A +6.00µC point charge is moving at a constant 8.00 * 106 m/s in the +y-direction, relative to a reference frame. At the instant when the point charge is at the origin of this reference frame, what is the magnetic-field vector BS 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, y= 0, z= +0.500 m;

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

Consider a charge q=+6.00 , it is moving at a constant velocity in the +y direction, relative to a reference frame.

First, we need to find the magnetic-field vector that this charge produces at z= 0.500 m, y= 0, and z = 0, at the instant when the point charge is at the origin of this reference frame. The magnetic field is given by.

$\overrightarrow{\mathrm{B}}=\frac{{\mu }_{o}q\overrightarrow{v}\times \hat{r}}{4\pi r^2}$

but $\hat{r}=\overrightarrow{r}/r$where $\overrightarrow{r}$is the vector from the charge to the point where the field is calculated. In our case , r = 0.500 m and$\overrightarrow{\mathrm{v}}=\left(8.00\times {10}^6\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}$, so,

$\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{r}}=\mathrm{vr}\overrightarrow{\mathrm{j}}\times \hat{\mathrm{i}}=-\mathrm{vr}\hat{\mathrm{k}}$

Thus,

$$\begin{gathered} \overrightarrow{\mathrm{B}}=-\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{qv}}{{\mathrm{r}}^2}\hat{\mathrm{k}} \\ =-\left(1\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)\frac{\left(6.00\times {10}^{-6}\mathrm{C}\right)\left(8.00\times {10}^6\mathrm{m}/\mathrm{s}\right)}{{\left(0.500\mathrm{m}\right)}^2}\hat{\mathrm{k}} \\ =-\left(1.92\times {10}^{-5}\mathrm{T}\right)\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{B}}=-\left(1.92\times {10}^{-5}\mathrm{T}\right)\hat{\mathrm{k}} \end{gathered}$$

Now at x=0, y =-0.500 m and z=0, we have ,$\overrightarrow{\mathrm{r}}=\left(0.500\mathrm{m}\right)\hat{\mathrm{j}},\mathrm{r}=0.500\mathrm{m}$

$\overrightarrow{v}\times \overrightarrow{\mathrm{r}}=v\hat{j}\times \hat{j}=0$

Thus,

$\overrightarrow{B}=0$

Now at z = 0, y = 0 and z = +0.500 m, we have ,$\overrightarrow{\mathrm{r}}=\left(0.500\mathrm{m}\right)\hat{\mathrm{k}},\mathrm{r}=0.500\mathrm{m}$

$\overrightarrow{v}\times \overrightarrow{\mathrm{r}}=v\hat{j}\times \hat{k}=-vr\hat{i}$

Thus,

$$\begin{gathered} \overrightarrow{\mathrm{B}}=-\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{qv}}{{\mathrm{r}}^2}\hat{\mathrm{k}} \\ =-\left(1\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)\frac{\left(6.00\times {10}^{-6}\mathrm{C}\right)\left(8.00\times {10}^6\mathrm{m}/\mathrm{s}\right)}{{\left(0.500\mathrm{m}\right)}^2}\hat{\mathrm{i}} \\ =\left(1.92\times {10}^{-5}\mathrm{T}\right)\hat{\mathrm{i}} \\ \overrightarrow{\mathrm{B}}=\left(1.92\times {10}^{-5}\mathrm{T}\right)\hat{\mathrm{i}} \end{gathered}$$

Finally at z = 0, y = -0.500 m and z=+0.500 m, so we have, $\overrightarrow{\mathrm{r}}=\left(0.500\mathrm{m}\right)\hat{\mathrm{j}}+\left(0.500\mathrm{m}\right)\hat{\mathrm{k}}$and,

$\mathrm{r}=\sqrt{{\left(0.500\mathrm{m}\right)}^2+{\left(0.500\mathrm{m}\right)}^2}=0.7071\mathrm{m}$

So we get,

$\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{r}}=\mathrm{v}\left(0.500\mathrm{m}\right)\left(-\hat{\mathrm{j}}\times \hat{\mathrm{j}}+\hat{\mathrm{j}}\times \hat{\mathrm{k}}\right)=\left(4.00\times {10}^6{\mathrm{m}}^2/\mathrm{s}\right)\hat{\mathrm{i}}$

Thus,

$$\begin{gathered} \overrightarrow{\mathrm{B}}=\frac{{\mathrm{\mu }}_0\mathrm{qv}}{4{\mathrm{\pi r}}^2}\hat{\mathrm{i}} \\ =\left(1\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)\frac{\left(6.00\times {10}^{-6}\mathrm{C}\right)\left(4.00\times {10}^6\mathrm{m}/\mathrm{s}\right)}{{\left(0.500\mathrm{m}\right)}^2}\hat{\mathrm{i}} \\ =+\left(6.79\times {10}^{-6}\mathrm{T}\right)\hat{\mathrm{i}} \\ \overrightarrow{\mathrm{B}}=+\left(6.79\times {10}^{-6}\mathrm{T}\right)\hat{\mathrm{i}} \end{gathered}$$