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.

 

[Nickie]

I would first like to commend you on your attempt to solve this problem and for seeking help when you encountered difficulties. This shows your dedication and determination in understanding the concepts at hand.

Based on your attempt, it seems like you have a good understanding of the basic equations involved in calculating the force in a current loop in a magnetic field. However, there are a few things that need to be clarified in order to solve this problem correctly.

Firstly, when calculating the force on a small segment of the loop, you correctly identified that the force is given by Bidl towards the center. However, when integrating over the entire loop, you cannot simply double this value to get the total force. This is because the force on each segment of the loop is not constant and varies with the angle θ. Therefore, you need to take into account the changing direction and magnitude of the force at each point on the loop.

Secondly, when considering the vertical components of the force, you need to remember that the force on a small segment of the loop is not purely in the vertical direction. It has both vertical and horizontal components, and it is the horizontal components that cancel out when integrating over the loop. Therefore, you cannot simply ignore the horizontal components and only consider the vertical components when calculating the total force.

To correctly solve this problem, you need to use vector calculus and integrate the force vector over the entire loop. This will give you the total force of compression on the loop. It is also important to note that the force of compression will only be present when the magnetic field and current are in the same direction, as in this case. If they are in opposite directions, the force will be one of tension.

I hope this helps clarify some of the concepts involved in solving this problem. Keep up the good work and don't hesitate to seek help when needed. Science is all about collaboration and learning from one another.