Total force made by the magnetic fields and the current, it should be zero but as I tried to derive it , the upward components only remains

Question

As shown in the above diagram , the same 2 coils of square are aligined with paralell of same plane .

$$ I:=\text{current which flows to each wire of the square } $$

What I am confusing to understand currently is about a force which acts against the wire A'B' made by BC and AD.

Using Biot–Savart law , the wire BC makes magnetic fields on points on A'B' . Each of the magnetic fields is in the same plane(this plane is perpendicualr to BC )

$$ \boldsymbol{ dF }_{}= I \left( \boldsymbol{ds}_{}\times \boldsymbol{B}_{} \right) \tag{1} $$

As we use right hand rule with the above formula , we can easily know that each cross product between \$~ I ~\$ of A'B' and that each magnetic field points upward of force vector .

Nextly we handle AD .

Each magnetic field is made on a point on A'B' by AD .

As we use eqn1 again , the vector of force points up upward again .

So by 2 straight wires of BC , AD , only the upward forces are made at A'B'

But the book states the below .

The currents of BC , AD are opposite hence the forces of 2 are opposite and be cancelled out

Where I've made (a) mistake(s)?

Answer 1

$$ H= \frac{ I }{ 4\pi } \left\{ \cos^{}\left(\theta_{1} \right) + \cos^{}\left(\theta_{2} \right) \right\} \tag{1} $$

$$ \boldsymbol{ dF }_{}= I \left( \boldsymbol{ds}_{}\times \boldsymbol{B}_{} \right) \tag{2} $$

As we mark a point on A'B' , then drop the position of that dot and we will get a point on C'D' .

Using eqn1 , we can get that those 2 dots have completely same magnetic field vector .

And, \$~ I ~\$ flows to A'B' , C'D' with opposite direction each other .

Hence , noticing \$~ \sin^{}\left(\theta_{} \right) = \sin^{}\left(\pi-\theta_{} \right) ~\$ and combining eqn2 , we can easily know that those points have opposite direction of same magnitude of vectors of force .