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)

Correct option is 1B3 > B1 > B2

Questions

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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B3 > B1 > B2

B2 > B3 > B1

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B1 > B2 > B3

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detailed solution

Correct option is A

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)

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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)