Question

A long, straight wire carries a current i. Let B₁ be the magnetic field at a point P at a distance d from the wire. Consider a section of length l of this wire such that the point P lies on a perpendicular bisector of the section B₂ be the magnetic field at this point due to this second only. Find the value of d/l so that B₂ differs from B₁ by 1%.

Answer 1

Given:
Magnitude of current = i
Separation of the point from the wire = d


Thus, the magnetic field due to current in the long wire is given by
$B_1=\frac{{\mu }_{0}i}{2\mathrm{\pi }d}$
Also, the magnetic field due to a section of length l on a perpendicular bisector is given by
$B_2=\frac{{\mu }_{0}i}{4\pi d}\frac{2l}{\sqrt{l^2+4d^2}}$

$$\begin{gathered} \Rightarrow \frac{{\mu }_{0}i\mathit{}l}{4\mathrm{\pi }d}\frac{2}{d\sqrt{{\displaystyle \frac{l^2}{d^2}}+4}} \\ \mathrm{Neglecting}\frac{l^2}{d^2}\left(\mathrm{very}\mathrm{small}\right),\mathrm{we}\mathrm{get} \\ {B}_2=\frac{{\mu }_{0}i\mathit{}l}{4\mathrm{\pi }{d}^2}\times \frac{2}{\sqrt{2}} \\ =\frac{\sqrt{2}{\mu }_{0}i\mathit{}l}{4\mathrm{\pi }{d}^2} \end{gathered}$$
Now,
B₁ > B₂
According to the question,
$$\begin{gathered} \frac{B_1\mathit{-}{B}_2}{B_1}\mathit{=}\frac{1}{100} \\ \Rightarrow B_2=0.99B_1 \\ \Rightarrow \frac{\sqrt{2}{\mu }_{0}il}{4\pi d^2}=0.99\times \frac{{\mu }_{0}i}{2\pi d} \\ \Rightarrow \frac{d}{l}=\frac{1.414}{1.98}=0.71 \end{gathered}$$