Question

In the below 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 $I=I_{0}cos\omega t$ $A$, but the directions 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. $\frac{{\mu }_{0}aln2I_0\omega }{\pi }sin\omega t$
• B. $\frac{2{\mu }_{0}aln2I_0\omega }{\pi }sin\omega t$
• C. $\frac{{\mu }_{0}aln2I_0\omega }{2\pi }cos\omega t$
• D. $\frac{{\mu }_{0}aln2I_0\omega }{\pi }cos\omega t$

Answer 1

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

Let us consider a small section as shown$d\varphi =B\cdot dA$
$d\varphi =\frac{{\mu }_{0}I}{2\pi x}a\cdot dx$
${\int }_0^{\varphi }d\varphi =\frac{{\mu }_{0}Ia}{2\pi }{\int }_a^{2a}\frac{dx}{x}$
$\varphi =\frac{{\mu }_{0}Ia}{2\pi }\ln 2$
Net Flux,
$\varphi =2\times \varphi$
$=\frac{2{\mu }_0\left(I_0\cos \omega t\right)a\ln 2}{2\pi }$
$=\frac{{\mu }_0\left(I_0\cos \omega t\right)a\ln 2}{\pi }$
$\varphi =\frac{{\mu }_0\left(I_0\cos \omega t\right)a\ln 2}{\pi }$
$\begin{array}{cc}\epsilon & =|-\frac{d\varphi }{dt}|\\ \frac{d\varphi }{dt}& =\frac{{\mu }_0{I}_{0}a\ln 2}{\pi }\frac{d\left(\cos \omega t\right)}{dt}\\ \epsilon & =\frac{{\mu }_0{I}_{0}a\ln 2\omega \cdot \sin \omega t}{\pi }\end{array}$