First slide

Magnetic field due to a straight current carrying wire

Question

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)

Difficult

A

B₃ > B₁ > B₂
B

B₂ > B₃ > B₁
C

B₂ > B₁ > B₃
D

B₁ > B₂ > B₃

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 $B_{3} > B_{1} .$ Also $B_{2} = \frac{1}{4} (B_{3} - B_{1}) .$ So $B_{2} h a s t h e l e a s t v a l u e .$

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