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

\[B_1  = \frac{\mu_0 i}{2\pi d}\] 

Also, the magnetic field due to a section of length l on a perpendicular bisector is given by

\[B_2 = \frac{\mu_0 i}{4\pi d}\frac{2l}{\sqrt{l^2 + 4 d^2}}\]

\[\Rightarrow \frac{\mu_0 i  l}{4\pi d}\frac{2}{d\sqrt{\frac{l^2}{d^2} + 4}}\] 

\[\text{ Neglecting  } \frac{l^2}{d^2}  \left( \text{ very  small } \right),  \text{ we  get } \] 

\[ B_2  = \frac{\mu_0 i  l}{4\pi d^2} \times \frac{2}{\sqrt{2}}\] 

\[         = \frac{\sqrt{2} \mu_0 i  l}{4\pi d^2}\] 

Now,
B₁ > B₂
According to the question,

\[\frac{B_1 - B_2}{B_1} = \frac{1}{100}\]
\[ \Rightarrow B_2 = 0 . 99 B_1 \]
\[ \Rightarrow \frac{\sqrt{2} \mu_0 il}{4\pi d^2} = 0 . 99 \times \frac{\mu_0 i}{2\pi d}\]
\[ \Rightarrow \frac{d}{l}=\frac{1 . 414}{1 . 98}=0.71\]