Magnetic flux + Calculus = MagaCulus

To find the magnetic flux passing through the X-Y plane when X<0 it is difficult to direct integrate it..!! So let's Do some Different.

Assume an Hypothetical current carrying wire which is Placed on entire Y-axis and whose ends are connected at infinity so that they form a Square Type loop. Also Let the current flowing in the square Loop is ${ I }_{ 1 }$

Now name the two loop. Call circular loop to be 2 and the square loop as 1.

And Let Mutual inductance between these Loops is " M "

$\displaystyle{{ \phi }_{ 2 }\quad =\quad M{ I }_{ 1 }\quad \longrightarrow (1)}$ .

Also

$\displaystyle{{ \phi }_{ 2 }\quad =\quad \int { { B }_{ 1 }{ dA }_{ 2 }\cos { \pi } } }$ .

now Consider an rectangular strip in the circular loop-2 of width dx which is at an distance of x From the loop-1 (From y-axis)

So According to figure :

Diagram

$\displaystyle{x\quad =\quad { x }_{ 0 }\quad -\quad R\sin { \theta } \quad \\ \\ dx\quad =\quad -R\cos { \theta } d\theta \\ \\ { B }_{ 1 }\quad =\quad \frac { { \mu }_{ 0 }{ I }_{ 1 } }{ 2\pi x } \\ \\ { dA }_{ 2 }\quad =\quad ldx\quad =\quad -(2R\cos { \theta } )R\cos { \theta } }$ .

So magnetic flux passing through Loop-2 is

$\displaystyle{{ \phi }_{ 2 }\quad =\quad \frac { { \mu }_{ 0 }{ I }_{ 1 }{ R }^{ 2 } }{ \pi } \int _{ \pi /2 }^{ -\pi /2 }{ \frac { \cos ^{ 2 }{ \theta } }{ \quad { x }_{ 0 }\quad -\quad R\sin { \theta } \quad } d\theta } \\ \\ { \phi }_{ 2 }\quad =\quad \frac { { \mu }_{ 0 }{ I }_{ 1 }{ R }^{ 2 } }{ \pi } ((2-\sqrt { 3 } )\pi )\quad \longrightarrow \quad (2)}$ .

Now comparing equation 1 and 2 we get value of M that is :

$\displaystyle{M\quad =\quad \frac { { \mu }_{ 0 }{ R }^{ 2 } }{ \pi } ((2-\sqrt { 3 } )\pi )}$ .

Now The Magnetic Flux Passing through The X-Y Plane where (X < 0 ) is equal to magnetic flux passing through the Loop-1

$\displaystyle{{ \phi }_{ (XY-Plane\quad X\le 0) }\quad ={ \quad \phi }_{ 1 }\quad =\quad M\quad { I }_{ loop-2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { { \mu }_{ 0 }{ R }^{ 2 }{ I }_{ 0 } }{ \pi } ((2-\sqrt { 3 } )\pi )\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 26.93\quad Ans}$ .

Consider the boundary of the region $X \leq 0$ as a rectangular loop. Let $\phi$ be the required flux.

$\phi = M I_{0}$ , where $M$ is the coefficient of mutual induction.

If we introduce current $I$ in the rectangular loop. Let $\phi'$ be the flux in the loop due to this current.

Clearly,

$\phi' = M I$

Clearly, $\phi'$ is due to $Y-axis$ only, because other wires are far away from the loop contributing to negligiblle magnetic flux.

Clearly,

$\phi' = \displaystyle \int_{1}^{3} \dfrac{\mu_{0} I}{2 \pi r} 2 \sqrt{1 - (2-r)^2} dr$

= $(8- 4\sqrt{3}) \pi I$ (Using the substitution $(2-r) = \sin \theta$ )

Clearly, $M = (8-4\sqrt{3}) \pi \text{ Henry }$

Hence, $\phi = M I_{0} = \boxed{26.94 {\text{Tm}}^2}$