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

- →μ2=0

- →μ2 and →μ2 are in the same direction

- |→μ1||μ2|=√2

- |→μ1||μ2|=(1√2)

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=M√2=μ12×√2=μ1√2

Initial magnetic moment = μ1=iL2

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

⇒ μ2=M√2=μ12×√2=μ1√2

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