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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 :

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Solution

The correct option is **A** 32h2r2

The field at center is given by Bc=μoi2r

and on its axis at a distance h is given by Ba=μoir22(r2+h2)3/2 when h<<r

then Ba=μoi2R(1+(h/r)2)3/2

By applying binomial approximation we get,

Ba=μoi(1−3h22r2)2r

Now Difference in the field at center and at distance h is ΔB=Bc−Ba

ΔB=μoi×3h24r3

Now ΔB/B=3h22r2

Hence option(A)

The field at center is given by Bc=μoi2r

and on its axis at a distance h is given by Ba=μoir22(r2+h2)3/2 when h<<r

then Ba=μoi2R(1+(h/r)2)3/2

By applying binomial approximation we get,

Ba=μoi(1−3h22r2)2r

Now Difference in the field at center and at distance h is ΔB=Bc−Ba

ΔB=μoi×3h24r3

Now ΔB/B=3h22r2

Hence option(A)

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