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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 magnitude \mathrm{I}=\mathrm{I}_0 \cos \omega \mathrm{t} but the direction of current in them are opposite.

 

Question

The instantaneous current in the circuit will be

 

Option: 1

\mathrm{\frac{2 \mu_0 \mathrm{a} \ln 2 \mathrm{I}_0 \omega}{\pi \sqrt{\mathrm{R}^2+\omega^2 \mathrm{~L}^2}} \sin (\omega \mathrm{t}-\phi)}


Option: 2

\mathrm{\frac{2 \mu_0 \mathrm{a} \ln 2 \mathrm{I}_0 \omega}{\pi \sqrt{\mathrm{R}^2+\omega^2 \mathrm{~L}^2}} \sin (\omega \mathrm{t}+\phi)}


Option: 3

\mathrm{\frac{\mu_0 \mathrm{a} \ln 2 \mathrm{I}_0 \omega}{\pi \sqrt{\mathrm{R}^2+\omega^2 \mathrm{~L}^2}} \sin \omega \mathrm{t}}


Option: 4

\mathrm{\frac{\mu_0 \mathrm{a} \ln 2 \mathrm{I}_0 \omega}{\pi \sqrt{\mathrm{R}^2+\omega^2 \mathrm{~L}^2}} \sin (\omega \mathrm{t}-\phi)}


Answers (1)

best_answer

a.c. current

\mathrm{I}=\frac{\mu_0 \mathrm{a} \ln 2 \mathrm{I}_0 \omega}{\pi \sqrt{\mathrm{R}^2+\omega^2 \mathrm{~L}^2}} \sin (\omega \mathrm{t}-\phi)

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Gunjita

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