Question

A straight wire carries a 10.0-A current (Fig. ). ABCDis a rectangle with point D n the middle of a 1.10-m segment of the wire and point C in the wire. Find the magnitude and direction of the magnetic field due to this segment at

(a) point A;

(b) point B;

(c) point C.

Short answer

1. $\left(4.40\times {10}^{-7}\mathrm{T}\right)\hat{\mathrm{k}}$
   
2. $\left(1.67\times {10}^{-8}\mathrm{T}\right)\hat{\mathrm{k}}$
   
3. 0 (zero).

Step-by-step solution

Solving part (a) of the problem.

The current element is shown in red in Fig. 1 and points in the +r-direction (the direction of the current), so$\mathrm{d}\overrightarrow{\mathrm{l}}=\left(1.10\mathrm{mm}\right)\hat{\mathrm{i}}$. 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

$$\begin{gathered} \overrightarrow{\mathrm{B}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{ld}\overrightarrow{\mathrm{l}}\times \hat{\mathrm{r}}}{{\mathrm{r}}^2}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{ld}\overrightarrow{\mathrm{l}}\left(\hat{\mathrm{\iota }}\right)\times \hat{\mathrm{j}}}{{\mathrm{r}}^2}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{ldl}}{4\mathrm{\pi }{\mathrm{r}}^2}\hat{\mathrm{k}} \\ =\left({10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)\frac{\left(10.0\mathrm{A}\right)\left(1.10\times {10}^{-3}\mathrm{m}\right)}{{\left(5.00\times {10}^{-2}\mathrm{m}\right)}^2}\hat{\mathrm{k}} \\ =\left(4.40\times {10}^{-7}\mathrm{T}\right)\hat{\mathrm{k}} \end{gathered}$$

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

Solving part (b) of the problem.

From Fig.1 the distance from D to B is

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

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

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

Solving part (c) of the problem.

At C,$\hat{r}=\hat{i}$ , so the cross product$d\overrightarrow{l}=dl\left(\hat{i}\right)\times \hat{i}=0$ and$\overrightarrow{B}=0$ 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 $\overrightarrow{B}$is given by the right-hand rule; coming out of the page.$\overrightarrow{B}$ has its greatest magnitude at points lying in the plane perpendicular to the segment length $d\overrightarrow{l}$(in red in Fig. 1) because there 0 = 90° and sin 0 = 1; indeed, B is greater at A (where $\theta$= 90°) than at B (where 0 <$\theta$ <90°) than at C (where $\theta$= 0°), as the answers of step (1). (2) and (3) show.