8

\vec{dB} = \frac{\vec{M}_0}{4\pi} \frac{\vec{I}_1}{\vec{r}^2} \vec{u}_T \times \vec{u}_r \ dC

Vavdindsystem:

Cylindriskt, origo i P.

$\vec{z}$ innat i papperet

a)

B-fält från kvartseinkel.

\hat{u}_r = -\hat{r} \quad \hat{u}_T = -\hat{p} \quad dC = r_1 \, dp\vec{dB} = \frac{\vec{M}_0}{4\pi} \frac{\vec{I}_1}{\vec{r}_2} (-\hat{p}) \times (-\hat{r}) r_1 \, dp = \frac{\vec{M}_0}{4\pi} \frac{\vec{I}_1}{\vec{r}_2} \hat{z} \, dp \vec{B}_1 = \int \vec{dB} = \int_0^{\pi/2} \frac{\vec{M}_0}{4\pi} \vec{I}_1 \hat{z} \, \frac{d\varphi}{r_1} = \frac{\vec{M}_0}{4\pi} \frac{\vec{I}_1 \hat{z}}{\vec{r}_2} \int_0^{\pi/2} \hat{z} \, dp = \frac{\vec{M}_0}{4\pi} \frac{\vec{I}_1 \hat{\pi}}{2 \vec{r}_1} \hat{z}

På samma sätt från andra kvartseinkeln

\vec{B}_2 = -\frac{\vec{M}_0}{4\pi} \frac{\vec{I}_1 \hat{\pi}}{2 \vec{r}_2} \hat{z}

Raka sektioner:

\hat{u}_T \parallel \hat{u}_r \Rightarrow \hat{u}_T \times \hat{u}_r = \vec{0} \Rightarrow \vec{B} = \vec{0}

Totala $\vec{B}$ -fältet: $\vec{B} = \frac{\vec{M}_0 \vec{I}_1}{\vec{B}} \left( \frac{1}{\vec{r}_1} - \frac{1}{\vec{r}_2} \right) \hat{z}$

$\vec{B} = \frac{\vec{M}_0 \vec{I}_1}{\vec{B}} \left( \frac{1}{\vec{r}_1} - \frac{1}{\vec{r}_2} \right) \hat{z}$