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Q.

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 M1  and M2 respectively denote the magnetic moments due to current loop before and after folding. Then

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a

M2=0

b

M1 and M2 are in the same direction 
 

c

M1/M2=2

d

M1/M2=1/2

answer is C.

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Detailed Solution

M1=l2i k^M2=l22i j^+l22i k^M2=(12×l×i)2+(12×l×i)2=l2i2M1M2=2

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