# H due to finite long straight conuctor

Consider a conductor of finite length placed along z-axis as shown in the figure

The conductor has a finite length  AB , where A and B are located at distances Z1 and Z2 above the origin with it’s upper and lower ends respectively subtending angles $\alpha&space;_{2}$ and $\alpha&space;_{1}$ at P.

P is the point at which  $\overrightarrow{H}$   is to be determined.

Consider a differential element $\overrightarrow{dl}$  along the Z-axis at a distance Z from the origin.

where $\overrightarrow{dl}&space;=dl&space;\&space;\overrightarrow{a_{z}}$ .

$\overrightarrow{R}&space;=&space;-Z&space;\overrightarrow{a_{z}}+\rho&space;\overrightarrow{a_{\rho&space;}}$ .

$\widehat{a_{R}}&space;=&space;\frac{\overrightarrow{R}&space;=&space;-Z&space;\overrightarrow{a_{z}}+\rho&space;\overrightarrow{a_{\rho&space;}}}{\sqrt{(Z^{2}+\rho&space;^{2})}}$ .

$\overrightarrow{dl}&space;X&space;\&space;\widehat{a_{R}}&space;=&space;\frac{\rho&space;dz&space;\&space;\overrightarrow{a_{\phi&space;}}}{\sqrt{(Z^{2}+\rho&space;^{2})}}$ .

$\overrightarrow{H}&space;=&space;\oint&space;\frac{&space;I&space;\&space;\rho&space;\&space;dz&space;\&space;\overrightarrow{a_{\phi&space;}}}{4\pi&space;(Z^{2}+\rho&space;^{2})^{\frac{3}{2}}}$ .

$\overrightarrow{H}&space;=&space;\oint_{Z_{1}}^{Z_{2}}&space;\frac{&space;I&space;\&space;\rho&space;\&space;dz&space;\&space;\overrightarrow{a_{\phi&space;}}}{4\pi&space;(Z^{2}+\rho&space;^{2})^{\frac{3}{2}}}$ .

as $z=\rho&space;\&space;cot\alpha$    ,  $Z_{2}&space;=&space;\rho&space;\&space;\cot&space;\alpha&space;_{2}$  and  $dz&space;=&space;-&space;\rho&space;\&space;cosec^{2}&space;\alpha&space;\&space;d\alpha$  .

$\overrightarrow{H}&space;=\frac{-I}{4\pi&space;}&space;\int_{\alpha&space;_{1}}^{\alpha&space;_{2}}&space;\frac{\rho&space;^{2}&space;\&space;cosec^{2}\alpha&space;\&space;d\alpha&space;\&space;\overrightarrow{a_{\phi&space;}}}{&space;(\rho&space;^{2}+\rho&space;^{2}&space;\cot&space;^{2}\alpha&space;)^{\frac{3}{2}}}$ .

$\overrightarrow{H}&space;=\frac{-I}{4\pi\&space;\rho&space;}&space;\int_{\alpha&space;_{1}}^{\alpha&space;_{2}}&space;\rho&space;^{2}&space;\&space;\sin&space;\alpha&space;\&space;d\alpha&space;\&space;\overrightarrow{a_{\phi&space;}}$ .

$\overrightarrow{H}&space;=\frac{I}{4\pi\&space;\rho&space;}&space;\left&space;[&space;\cos&space;\alpha&space;_{2}&space;-\cos&space;\alpha&space;_{1}\right&space;]\overrightarrow{a_{\phi&space;}}$ .

Case 1 :-

when the conductor is semi-finite   that is A is located at origin and B at $\infty$ .

i.e,    $Z_{1}&space;=0&space;\&space;\Rightarrow&space;\&space;\alpha&space;_{1}&space;=&space;90^{o}$     and    $Z_{2}&space;=\infty&space;\&space;\Rightarrow&space;\&space;\alpha&space;_{2}&space;=&space;0^{o}$ .

then  $\overrightarrow{H}&space;=&space;\frac{I}{4\pi&space;\&space;\rho&space;}&space;\overrightarrow{a_{\phi&space;}}$ .

Case 2:-

when conductor is infinite in length   A is at  $-\infty$  and B at $\infty$  implies  $Z_{1}&space;=&space;-\infty&space;\Rightarrow&space;\alpha&space;_{1}&space;=&space;180&space;^{o}$  and  $Z_{2}&space;=&space;\infty&space;\Rightarrow&space;\alpha&space;_{2}&space;=&space;0&space;^{o}$ .

$\overrightarrow{H}&space;=&space;\frac{I}{2\pi&space;\&space;\rho&space;}&space;\overrightarrow{a_{\phi&space;}}$ .

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