A -4.80@mC charge is moving at a constant speed of 6.80 * 105 m>s in the +x-direction relative to a reference frame. At the instant when the point charge is at the origin, what is the magnetic-field vector it produces at the following points:

(a) x= 0.500 m, y= 0, z= 0;

(b) x= 0, y= 0.500 m, z= 0;

(c) x= 0.500 m, y= 0.500 m, z= 0;

(d) x= 0, y= 0, z= 0.500 m?

Consider a charge (q=-4.80) which is moving at a constant speed of$v=6.80\times {10}^{5}m/s$ in the +x direction relative to a reference frame. The magnetic field produced by the moving charge is given by,

$\overrightarrow{B}=\frac{{\mu }_o}{4\pi }\frac{q\overrightarrow{v}\times \hat{r}}{r^2}$ (1)

We need to find the magnetic field vector that produces at the point = 0.500 m, y =0, z = 0 at the instant when the point charge is at the origin. In this ca have $\overrightarrow{v}=v\widehat{i},\overrightarrow{r}=r\widehat{i}$so,

$\overrightarrow{v}\times \overrightarrow{r}=0$

thus,

$\overrightarrow{B}=0$

Now we need to find the magnetic field vector that produces at the point z = 0,y = 0.500 m, and z =0 at the instant when the point charge is at the origin. In this $\overrightarrow{v}=v\widehat{i},\overrightarrow{r}=r\widehat{j}$so,

$\overrightarrow{v}\times \overrightarrow{r}=v\cdot \widehat{k}$

where $r=\sqrt{x^2+y^2+z^2}=y=0.500$Substitute into (1) we get,

$\begin{array}{c}\overrightarrow{B}=\left(\frac{{\mu }_0}{4\pi }\right)\frac{qv}{r^2}\widehat{k}\\ =-\frac{\left(1.0\times {10}^{-7}\mathrm{N}\cdot {\mathrm{s}}^2/{\mathrm{C}}^2\right)\left(4.80\times {10}^{-6}\mathrm{C}\right)\left(6.80\times {10}^5\mathrm{m}/\mathrm{s}\right)}{(0.500\mathrm{m}{)}^2}\widehat{k}\\ =-\left(1.31\times {10}^{-6}\mathrm{T}\right)\widehat{k}\\ \overrightarrow{B}=-\left(1.31\times {10}^{-6}\mathrm{T}\right)\widehat{k}\end{array}$

Now we need to find the magnetic field vector that produces at the point z= 0.500 m, y = 0.500 m, and z = 0 at the instant when the point charge is at the origin. In this case, we have $\overrightarrow{v}=v\widehat{i},\overrightarrow{r}=\left(0.500\right)(\widehat{i}+\widehat{j})$so,

role="math" localid="1668229000516" $\overrightarrow{v}\times \overrightarrow{r}=\left(0.500\right)v\widehat{k}$

where m $r=\sqrt{x^2+y^2+z^2}=\sqrt{x^2+y^2}=0.7071\mathrm{m}$Substitute into (1) we get

$\begin{array}{c}\overrightarrow{B}=\left(\frac{{\mu }_0}{4\pi }\right)\frac{q\left(0.500\right)v}{r^3}\widehat{k}\\ =-\frac{\left(1.0\times {10}^{-7}\mathrm{N}\cdot \frac{{\mathrm{s}}^2}{{\mathrm{C}}^2}\right)\left(4.80\times {10}^{-6}\mathrm{C}\right)\left(0.500\right)\left(6.80\times {10}^5\mathrm{m}/\mathrm{s}\right)}{(0.500\mathrm{m}{)}^3}\widehat{k}\\ =-\left(4.62\times {10}^{-7}\mathrm{T}\right)\widehat{k}\\ \overrightarrow{B}=-\left(4.62\times {10}^{-7}\mathrm{T}\right)\widehat{k}\end{array}$

Finally, we need to find the magnetic field vector that produces at the point r = 0, y =0, z =0.500 m at the instant when the point charge is at the origin. In this case, we have $\overrightarrow{v}=v\widehat{i},\overrightarrow{r}=r\widehat{k}$ so,

$\overrightarrow{v}\times \overrightarrow{r}=-v\cdot \widehat{k}$

thus,

$\begin{array}{c}\overrightarrow{B}=\left(\frac{{\mu }_0}{4\pi }\right)\frac{q\left(0.500\right)v}{r^2}\widehat{k}\\ =-\frac{\left(1.0\times {10}^{-7}\mathrm{N}\cdot {\mathrm{s}}^2/{\mathrm{C}}^2\right)\left(4.80\times {10}^{-6}\mathrm{C}\right)\left(6.80\times {10}^5\mathrm{m}/\mathrm{s}\right)}{(0.500\mathrm{m}{)}^2}\widehat{k}\\ =-\left(1.31\times {10}^{-6}\mathrm{T}\right)\widehat{k}\\ \overrightarrow{B}=-\left(1.31\times {10}^{-6}\mathrm{T}\right)\widehat{k}\end{array}$