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 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%.

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Solution

$B_{1} = \frac{μ_{0} i}{2 π d}$ and $B_{2} = \frac{μ_{0} i l}{4 π d^{2}} \times \frac{2}{\sqrt{2}}$

Given $B_{2} - B_{1} = B_{1} \times \frac{1}{100}$

$\Rightarrow B_{2} = 1.01 B_{2}$

$\Rightarrow \frac{\sqrt{2} μ_{0} i l}{4 π d^{2}} = (1.01) \frac{μ_{0} i}{2 π d}$

$\frac{d}{l} = \frac{1.414}{2.02} = 0.7$

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