# 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

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 .

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

• I tried to prove it ,but I have been failing it . From definition of Biot-Savart law, a direction of mangenitc field is perpendicular to the plane which is determined by small line displacement and the displacement vector . Hence , my assumption which you said seemingly still be held I think . Commented Oct 3, 2021 at 23:50
• I'll delete my previous comment, your conclusion that they span a plane is correct but in biot savart law B is proportional to dl × r but dl vector is in opposite directions in both braches because current in both branch (BC and AD) are in opposite direction and for same r vector you'll always get B in opposite direction ex. If we take dl vector at C and calculate magnetic field at B' and if we take dl vector at D and calculate magnetic field at A' we'll get same magnitude of dB(magnetic field ) because r vector is same in both cases but direction will be opposite due to opposite sign of dl Commented Oct 4, 2021 at 14:16
• I think I derived the thing . Commented Oct 6, 2021 at 0:23

$$H= \frac{ I }{ 4\pi } \left\{ \cos^{}\left(\theta_{1} \right) + \cos^{}\left(\theta_{2} \right) \right\} \tag{1}$$
$$\boldsymbol{ dF }_{}= I \left( \boldsymbol{ds}_{}\times \boldsymbol{B}_{} \right) \tag{2}$$
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 .