91影视

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

$\stackrel{鈫�}{B}=\frac{{渭}_o}{4蟺}\frac{q\stackrel{鈫�}{v}脳\hat{r}}{r^2}$ (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 $\stackrel{鈫�}{v}=v\widehat{i},\stackrel{鈫�}{r}=r\widehat{i}$so,

$\stackrel{鈫�}{v}脳\stackrel{鈫�}{r}=0$

thus,

$\stackrel{鈫�}{B}=0$

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

$\stackrel{鈫�}{v}脳\stackrel{鈫�}{r}=v鈰�\widehat{k}$

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

$\begin{array}{c}\stackrel{鈫�}{B}=\left(\frac{{渭}_0}{4蟺}\right)\frac{qv}{r^2}\widehat{k}\\ =鈭�\frac{\left(1.0脳{10}^{鈭�7}\mathrm{N}鈰�{\mathrm{s}}^2/{\mathrm{C}}^2\right)\left(4.80脳{10}^{鈭�6}\mathrm{C}\right)\left(6.80脳{10}^5\mathrm{m}/\mathrm{s}\right)}{(0.500\mathrm{m}{)}^2}\widehat{k}\\ =鈭�\left(1.31脳{10}^{鈭�6}\mathrm{T}\right)\widehat{k}\\ \stackrel{鈫�}{B}=鈭�\left(1.31脳{10}^{鈭�6}\mathrm{T}\right)\widehat{k}\end{array}$

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

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

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

$\begin{array}{c}\stackrel{鈫�}{B}=\left(\frac{{渭}_0}{4蟺}\right)\frac{q\left(0.500\right)v}{r^3}\widehat{k}\\ =鈭�\frac{\left(1.0脳{10}^{鈭�7}\mathrm{N}鈰�\frac{{\mathrm{s}}^2}{{\mathrm{C}}^2}\right)\left(4.80脳{10}^{鈭�6}\mathrm{C}\right)\left(0.500\right)\left(6.80脳{10}^5\mathrm{m}/\mathrm{s}\right)}{(0.500\mathrm{m}{)}^3}\widehat{k}\\ =鈭�\left(4.62脳{10}^{鈭�7}\mathrm{T}\right)\widehat{k}\\ \stackrel{鈫�}{B}=鈭�\left(4.62脳{10}^{鈭�7}\mathrm{T}\right)\widehat{k}\end{array}$

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

$\stackrel{鈫�}{v}脳\stackrel{鈫�}{r}=鈭�v鈰�\widehat{k}$

thus,

$\begin{array}{c}\stackrel{鈫�}{B}=\left(\frac{{渭}_0}{4蟺}\right)\frac{q\left(0.500\right)v}{r^2}\widehat{k}\\ =鈭�\frac{\left(1.0脳{10}^{鈭�7}\mathrm{N}鈰�{\mathrm{s}}^2/{\mathrm{C}}^2\right)\left(4.80脳{10}^{鈭�6}\mathrm{C}\right)\left(6.80脳{10}^5\mathrm{m}/\mathrm{s}\right)}{(0.500\mathrm{m}{)}^2}\widehat{k}\\ =鈭�\left(1.31脳{10}^{鈭�6}\mathrm{T}\right)\widehat{k}\\ \stackrel{鈫�}{B}=鈭�\left(1.31脳{10}^{鈭�6}\mathrm{T}\right)\widehat{k}\end{array}$