Question

A long, straight wire carries a current $i$. Let $B_1$ 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. Let $B_2$ be the magnetic field at this point due to this section only. Find the value of $d/l$ so that $B_2$ differs from $B_1$ by $1\%$.

Answer 1

$B_1=\frac{{\mu }_{0}I}{2\pi d}$

$B_2=\frac{{\mu }_{0}I}{4\pi d}(2\times sin\theta )=\frac{{\mu }_{0}I}{4\pi d}$

$=\frac{2\times l}{2\sqrt{d^2+\frac{l^2}{4}}}=\frac{{\mu }_{0}Il}{4\pi d\sqrt{d^2+\frac{l^2}{4}}}$

$B_1-B_2=\frac{1}{100B_2}$

$\Rightarrow \frac{{\mu }_{0}I}{2\pi d}-\frac{{\mu }_{0}Il}{4\pi d\sqrt{d^2+\frac{l^2}{4}}}=\frac{{\mu }_{0}I}{200\pi d}$

$\Rightarrow \frac{{\mu }_{0}I}{4\pi d\sqrt{d^2+\frac{l^2}{4}}}=\frac{{\mu }_{0}Il}{\pi d}(\frac{1}{2}-\frac{1}{200})$

$\Rightarrow \frac{l}{d^2+\frac{l^2}{4}}={\left(\frac{99\times 4}{200}\right)}^2=\frac{156816}{40000}$

$=3.92$

$\Rightarrow l^2=3.92d^2+\frac{3.92}{4}{l}^2\left(\frac{1-3.92}{4}\right)l^2=3.92d^2$

$\Rightarrow 0.02l^2=3.92d^2$

$\Rightarrow \frac{d^2}{l^2}=\frac{0.02}{3.92}=\frac{d}{l}=\sqrt{\frac{0.02}{3.92}}=0.07$