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A thin flexible wire of length L is connected to two adjacent fixed points and the wire carries a current I in the clockwise direction. When the system is put in a uniform magnetic field with a strength of B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is?

F.I. Mazumder , 6 Years ago
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anser 1 Answers
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
 
idLB = iRdθB, as is clear from the figure.
 
The tension 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 is related to the tension T as
 
iRdθB = 2T sin(dθ/2)
 
Since  is a very small angle, sin(dθ/2) ≈ dθ/2
 
Therefore, the above equation gives T = iRB
 
But R = L /
 
Therefore, T = iBL /

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