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
$($ where $\tan \varphi =\frac{\omega L}{R})$

• A. $\frac{2{\mu }_{0}a\ln 2I_0\omega }{\pi \sqrt{R^2+{\omega }^2{L}^2}}\sin (\omega t-\varphi )$
• B. $\frac{2{\mu }_{0}a\ln 2I_0\omega }{\pi \sqrt{R^2+{\omega }^2{L}^2}}\sin (\omega t+\varphi )$
• C. $\frac{{\mu }_{0}a\ln 2I_0\omega }{\pi \sqrt{R^2+{\omega }^2{L}^2}}\sin \omega t$
• D. $\frac{{\mu }_{0}a\ln 2I_0\omega }{\pi \sqrt{R^2+{\omega }^2{L}^2}}\sin (\omega t-\varphi ]$

Answer 1

Answer: (D)

$d\varphi =BdA$
$d\varphi =\left[\frac{{\mu }_{0}I}{2\pi x}+\frac{{\mu }_{0}I}{2\pi (3a-x)}\right]adx$

$\varphi =\frac{{\mu }_{0}I}{2\pi }\left[\underset{a}{\overset{2a}{\int }}\frac{dx}{x}+\underset{a}{\overset{2a}{\int }}\frac{dx}{(3a-x)}\right]a;\varphi =\frac{{\mu }_{0}Ia}{\pi }\ln 2$
Magnitude of emf in this circuit:
$\epsilon =\left|\frac{d\varphi }{dt}\right|=\frac{{\mu }_{0}a(\ln 2)}{\pi }\left|\frac{dI}{dt}\right|$
$=\frac{{\mu }_{0}a\ln 2}{\pi }{I}_0\omega \sin \omega t$
ac current, $I=\frac{{\mu }_{0}a\ln 2I_0\omega }{\pi \sqrt{R^2+{\omega }^2{L}^2}}\sin (\omega t-\varphi )$