91Ó°ÊÓ

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.

$\stackrel{→}{\mathrm{B}}=\frac{{\mathrm{μ}}_{o}q\stackrel{→}{v}×\hat{r}}{4\mathrm{π}{r}^2}$

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

$\stackrel{→}{\mathrm{v}}×\stackrel{→}{\mathrm{r}}=\mathrm{vr}\stackrel{→}{\mathrm{j}}×\hat{\mathrm{i}}=-\mathrm{vr}\hat{\mathrm{k}}$

Thus,

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

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

$\stackrel{→}{v}×\stackrel{→}{\mathrm{r}}=v\hat{j}×\hat{j}=0$

Thus,

$\stackrel{→}{B}=0$

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

$\stackrel{→}{v}×\stackrel{→}{\mathrm{r}}=v\hat{j}×\hat{k}=-vr\hat{i}$

Thus,

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

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

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

Thus,

$$\begin{gathered} \stackrel{→}{\mathrm{B}}=\frac{{\mathrm{μ}}_0\mathrm{qv}}{4{\mathrm{Ï€°ù}}^2}\hat{\mathrm{i}} \\ =\left(1×{10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)\frac{\left(6.00×{10}^{-6}\mathrm{C}\right)\left(4.00×{10}^6\mathrm{m}/\mathrm{s}\right)}{{\left(0.500\mathrm{m}\right)}^2}\hat{\mathrm{i}} \\ =+\left(6.79×{10}^{-6}\mathrm{T}\right)\hat{\mathrm{i}} \\ \stackrel{→}{\mathrm{B}}=+\left(6.79×{10}^{-6}\mathrm{T}\right)\hat{\mathrm{i}} \end{gathered}$$