Question

A long, straight wire lies along the z-axis and carries a 4.00-A current in the +z-direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.500-mm segment of the wire cantered at the origin: (a) x = 2.00 m, y = 0, z = 0; (b) x = 0, y = 2.00 m, z = 0; (c) x = 2.00 m, y = 2.00 m, z = 0; (d) x = 0, y = 0, z = 2.00 m,

Short answer

(a) the magnetic field (magnitude and direction) produced at the following points by a 0.500 - mm segment of the wire cantered at the origin $\mathrm{x}=2.00\mathrm{m},\mathrm{y}=0,\mathrm{z}=0\mathrm{is}\left(50.0\mathrm{pT}\right)\hat{\mathrm{i}}$

(b) the magnetic field (magnitude and direction) produced at the following points by a 0.500-mm segment of the wire cantered at the origin $\mathrm{x}=0,\mathrm{y}=0,2.00\mathrm{m},\mathrm{z}=0\mathrm{is}\left(50.0\mathrm{pT}\right)\hat{\mathrm{i}}$

(c) the magnetic field (magnitude and direction) produced at the following points by a 0.500 - mm segment of the wire cantered at the origin $\mathrm{x}=2.00\mathrm{m},\mathrm{y}=0,2.00\mathrm{m},\mathrm{z}=0\mathrm{is}\left(17.7\mathrm{pT}\right)\left(-\hat{\mathrm{i}}+\hat{\mathrm{j}}\right)$

(d) the magnetic field (magnitude and direction) produced at the following points by a segment of the wire cantered at the origin$\mathrm{x}=0,\mathrm{y}=0,\mathrm{z}=2.00\mathrm{m}\mathrm{is}0$

Step-by-step solution

Direction of at point in the - x direction

Given the field point having co-ordinates (2.00,0,0)m$so\hat{r}=\left(2.00m\right)\hat{i}$ and

$$\begin{gathered} \hat{B}=\frac{{\mu }_0}{4\pi }\frac{ld\hat{l}\times \hat{r}}{r^2} \\ =\frac{{\mu }_0}{4\pi }ldl{r}^2\hat{j} \\ =\left({10}^{-7}T.m/A\right)\frac{\left(4.00A\right)\left(0.500\times {10}^{-3}m\right)}{2.00m{)}^2\hat{j}} \\ =\left(50.0pT\right)\hat{j} \end{gathered}$$

The direction of at the point in the direction is

For the points (0,2.00,0)m

For the field point has the coordinates $\left(0,2.00,0\right)mso\hat{r}=\left(2.00m\right)\hat{j}$

Now, $$\begin{gathered} \hat{B}=\frac{{\mu }_0}{4\pi }\frac{ld\hat{l}}{r^2} \\ =-\frac{{\mu }_0}{4\pi }\frac{lD\hat{l}}{r^2}\hat{l} \end{gathered}$$

$$\begin{gathered} =-\left({10}^{-7}T.m/A\right)\frac{4.00A\left(0.500\times {10}^{-3}m\right)}{{\left(2.00\right)}^2}\hat{i} \\ =-\left(50.0pT\right)\hat{i} \end{gathered}$$

The direction of $\hat{B}$at this point is in the - x direction

For xy-planes

For the field point the radial position

$$\begin{gathered} \hat{\mathrm{r}}=\left(2\mathrm{m}\right)\left(\hat{\mathrm{i}}+\mathrm{j}\right){\mathrm{andr}}^2=8.00{\mathrm{m}}^2 \\ \hat{\mathrm{B}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{ld}\hat{\mathrm{l}}\times \hat{\mathrm{r}}}{{\mathrm{r}}^2} \\ =\frac{\left({10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)\left(4.00\mathrm{A}\right)\left(0.500\times {10}^{-3}\mathrm{m}\right)}{8.00{\mathrm{m}}^2}\left(\widehat{-\mathrm{i}}+\mathrm{j}\right) \\ =\left(17.7\mathrm{pT}\right)\left(-\hat{\mathrm{i}}+\hat{\mathrm{j}}\right)) \end{gathered}$$

The vector$\hat{B}$ lies in the xy planes and makes an angle of$45.0°$ with the -x direction.

The field point has the coordinates (0,02.00) m , hence the cross products

$$\begin{gathered} \mathrm{d}\hat{\mathrm{l}}\times \hat{\mathrm{r}}=\mathrm{dl}\left(\hat{\mathrm{k}}\right)\times \hat{\mathrm{k}} \\ =0\mathrm{and}\mathrm{thus}\hat{\mathrm{B}} \\ =0 \end{gathered}$$