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Question

# The magnetic field normal to the plane of a circular coil of n turns and radius r which carries a current i is measured on the axis of coil at a very small distance h from the centre of coil. The measured field is smaller than the magnetic field at centre by a fraction

A
2r23h2
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B
3r22h2
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C
2h23r2
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D
3h22r2
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Solution

## The correct option is D 3h22r2At the centre of coil the magnetic field is, B1=μ0ni2r On the axis of the coil the value of magnetic field is, B2=nμ0ir22(r2+h2)3/2 Fractional decrease in magnetic field will be: ΔBB=B1−B2B1 ΔBB=1−B2B1 ΔBB=1−r3(r2+h2)3/2 ΔBB=[1−(r2)3/2(r2+h2)3/2] ΔBB=⎡⎢ ⎢ ⎢⎣1−(r2+h2r2)−32⎤⎥ ⎥ ⎥⎦ ΔBB=[1−(1+h2r2)−3/2] Since h<<r, neglecting the higher order terms in the expansion, we can write (1+h2r2)−3/2=1−32(h2r2) [(1+nx)−1=1−nx+n(n−1)2x2+.......] ⇒ΔBB=1−[1−32(h2r2)] or, ΔBB=32(h2r2) ∴ΔBB=3h22r2 Why this Question ? Note: The magnetic field due to a circular current carrying loop having n turns on an axial point is given by, Baxis=nμ0ir22(r2+x2)3/2

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