Compression force in current loop in magnetic field

[amal]

Homework Statement

Please check the enclosed figure.
Find the force of compression in the wire loop.
Magnetic field B is directed into the page and current i is flowing anti-clockwise. The radius of the wire loop is 'a'.

Homework Equations

$\vec{F}=i\vec{l}\times\vec{B}$

The Attempt at a Solution

I first took a component $dl$ and calculated force on it. It came out to be $Bidl$ towards center. That, in angular form, is $Biadθ$.
Now, as force of compression is asked, I thought that I will have to consider vertical components of that force only. So, I took two $dl$s and I resolved forces on them. I have elaborated in the figure. Horizontals (I am saying horizontals and verticals because they appear like that in the figure) cancel out and what I get is $2Biacosθdθ$. Integrating over $0$ to $2π$ gets you zero.
I gave it a thought and it occurred to me that this might be so because net force is zero, just like when you press a pen from both ends towards its center. Then tried it by cutting the wire in semicircles and trying to find effective force towards the center and doubling it (again thinking of a pen, like if you apply 5N from each end, net compressive force is 10N). I did it pretty much the same way, only in vain. Now, something is going wrong in my basic assumptions, I think, but what exactly I don't know. Help me.

 

[andrien]

the compression force is simply T=iaB because
Tdθ=iaBdθ

 

[amal]

Could you please explain in more detail?

 

[andrien]

sorry for late reply,but there is a thing which you can derive.whenever there is a motion of a chain or string in any curved path then you can resolve the forces along normal and perpendicular to the normal ,these are Tdθ and dT respectively also accompanied by other forces to obtain eqn of motion.In this case compressive force is uniform because there is no tangential force so dT=0,and along the normal it is 2TSindθ/2 which is Tdθ.if the sense of current is changed then force is tensile in nature.