A straight wire piece of length x carries a current i . A point P...

The magnetic field at a point on the perpendicular bisector of a finite straight wire of length $x$ carrying current $i$ at a distance $d$ is given by Biot-Savart law:

$B = \dfrac{\mu_0 i}{4\pi d} (\sin\theta_1 + \sin\theta_2)$

Since the point is on the perpendicular bisector, $\theta_1 = \theta_2 = \theta$. From the geometry of the figure, $\sin\theta = \dfrac{x/2}{\sqrt{d^2 + \dfrac{x^2}{4}}}$.

Substituting this into the magnetic field expression:

B = \dfrac{\mu_0 i}{4\pi d} \left(\dfrac{x}{\sqrt{d^2 + \dfrac{x^2}{4}}} \right)

For the limit $d \gg x$, the term \dfrac{x^2}{4} can be neglected compared to $d^2$. Thus, \sqrt{d^2 + \dfrac{x^2}{4}} \approx$d$.

$B \approx$\dfrac{\mu_0 i x}{4\pi d^2}

Therefore, $B \propto$\dfrac{1}{d^2} for $d \gg x$.

For the limit $d \ll x$, the term $d^2$ can be neglected compared to \dfrac{x^2}{4}. Thus, \sqrt{d^2 + \dfrac{x^2}{4}} \approx \dfrac{x}{2}.

$B \approx$\dfrac{\mu_0 i x}{4\pi d \left(\dfrac{x}{2}\right)} = \dfrac{\mu_0 i}{2\pi d}

Therefore, $B \propto$\dfrac{1}{d} for $d \ll x$.