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 $\frac{x}{2} \frac{h^{2}}{r^{2}}$ . The value of $x$ is ______

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Detailed Solution

Field at the centre is $B_{1} = \frac{u_{0}}{4 π} \times \frac{2 π I n}{r} = \frac{u_{0}}{2} \frac{n I}{r}$
Field at a distance ‘L’ from the centre is $B_{2} = \frac{u_{0}}{4 π} \times \frac{2 π n I r^{2}}{{(r^{2} + h^{2})}^{3 / 2}} =$
$∴ B_{2} = \frac{u_{0}}{2} \times \frac{n I r^{2}}{r^{3} {(1 + \frac{h^{2}}{r^{2}})}^{3 / 2}} B_{2} = B_{1} {(1 + \frac{h^{2}}{r^{2}})}^{- 3 / 2}$

$∴ B_{2} = B_{1} {(1 - \frac{3}{2} \frac{h^{2}}{r^{2}})}^{}$ (by binomial theorem)
Hence B₂ is less than B₁ by a fraction = $\frac{3}{2} \frac{h^{2}}{r^{2}}$