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The ratio of magnetic induction on the axis of a circular current carrying coil of radius r to the magnetic induction at its centre will be

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11+x2r23/2

11+x2r21/2

11−X2r2−3/2

11+X2r2−1/2

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detailed solution

Correct option is A

We know that Bx=μ0Nir22r2+x23/2 and B0=μ0Ni2r∴ BXB0=r3r2+x23/2=11+x2/r23/2

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