question_answer

A loop of flexible conducting wire of length \[l\] lies in magnetic field B which is normal to the plane of loop. A current is passed through the loop. The tension I developed in the wire to open up is

A) \[\frac{\pi }{2}BIl\]        

B) \[\frac{BIl}{2}\]

C) \[\frac{BIl}{2\pi }\]        

D) \[BIl\]  

Correct Answer: C

Solution :

[c] \[BI\,(dl)=2T\sin (d\theta /2)\] \[\Rightarrow BI(rd\theta )=2T(d\theta \text{/}2)\] (\[\theta \]is small, \[(\sin \theta =\theta )\] \[\Rightarrow T=BIr=BIl/2\pi \]