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 direction of current in them are opposite.

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

• A. $\frac{{\mu }_{0}aln2I_0\omega }{\pi }sin\left(\omega t\right)$
• B. $\frac{2{\mu }_{0}aln2I_0\omega }{\pi }sin\left(\omega t\right)$
• C. $\frac{{\mu }_{0}aln2I_0\omega }{2\pi }cos\left(\omega t\right)$
• D. $\frac{{\mu }_{0}aln2I_0\omega }{\pi }cos\left(\omega t\right)$

Answer 1

The correct option is A $\frac{{\mu }_{0}aln2I_0\omega }{\pi }sin\left(\omega t\right)$

Magnitude of emf in this circuit
$ϵ=\left|\frac{d\varphi }{dt}\right|=\frac{{\mu }_0\left(ln2\right)}{\pi }\left|\frac{dI}{dt}\right|$
$ϵ=\frac{{\mu }_0\left(ln2\right)}{\pi }{I}_0\omega sin\omega t$
a.c. current is
$I=\frac{{\mu }_{0}aln2I_0\omega }{\pi \sqrt{R^2+{\omega }^2{L}^2}}sin(\omega t-\varphi )$