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 _{2} and \alpha _{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} =dl \ \overrightarrow{a_{z}} .

\overrightarrow{R} = -Z \overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }} .

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

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

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

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

as Z_{1} = \rho \ \cot \alpha _{1}   ,  Z_{2} = \rho \ \cot \alpha _{2}  and  dz = - \rho \ cosec^{2} \alpha \ d\alpha  .

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

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

\overrightarrow{H} =\frac{I}{4\pi\ \rho } \left \overrightarrow{a_{\phi }} .

Case 1 :- 

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

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

then  \overrightarrow{H} = \frac{I}{4\pi \ \rho } \overrightarrow{a_{\phi }} .

Case 2:-

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

\overrightarrow{H} = \frac{I}{2\pi \ \rho } \overrightarrow{a_{\phi }} .

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Author: Lakshmi Prasanna Ponnala

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.

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