91影视

The radius of the wire is$鈭�$.

Let鈥檚 assume the lop carrying the current and having the length .

Consider the diagram shows current carrying loop as,

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

$\mathbf{鈭�}\mathbf{B}\mathbf{脳}\mathbf{ds}\mathbf{=}{\mathbf{渭}}_{\mathbf{0}}\mathbf{I}$

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

$B=\frac{{渭}_{0}I}{2蟺s}$

Here s is the distance.

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

$B_R=2B$

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

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

The magnetic flux is given by,

$蠒={鈭�}_{鈭�}^{d-鈭�}{B}_R.da$

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

$$\begin{gathered} 蠒={鈭�}_{鈭�}^{d-鈭�}\frac{{渭}_{0}I}{蟺s}.Ids \\ 蠒\frac{{渭}_{0}Il}{蟺}{鈭�}_{鈭�}^{d-鈭�}\frac{1}{S}.ds \\ 蠒\frac{{渭}_{0}Il}{蟺}{\left(In\right)}_{鈭�}^{d-鈭�} \\ 蠒=\frac{{渭}_{0}Il}{蟺}In\left(\frac{d-鈭�}{鈭�}\right) \end{gathered}$$

Since$d鈮�鈭�$, therefore $d-鈭�鈮�d$

$蠒=\frac{{渭}_{0}Il}{蟺}In\left(\frac{d}{鈭�}\right)$ 鈥�.. (1)

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

$蠒=LI....\left(2\right)$

Equate the equation (1) and (2),

$$\begin{gathered} LI=\frac{{渭}_{0}Il}{蟺}In\left(\frac{d}{鈭�}\right) \\ L=\frac{{渭}_{0}I}{蟺}In\left(\frac{d}{鈭�}\right) \end{gathered}$$

Hence the self-inductance of the hairpin loop is$\frac{{渭}_{0}I}{蟺}In\left(\frac{d}{鈭�}\right)$.