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Question

# The field normal to the plane of a coil of n turns and radius r carrying a current I is measured on the axis of the coil at a small horizontal distance h from the centre of the coil. This is smaller than the field at the centre by the fraction :

A
32h2r2
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B
23h2r2
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C
32r2h2
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D
23r2h2
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Solution

## The correct option is A 32h2r2The field at center is given by Bc=μoi2rand on its axis at a distance h is given by Ba=μoir22(r2+h2)3/2 when h<<rthen Ba=μoi2R(1+(h/r)2)3/2By applying binomial approximation we get,Ba=μoi(1−3h22r2)2rNow Difference in the field at center and at distance h is ΔB=Bc−BaΔB=μoi×3h24r3Now ΔB/B=3h22r2Hence option(A)

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