Try to compute the self-inductance of the "hairpin" loop shown in Fig. 7.38. (Neglect the contribution from the ends; most of the flux comes from the long straight section.) You'll run into a snag that is characteristic of many self-inductance calculations. To get a definite answer, assume the wire has a tiny radius$\mathbf{\in }$, and ignore any flux through the wire itself.

The radius of the wire is$\in$.

Let’s assume the lop carrying the current and having the length .

Consider the diagram shows current carrying loop as,

According to the Ampere’s law, the magnetic field along a closed loop is given by,

$\mathbf{\oint }\mathbf{B}\mathbf{\times }\mathbf{ds}\mathbf{=}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{I}$

The expression for the magnetic field due to the wire is given as,

$B=\frac{{\mu }_{0}I}{2\pi s}$

Here s is the distance.

The resultant field due to both the wire is given by,

$B_R=2B$

Substitute $\frac{{\mu }_{0}I}{2\pi s}$for B into above equation.

role="math" localid="1658138349161" $$\begin{gathered} B_R=2\frac{{\mu }_{0}I}{2\pi s} \\ {B}_R=\frac{{\mu }_{0}I}{\pi s} \end{gathered}$$

The magnetic flux is given by,

$\varphi ={\int }_{\in }^{d-\in }{B}_R.da$

Substitute $\frac{{\mu }_{0}I}{\pi s}for{B}_{R}andIdsforda$into above equation.

$$\begin{gathered} \varphi ={\int }_{\in }^{d-\in }\frac{{\mu }_{0}I}{\pi s}.Ids \\ \varphi \frac{{\mu }_{0}Il}{\pi }{\int }_{\in }^{d-\in }\frac{1}{S}.ds \\ \varphi \frac{{\mu }_{0}Il}{\pi }{\left(In\right)}_{\in }^{d-\in } \\ \varphi =\frac{{\mu }_{0}Il}{\pi }In\left(\frac{d-\in }{\in }\right) \end{gathered}$$

Since$d\gg \in$, therefore $d-\in \approx d$

$\varphi =\frac{{\mu }_{0}Il}{\pi }In\left(\frac{d}{\in }\right)$ ….. (1)

The magnetic flux in term of inductance and current is given by,

$\varphi =LI....\left(2\right)$

Equate the equation (1) and (2),

$$\begin{gathered} LI=\frac{{\mu }_{0}Il}{\pi }In\left(\frac{d}{\in }\right) \\ L=\frac{{\mu }_{0}I}{\pi }In\left(\frac{d}{\in }\right) \end{gathered}$$

Hence the self-inductance of the hairpin loop is$\frac{{\mu }_{0}I}{\pi }In\left(\frac{d}{\in }\right)$.