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  • In figure, a square loop consisting of an inductor of inductance L and resistor of resistance R is placed between two long parallel wires. The two long straight wires have time varying current of m

In figure, a square loop consisting of an inductor of inductance L and resistor of resistance R is placed between two long parallel wires. The two long straight wires have time varying current of magnitude \mathrm{I}=\mathrm{I}_0 \cos \omega \mathrm{t} but the direction of current in them are opposite.

Question

Magnitude of emf in this circuit only due to flux change associated with two long straight current-carrying wires will be

 

Option: 1

\mathrm{\frac{\mu_0 a \ln 2 I_0 \omega}{\pi} \sin \omega t}


Option: 2

\mathrm{\frac{2 \mu_0 \mathrm{a} \ln 2 \mathrm{I}_0 \omega}{\pi} \sin \omega \mathrm{t}}


Option: 3

\mathrm{\frac{\mu_0 \mathrm{a} \ln 2 \mathrm{I}_0 \omega}{2 \pi} \cos \omega \mathrm{t}}


Option: 4

\mathrm{\frac{\mu_0 \mathrm{a} \ln 2 \mathrm{I}_0 \omega}{\pi} \cos \omega \mathrm{t}}


Answers (1)

best_answer

Magnitude of emf in this circuit

\varepsilon=\left|\frac{\mathrm{d} \phi}{\mathrm{dt}}\right|=\frac{\mu_0 \mathrm{a}(\ln 2)}{\pi}\left|\frac{\mathrm{dI}}{\mathrm{dt}}\right| \quad \varepsilon=\frac{\mu_0 \mathrm{a} \ln 2}{\pi} \mathrm{I}_0 \omega \sin \omega \mathrm{t} .

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shivangi.shekhar

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