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.

Total magnetic flux in this loop is

• A. $\frac{{\mu }_{0}a{I}_{0}cos\omega t}{\pi }ln2$
• B. $\frac{2{\mu }_{0}a{I}_{0}cos\omega t}{\pi }ln2$
• C. $\frac{4{\mu }_{0}a{I}_{0}cos\omega t}{\pi }ln2$
• D. $\frac{{\mu }_{0}a{I}_{0}cos\omega t}{2\pi }ln2$

Answer 1

The correct option is A $\frac{{\mu }_{0}a{I}_{0}cos\omega t}{\pi }ln2$

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,
${\varphi }_{net}=2\times \varphi$
$=\frac{2{\mu }_0\left(I_0\cos \omega t\right)a\ln 2}{2\pi }$
$=\frac{{\mu }_0\left(I_0\cos \omega t\right)a\ln 2}{\pi }$