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Consider following coils each of one turn carrying current I. The magnitude of the magnetic induction at X, Y, Z are B1 , B2 and B3 , respectively. Then (assume side of square to be same in each case)

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a

B3 > B1 > B2

b

B2 > B3 > B1

c

B2 > B1 > B3

d

B1 > B2 > B3

answer is A.

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Detailed Solution

If L be the length of side of the current carrying square , then magnetic field at its centre B is inversely proportional to L.Therefore B3>B1.Also B2=14(B3-B1).So B2 has the least value .

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Consider following coils each of one turn carrying current I. The magnitude of the magnetic induction at X, Y, Z are B1 , B2 and B3 , respectively. Then (assume side of square to be same in each case)