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.

Total magnetic flux in this loop is

• A. $\frac{{\mu }_0{I}_a}{\pi }ln2$
• B. $\frac{2{\mu }_0{I}_a}{\pi }ln2$
• C. $\frac{4{\mu }_0{I}_a}{\pi }ln2$
• D. $\frac{{\mu }_0{I}_a}{2\pi }ln2$

Answer 1

The correct option is A $\frac{{\mu }_0{I}_a}{\pi }ln2$

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

$\varphi =\frac{{\mu }_{0}I}{2\pi x}\left[{\int }_a^{2a}\frac{dx}{x}+{\int }_a^{2a}\frac{dx}{(3a-x)}\right]a$
$\varphi =\frac{{\mu }_0{I}_a}{\pi }ln2$.