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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

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Solution

The correct option is **D** 3h22r2

At 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

At 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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