91Ó°ÊÓ

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

$\stackrel{→}{B}=\frac{{\mathrm{μ}}_o}{4\mathrm{π}}\frac{q\stackrel{→}{v}×\hat{r}}{r^2}$ (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 $\stackrel{→}{v}=v\widehat{i},\stackrel{→}{r}=r\widehat{i}$so,

$\stackrel{→}{v}×\stackrel{→}{r}=0$

thus,

$\stackrel{→}{B}=0$

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 $\stackrel{→}{v}=v\widehat{i},\stackrel{→}{r}=r\widehat{j}$so,

$\stackrel{→}{v}×\stackrel{→}{r}=vâ‹\dots \widehat{k}$

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

$\begin{array}{c}\stackrel{→}{B}=\left(\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\right)\frac{qv}{r^2}\widehat{k}\\ =−\frac{\left(1.0×{10}^{−7}\mathrm{N}â‹\dots {\mathrm{s}}^2/{\mathrm{C}}^2\right)\left(4.80×{10}^{−6}\mathrm{C}\right)\left(6.80×{10}^5\mathrm{m}/\mathrm{s}\right)}{(0.500\mathrm{m}{)}^2}\widehat{k}\\ =−\left(1.31×{10}^{−6}\mathrm{T}\right)\widehat{k}\\ \stackrel{→}{B}=−\left(1.31×{10}^{−6}\mathrm{T}\right)\widehat{k}\end{array}$

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 $\stackrel{→}{v}=v\widehat{i},\stackrel{→}{r}=\left(0.500\right)(\widehat{i}+\widehat{j})$so,

role="math" localid="1668229000516" $\stackrel{→}{v}×\stackrel{→}{r}=\left(0.500\right)v\widehat{k}$

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

$\begin{array}{c}\stackrel{→}{B}=\left(\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\right)\frac{q\left(0.500\right)v}{r^3}\widehat{k}\\ =−\frac{\left(1.0×{10}^{−7}\mathrm{N}â‹\dots \frac{{\mathrm{s}}^2}{{\mathrm{C}}^2}\right)\left(4.80×{10}^{−6}\mathrm{C}\right)\left(0.500\right)\left(6.80×{10}^5\mathrm{m}/\mathrm{s}\right)}{(0.500\mathrm{m}{)}^3}\widehat{k}\\ =−\left(4.62×{10}^{−7}\mathrm{T}\right)\widehat{k}\\ \stackrel{→}{B}=−\left(4.62×{10}^{−7}\mathrm{T}\right)\widehat{k}\end{array}$

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 $\stackrel{→}{v}=v\widehat{i},\stackrel{→}{r}=r\widehat{k}$ so,

$\stackrel{→}{v}×\stackrel{→}{r}=−vâ‹\dots \widehat{k}$

thus,

$\begin{array}{c}\stackrel{→}{B}=\left(\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\right)\frac{q\left(0.500\right)v}{r^2}\widehat{k}\\ =−\frac{\left(1.0×{10}^{−7}\mathrm{N}â‹\dots {\mathrm{s}}^2/{\mathrm{C}}^2\right)\left(4.80×{10}^{−6}\mathrm{C}\right)\left(6.80×{10}^5\mathrm{m}/\mathrm{s}\right)}{(0.500\mathrm{m}{)}^2}\widehat{k}\\ =−\left(1.31×{10}^{−6}\mathrm{T}\right)\widehat{k}\\ \stackrel{→}{B}=−\left(1.31×{10}^{−6}\mathrm{T}\right)\widehat{k}\end{array}$