The field normal to the plane of a wire of n turns and radius r which carries a current i is measured on the axis of the coil at a small distance h from the centre of the coil. This is smaller than the field at the centre by the fraction
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a
32 h2r2
b
23 h2r2
c
32 r2h2
d
23 r2h2
answer is A.
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Detailed Solution
Field at the centre, B1 =μ04π × 2πinr = μ02 . nirField at a distance h from the centre,B2 =μ04π . 2πnir2r2 + h23/2 = μ02 . nir2r3 1+h2r23/2=B1 1+h2r2−3/2 = B1 1−32.h2r2 (By binomial theorem)Hence B2 is less than B1 by a fraction = 32 h2r2