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(B) For a finite wire segment of length $L$ carrying current $I$, the magnetic field $B$ at a distance $r$ on its perpendicular bisector is given by t

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decreases as $\frac{1}{r^{2}}$

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(B) For a finite wire segment of length $L$ carrying current $I$, the magnetic field $B$ at a distance $r$ on its perpendicular bisector is given by the Biot-Savart law integration:$B = \frac{\mu_{0} I}{4 \pi r} (\sin 	heta_{1} + \sin 	heta_{2})$For a segment of length $L$, $\sin 	heta_{1} = \sin 	heta_{2} = \frac{L/2}{\sqrt{r^{2} + (L/2)^{2}}}$.Substituting this, we get $B = \frac{\mu_{0} I}{4 \pi r} \cdot \frac{L}{\sqrt{r^{2} + (L/2)^{2}}}$.Since $r >> L$, the denominator $\sqrt{r^{2} + (L/2)^{2}} \approx r$.Thus, $B \approx \frac{\mu_{0} I L}{4 \pi r^{2}}$.Therefore, the magnetic field decreases as $\frac{1}{r^{2}}$.

The magnetic field due to a current in a straight wire segment of length $L$ at a point on its perpendicular bisector at a distance $r$ $(r >> L)$ is:

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