Question

For the situation shown in the figure, flux through the square loop is

A) $(\dfrac{{{\mu _0}ia}}{{2\pi }})\ln (\dfrac{a}{{2a - b}})$
B) $(\dfrac{{{\mu _0}ib}}{{2\pi }})\ln (\dfrac{a}{{2b - b}})$
C) $(\dfrac{{{\mu _0}ib}}{{2\pi }})\ln (\dfrac{a}{{b - a}})$
D) $(\dfrac{{{\mu _0}ia}}{{2\pi }})\ln (\dfrac{{2a}}{{b - a}})$

Answer 1

Due to symmetry and opposite direction of magnetic field just below and above the wire, net flux in the region $(b - a)$ length above and below the wire will be zero. You only need to calculate the flux below this length.

Formula Used:
The magnetic flux of small area at any distance $x$ from the wire below it is given by $d\phi = B(ldx) = \dfrac{{{\mu _0}I}}{{2\pi x}}ldx$
Where, $B$ is the magnitude of magnetic field and $l$ is the length of the square loop.

Complete step by step solution:
First find the direction of the magnetic field around the wire using the right hand rule. Point your right-hand thumb in the direction of propagation of current in the wire. Then your curled fingers will indicate the direction of the magnetic field due to the wire.
For this case, the magnetic field above the wire is directed out of the plane of paper/screen and below the wire, it is directed inside the plane of the paper/screen.
Now we know that the magnetic flux of small area at any distance $x$ from the wire below it is given by $d\phi = B(ldx) = \dfrac{{{\mu _0}I}}{{2\pi x}}ldx$
Where, $B$ is the magnitude of magnetic field and $l$ is the length of the square loop.
Therefore, magnetic flux above the wire, in the region of the square loop will be,
$d\phi = \dfrac{{{\mu _0}I}}{{2\pi (b - a)}}bdx$ (since above the wire only $(b - a)$ length of the square loop is being considered)
And below the wire, till a distance of $(b - a)$ the magnetic flux will be
$d\phi = - \dfrac{{{\mu _0}I}}{{2\pi (b - a)}}bdx$ (since direction of magnetic field is inverted, magnetic field will be negative)
Hence, in a distance of $(b - a)$ above and below the wire, the net magnetic flux will be zero as the magnetic flux above and below the wire will get cancelled out.
So, to get the total flux in the square loop, we only need to calculate the magnetic flux from a distance of $(b - a)$ below the wire to the end of the square loop.
Therefore, ${\phi _{net}} = \int\limits_{b - a}^a {\dfrac{{{\mu _0}Ib}}{{2\pi }}} \times \dfrac{{dx}}{x}$ (integrating from $(b - a)$ to $a$ as the rest flux will get cancelled out)
Which gives, ${\phi _{net}} = \dfrac{{{\mu _0}Ib}}{{2\pi }}({\log _e}x)_{b - a}^a$ (integrating $\dfrac{1}{x}$ )
On solving, ${\phi _{net}} = \dfrac{{{\mu _0}Ib}}{{2\pi }}{\log _e}(\dfrac{a}{{b - a}})$ or ${\phi _{net}} = \dfrac{{{\mu _0}Ib}}{{2\pi }}\ln (\dfrac{a}{{b - a}})$

Therefore the correct answer is option B

Note: If you do not get the fact that magnetic flux above and below the wire will get cancelled out over a distance of $(b - a)$, you can still get the right answer as it will later on get subtracted. However, the calculation will be hefty and tiring. So, to save time this technique is preferred.