) \mathrm{B} \omega}

\mathrm{=20 \times \pi \times\left(6 \times 10^{-2}\right)^2 \times 2 \times 10^{-2} \times 40=0.18 \mathrm{~V}}

Average value of emf induced over a full coil, \mathrm{I=\frac{\varepsilon}{R}=\frac{0.18}{8}=0.023 \mathrm{~A}}

Maximum value of current in the coil , \mathrm{I=\frac{\varepsilon}{R}=\frac{0.18}{8}=0.023 \mathrm{~A}}

Average power dissipated, \mathrm{\mathrm{P}=\frac{\varepsilon \mathrm{I}}{2}=\frac{0.18 \times 0.023}{2}=2.07 \times 10^{-3} \mathrm{~W}}

 

 

 

Posted by

shivangi.shekhar

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