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Question

A steady current i flows in a small square loop of wire of side L in a horizontal plane. The loop is now folded about its middle such that half of it lies in a vertical plane. Let μ1 and μ2 respectively denote the magnetic moments due to the current loop before and after folding. Then
  1. μ2=0
     
  2. μ2 and μ2 are in the same direction
     
  3. |μ1||μ2|=2
     
  4. |μ1||μ2|=(12)


Solution

The correct option is C |μ1||μ2|=2
 
Initial magnetic moment = μ1=iL2


After folding the loop, M = magnetic moment due to each part =i(L2)×L=iL22=μ12
 μ2=M2=μ12×2=μ12

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