91Ó°ÊÓ

The radius of the wire is$∈$.

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,

$\mathbf{âˆ\circledR }\mathbf{B}\mathbf{×}\mathbf{ds}\mathbf{=}{\mathbf{μ}}_{\mathbf{0}}\mathbf{I}$

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

$B=\frac{{\mathrm{μ}}_{0}I}{2\mathrm{π}s}$

Here s is the distance.

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

$B_R=2B$

Substitute $\frac{{\mathrm{μ}}_{0}I}{2\mathrm{π}s}$for B into above equation.

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

The magnetic flux is given by,

$\mathrm{ϕ}={∫}_{∈}^{d-∈}{B}_R.da$

Substitute $\frac{{\mathrm{μ}}_{0}I}{\mathrm{π}s}for{B}_{R}andIdsforda$into above equation.

$$\begin{gathered} \mathrm{ϕ}={∫}_{∈}^{d-∈}\frac{{\mathrm{μ}}_{0}I}{\mathrm{π}s}.Ids \\ \mathrm{ϕ}\frac{{\mathrm{μ}}_{0}Il}{\mathrm{π}}{∫}_{∈}^{d-∈}\frac{1}{S}.ds \\ \mathrm{ϕ}\frac{{\mathrm{μ}}_{0}Il}{\mathrm{π}}{\left(In\right)}_{∈}^{d-∈} \\ \mathrm{ϕ}=\frac{{\mathrm{μ}}_{0}Il}{\mathrm{π}}In\left(\frac{d-∈}{∈}\right) \end{gathered}$$

Since$d≫∈$, therefore $d-∈≈d$

$\mathrm{ϕ}=\frac{{\mathrm{μ}}_{0}Il}{\mathrm{π}}In\left(\frac{d}{∈}\right)$ ….. (1)

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

$\mathrm{Ï•}=LI....\left(2\right)$

Equate the equation (1) and (2),

$$\begin{gathered} LI=\frac{{\mathrm{μ}}_{0}Il}{\mathrm{π}}In\left(\frac{d}{∈}\right) \\ L=\frac{{\mathrm{μ}}_{0}I}{\mathrm{π}}In\left(\frac{d}{∈}\right) \end{gathered}$$

Hence the self-inductance of the hairpin loop is$\frac{{\mathrm{μ}}_{0}I}{\mathrm{π}}In\left(\frac{d}{∈}\right)$.