91Ó°ÊÓ

We have given, a current carrying wire as shown in the above diagram.

We have to find the line integral of the magnetic field along the current carrying wire.

Using ampere's law, for any path is given by,

${∫}_i^{f}B.\stackrel{→}{dS}={\mathrm{μ}}_{0}I$

Where I is a total current in the magnetic field curve.

Since two straight component of wire is perpendicular to the magnetic field then,

${∫}_i^f\stackrel{→}{B}.\stackrel{→}{dS}=\mathrm{B}.\mathrm{dS}\cos {\left(90\right)}^0=0$

We need to find the integral of magnetic field only along the curve path.

$$\begin{gathered} {∫}_i^{f}B.\stackrel{→}{dS}=\mathrm{B}{∫}_{\mathrm{i}}^{\mathrm{f}}\mathrm{dS} \\ {∫}_i^{f}B.\stackrel{→}{dS}=\mathrm{B}.\mathrm{L} \\ {∫}_i^{f}B.\stackrel{→}{dS}=\mathrm{Ï€¸é\mu þ} \\ {∫}_i^{f}B.\stackrel{→}{dS}=\mathrm{Ï€}×(0.01\mathrm{m})×\mathrm{B}............................\left(1\right) \end{gathered}$$

we know the magnetic field due to half circular ring at its center using Biot's savert law.

$$\begin{gathered} \mathrm{B}=\frac{{\mathrm{μ}}_0\mathrm{i}}{2\mathrm{Ï€¸é}} \\ \mathrm{B}=\frac{(4\mathrm{Ï€}×{10}^{-7})×2\mathrm{A}}{2\mathrm{Ï€}×(0.01\mathrm{m})} \\ \mathrm{B}=4×{10}^{-5}\mathrm{T} \end{gathered}$$

put this value in equation (1),

$$\begin{gathered} {∫}_i^{f}B.\stackrel{→}{dS}=\mathrm{Ï€}×(0.01\mathrm{m})×\mathrm{B} \\ {∫}_i^{f}B.\stackrel{→}{dS}=\mathrm{Ï€}×(0.01\mathrm{m})×(4×{10}^{-5})\mathrm{T} \\ {∫}_i^{f}B.\stackrel{→}{dS}=12.56×{10}^{-7}\mathrm{T}.\mathrm{m} \\ {∫}_i^{f}B.\stackrel{→}{dS}=1.25\mathrm{μ°Õ}.\mathrm{m} \end{gathered}$$