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The current element is shown in red in Fig. 1 and points in the +r-direction (the direction of the current), so$\mathrm{d}\stackrel{→}{\mathrm{l}}=\left(1.10\mathrm{mm}\right)\hat{\mathrm{i}}$. The unit vector $\hat{r}$for each field point is directed from the current element (at D) toward that point: f is in the +y-direction for point A, at an angle 0 above the +r-direction for point B, and in the +r-direction for point C.

At point A,$\hat{r}=\hat{j}$, so

$$\begin{gathered} \stackrel{→}{\mathrm{B}}=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{\mathrm{ld}\stackrel{→}{\mathrm{l}}×\hat{\mathrm{r}}}{{\mathrm{r}}^2}=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{\mathrm{ld}\stackrel{→}{\mathrm{l}}\left(\widehat{\mathrm{ι}}\right)×\hat{\mathrm{j}}}{{\mathrm{r}}^2}=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{\mathrm{ldl}}{4\mathrm{π}{\mathrm{r}}^2}\hat{\mathrm{k}} \\ =\left({10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)\frac{\left(10.0\mathrm{A}\right)\left(1.10×{10}^{-3}\mathrm{m}\right)}{{\left(5.00×{10}^{-2}\mathrm{m}\right)}^2}\hat{\mathrm{k}} \\ =\left(4.40×{10}^{-7}\mathrm{T}\right)\hat{\mathrm{k}} \end{gathered}$$

The direction of at A is out of the x-y-plane of Fig.1

From Fig.1 the distance from D to B is

$\hat{r}=\stackrel{→}{r}/r\left[\left(14.0\right)\hat{i}+\left(5.00\right)\hat{j}\right]/\left(14.9cm\right)=0.940\hat{i}+0.336\hat{j}$From Eq

$$\begin{gathered} \stackrel{→}{\mathrm{B}}=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{\mathrm{ld}\stackrel{→}{\mathrm{l}}}{{\mathrm{r}}^2}=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{\mathrm{ldl}\left(\hat{\mathrm{i}}\right)×\left(0.940\hat{\mathrm{i}}+0.336\hat{\mathrm{j}}\right)}{{\mathrm{r}}^2} \\ =\frac{{\mathrm{μ}}_0\mathrm{l}}{4\mathrm{Ï€}}\frac{\mathrm{dl}\left(0.336\right)}{{\mathrm{r}}^2}\hat{\mathrm{k}} \\ =\left({10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)\frac{\left(10.0\mathrm{A}\right)\left(1.10Â\pm {10}^{-3}\mathrm{m}\right)\left(0.336\right)}{{\left(\sqrt{221}×{10}^{-2}\mathrm{m}\right)}^2}\hat{\mathrm{k}} \\ =\left(1.67×{10}^{-8}\mathrm{T}\right)\hat{\mathrm{k}} \end{gathered}$$

The direction of $\stackrel{→}{\mathrm{B}}$at B is out of the x-y plane of Fig.1

At C,$\hat{r}=\hat{i}$ , so the cross product$d\stackrel{→}{l}=dl\left(\hat{i}\right)×\hat{i}=0$ and$\stackrel{→}{B}=0$ thus.

The magnetic field lines caused by a straight wire are circles centered on the wire and lying in a plane perpendicular to it, and the direction of $\stackrel{→}{B}$is given by the right-hand rule; coming out of the page.$\stackrel{→}{B}$ has its greatest magnitude at points lying in the plane perpendicular to the segment length $d\stackrel{→}{l}$(in red in Fig. 1) because there 0 = 90° and sin 0 = 1; indeed, B is greater at A (where $\mathrm{θ}$= 90°) than at B (where 0 <$\mathrm{θ}$ <90°) than at C (where $\mathrm{θ}$= 0°), as the answers of step (1). (2) and (3) show.