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  • circular loop of area A and resistance R rotates with an angular velocity about an axis through its diameter as s

circular loop of area A and resistance R rotates with an angular velocity \omega about an axis through its diameter as shown in Fig. 9. The plane of the loop is initially perpendicular to a constant magnetic field B. Find the induced current in the loop.

Option: 1

\mathrm{\frac{B A \omega}{2 R}}


Option: 2

\mathrm{\frac{2 B A \omega}{R}}


Option: 3

\mathrm{\frac{B A \omega}{3 R}}


Option: 4

\mathrm{\frac{B A \omega}{R}}


Answers (1)

best_answer

The instantaneous magnetic flux through the loop is

\mathrm{\Phi_B=B A \cos \theta}

Since \mathrm{\theta=\omega \mathrm{t} \text {, therefore, } \Phi_B=B A \cos \omega t}

From Faraday's law, equation (3) is

\mathrm{\boldsymbol{E}=-\frac{d \Phi_B}{d t}=-\frac{d}{d t}[B A \cos \omega t] \quad \text { or } \quad \boldsymbol{E}=B A \omega \sin \omega \mathrm{f}}

The induced current is

\mathrm{I=\frac{\mathrm{E}}{R}=\left(\frac{B A \omega}{R}\right) \sin \omega t}

Both the induced emf and the current vary sinusoidally. The amplitude of the emf is \mathrm{B A \omega} and that of the current is \mathrm{\frac{B A \omega}{R} .}

Posted by

Ritika Kankaria

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