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Q.

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

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a

π2BIl

b

BIl2

c

BIl2π

d

BIl

answer is C.

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Detailed Solution

BI(dl)=2Tsin(dθ/2)  ⇒BI(rdθ)=2T(dθ/2) ( θ is small, sinθ=θ) ⇒T=Bir=Bil/2π
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