Question

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 $I=I_{0} \cos \omega t$ but the directions of current in them are opposite.

The instantaneous current in the circuit will be
$\left(\right.$ where $\left.\tan \phi=\frac{\omega L}{R}\right)$

• A. $\frac{2 \mu_{0} a \ln 2 I_{0} \omega}{\pi \sqrt{R^{2}+\omega^{2} L^{2}}} \sin (\omega t-\phi)$
• B. $\frac{2 \mu_{0} a \ln 2 I_{0} \omega}{\pi \sqrt{R^{2}+\omega^{2} L^{2}}} \sin (\omega t+\phi)$
• C. $\frac{\mu_{0} a \ln 2 I_{0} \omega}{\pi \sqrt{R^{2}+\omega^{2} L^{2}}} \sin \omega t$
• D. $\frac{\mu_{0} a \ln 2 I_{0} \omega}{\pi \sqrt{R^{2}+\omega^{2} L^{2}}} \sin (\omega t-\phi]$

Answer 1

Answer: (D)

$d \phi=B d A$
$d \phi=\left[\frac{\mu_{0} I}{2 \pi x}+\frac{\mu_{0} I}{2 \pi(3 a-x)}\right] a d x$

$\phi=\frac{\mu_{0} I}{2 \pi}\left[\int\limits_{a}^{2 a} \frac{d x}{x}+\int\limits_{a}^{2 a} \frac{d x}{(3 a-x)}\right] a ; \phi=\frac{\mu_{0} I a}{\pi} \ln 2$
Magnitude of emf in this circuit:
$\varepsilon =\left|\frac{d \phi}{d t}\right|=\frac{\mu_{0} a(\ln 2)}{\pi}\left|\frac{d I}{d t}\right| $
$=\frac{\mu_{0} a \ln 2}{\pi} I_{0} \omega \sin \omega t$
ac current, $I=\frac{\mu_{0} a \ln 2 I_{0} \omega}{\pi \sqrt{R^{2}+\omega^{2} L^{2}}} \sin (\omega t-\phi)$