Arun
Last Activity: 6 Years ago
Dear student
you have not attached an image.
but I remember that similiar question has been asked in IIT in 2010 or 2011.
So I am answering it from that context.
Considering an elemental length dL of the wire, the magnetic force F acting on the element is given by
F = idLB = iRdθB, as is clear from the figure.
The tension T developed in the wire is resolved into rectangular components T sin(dθ/2) and Tcos(dθ/2) as indicated in the figure.
The magnetic force F is related to the tension T as
F = iRdθB = 2T sin(dθ/2)
Since dθ is a very small angle, sin(dθ/2) ≈ dθ/2
Therefore, the above equation gives T = iRB
But R = L /2π
Therefore, T = iBL /2π