Calculation of tension in a loop with force acting outwards I'm having a few problems with understanding how to calculate tension in a loop.
If I have a circular loop, and some force is applied uniformly radially outwards in such a way that the force acting on each element of the string is normal to the string at that point, then how will the string develop a tension? 
My intuition tells me that some force must act by the string to prevent the string from expanding infinitely. However, how can the string apply such a force, when its only means is tension which acts perpendicular to the force always?
Any answers will be appreciated.
I suppose that this is analogous to asking why pumping air into a soap bubble will make it expand a certain amount (until the external pressure is equivalent to the excess internal pressure).
EDIT:
So, I can simply take the semi-circular half and apply Newton's Laws?

So $2 T = F $
Therefore, tension developed $= \frac F2$
Just want to clarify that by F I mean the total horizontal Force acting on the semicircle, not just the central element.
Is this correct? Just want to verify if I understood the concept. 
 A: The loop has a curvature. As a result, when you take a small element of the loop, the tension force applied to it from both the sides is at an angle, which results in a force component to the inside.

Since your question asks for a rationale behind the tension force, I'll leave it here. You can now try yourself to find the tension by equating the outward force with the inward force. Or, if you are given the total outward force, you can integrate the tension force over the loop and then equate the total outward and inward forces.
A: Alright I'm a bit late here, but I want to give an answer for all who Google this.
Suppose we approximate the loop to n particles joined by n light, inextensible and rigid rods on the vertices of a regular n-gon. 
I can't give a diagram here but if you look at the previous answer, you have a radial force of magnitude R/n where R is the total force, and a tension T. If the angle between vertices is θ, then the angle between T and the "horizontal" is θ/2. 
θ = 2π/n because 2π in  circle and n vertices.
By N2, R/n = 2Tsin(θ/2)
T = R/(2nsin(π/n))
As n tends to infinity and n >> π, sin(π/n) = π/n. So T = R/(2nπ/n) = R/2π. 
Surprisingly this seems to be independent of the radius of the circle, but I suppose with dimensional analysis you can't have distances involved in an equation relating two forces.
A: I like to challenge the answers above. Instead of looking at local equilibrium of forces, I'm checking the energy balance, meaning work input and output must be equal. Check out the sketch below, assuming the depth of loop is a unit length, not shown here.

