Exemplos
I. Fio levando uma corrente I uniforme
• d\ell x \hat{x} = sen \alpha d\ell \hat{z}
= sen(\frac{\pi}{2} + \theta) d\ell \hat{z}
= cos \theta d\ell \hat{z}
= \frac{s}{\sqrt{s^{2}+l^{2}}} d\ell \hat{z}
Assim,
\mathbf{B} = \frac{\mu_{0} I}{4\pi} \int \frac{s}{\sqrt{s^{2}+l^{2}}} d\ell = \frac{\mu_{0} I s}{4\pi} \int \frac{d\ell}{(s^{2}+l^{2})^{3/2}} \hat{z}
\Gamma por substituição trigonométrica:
s = s tan \theta
d\ell = s sec^{2} \theta d\theta
\Rightarrow \int \frac{d\ell}{(s^{2}+l^{2})^{3/2}} = \int \frac{s sec^{2}\theta d\theta}{s^{3} sec^{3}\theta} = \frac{1}{s^{2}} \int d\theta = sen \frac{s}{2}
Aproveitando os ângulos da figura:
\mathbf{B} = \frac{\mu_{0} I}{4\pi} sen \theta_{2} \hat{z} = \frac{\mu_{0} I}{4\pi} (sen \theta_{2} - sen \theta_{1}) \hat{z}
Para o fio infinito \theta_{1} = -\frac{\pi}{12}, \theta_{2} = \frac{\pi}{2}, temos.
\mathbf{B} = \frac{\mu_{0} I}{2\pi s} \hat{z}
