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$\begin{array}{l}{\text { A short current element } d \vec{l}=(0.500 \mathrm{mm}) \hat{\jmath} \text { carries a cur- }} \\ {\text { rent of } 8.20 \mathrm{A} \text { in the same direction as } d \vec{l} . \text { Point } P \text { is located at }} \\ {\vec{r}=(-0.730 \mathrm{m}) \hat{\imath}+(0.390 \mathrm{m}) \hat{k} . \text { Use unit vectors to express the }} \\ {\text { magnetic field at } P \text { produced by this current element. }}\end{array}$

$d \vec{B}=\frac{M_{0}}{4 \pi} \frac{q d \vec{L} \times \vec{r}}{r^{3}}$

$r=\sqrt{(-0.73)^{2}+(0.39)^{2}}=0.8267 \mathrm{m}$

$d \vec{L} \times \vec{r}=\left[0.5 * 10^{-3}\right] \mathrm{\hat j} \times[-0.73 \hat{\imath}+0.39 \hat{\mathrm{k}}]$

$=+3.65 * 10^{-4}\hat k+1.95*10^{-4}\hat i$

$d \vec{B}=\left(1 * 10^{-7}\right) \frac{8 \cdot 2}{(0.8276)^{3}}\left[3.65 * 10^{-4}\hat k+1.95 * 10^{-4}\hat j\right]$

$\rightarrow d \vec{B}=2.83*10^{-10} \hat{\imath}+5.28 * 10^{-10} \mathrm{k}$

$\begin{array}{l}{\text { A long, straight wire lies along the } z \text { -axis and carries a }} \\ {4.00 \text { -A current in the }+z \text { -direction. Find the magnetic field (magni) }} \\ {\text { tude and direction) produced at the following points by a } 0.500 \text { -mm }}\end{array}$

$\begin{array}{l}{\text { segment of the wire centered at the origin: (a) } x=2.00 \mathrm{m}, y=0 \text { , }} \\ {z=0 ;(\mathrm{b}) x=0, y=2.00 \mathrm{m}, z=0 ;(\mathrm{c}) x=2.00 \mathrm{m}, y=2.00 \mathrm{m},} \\ {z=0 ;(\mathrm{d}) x=0, y=0, z=2.00 \mathrm{m}}\end{array}$

$d \vec{B}=\frac{M_{0}}{4 \pi}$$\frac{I \vec L \times \hat r}{r^{2}}$$\longrightarrow \vec{B}=\frac{m_{0}}{4 \pi}$$\frac{I\Delta \vec L \times \hat{r}}{r^{2}}$

$=\frac{m_{0}}{4 \pi} \frac{I \Delta \vec{L} \times \vec{r}}{r{3}}$

(a) $x=2 m, y=0, z=0$

$∴ \vec{r}=2 \hat i$

$\Delta \vec{L} \times \vec{r}=$$=0.5 * 10^{-3}(2) \mathrm{\hat k x \hat i}=$$=1*10^{-3} \hat{j}$

$\rightarrow \vec{B}=$$\frac{1*10^{-7} * 4 * 1 * 10^{-3}}{2^{3}} \hat{\jmath}=$$5 * 10^{-11} \mathrm{T} \hat j$

(b) $\vec{r}=2 \hat{j}\quad,\quad r=2 m$

$\Delta \vec{L} \times \vec{r}$$=0.5*10^{-3}(2) \mathrm{\hat k x \hat j}$$=-1 * 10^{-3} \hat{i}$

$\rightarrow \vec{B}= \frac{1*10^{-7} * 4 *-1 * 10^{-3}}{2^{3}}\hat i=$$-5 * 10^{-11} \mathrm{T} \hat{\imath}$

(c) $\vec{r}=2(\hat i+\hat{j}) \quad, \quad r=\sqrt{2}(2) m$

$\Delta \vec{L} \times \vec{r}=$$\left(0.5 * 10^{-3}\right)(2) \quad \hat{k} \times(\hat i+\hat j)$$=1 * 10^{-3}(\hat{j}-\hat i)$

$\rightarrow \vec{B}=\frac{1 * 10^{-7} * 4 * 1*10^{-3}}{(\sqrt{2}(2))^{3}}$$(\hat{j}-\hat i)=\left(-1.77 * 10^{-11} T\right)(\hat i-\hat j)$

(d) $\vec{r}=2 \hat{k}, \quad r=2 m$

$\Delta \vec{\imath} \times \vec{r}=0.5 * 10^{-3}(2) \mathrm{\hat k} \times \hat{k}=0$

$\rightarrow \vec{B}=0$