91Ó°ÊÓ

Consider a negative charge$q=3.60×{10}^{−6}\mathrm{C}$ , which is located at the origin and has a velocity of$\stackrel{→}{v}=\left(7.5×{10}^4\mathrm{m}/\mathrm{s}\right)\widehat{i}+\left(−4.90×{10}^4\mathrm{m}/\mathrm{s}\right)\widehat{j}$ The magnetic field due to this moving electron at the nucleus, is given by,

$\stackrel{→}{B}=\frac{{\mathrm{μ}}_o}{4\mathrm{π}}\frac{q\stackrel{→}{v}×\stackrel{→}{r}}{r^3}$ (1)

we need to find the magnetic field produced by this charge at the point x = 0.200 m, y= -0.300 m, and z =0. So we have $\stackrel{→}{r}=\left(0.200\right)\widehat{i}+(−0.300)\widehat{j}$, thus,

$\begin{array}{c}\stackrel{→}{v}×\stackrel{→}{r}=\left[\left(7.50×{10}^4\mathrm{m}/\mathrm{s}\right)\widehat{i}+\left(−4.90×{10}^4\mathrm{m}/\mathrm{s}\right)\widehat{j}\right]\\ ×\left[\right(0.200\mathrm{m})\widehat{i}+(−0.300\mathrm{m}\left)\widehat{j}\right]\\ =\left(−2.25×{10}^4{\mathrm{m}}^2/\mathrm{s}\right)\widehat{k}+\left(9.80×{10}^3{\mathrm{m}}^2/\mathrm{s}\right)\widehat{k}\\ =\left(−1.27×{10}^4{\mathrm{m}}^2/\mathrm{s}\right)\widehat{k}\end{array}$

now we need to substitute into (1) but before we need to find r, as,

$\begin{array}{c}r=\sqrt{x^2+y^2+z^2}=\sqrt{(0.200\mathrm{m}{)}^2+(−0.300\mathrm{m}{)}^2}=0.3606\mathrm{m}\\ \stackrel{→}{B}=\frac{\left(1.00×{10}^{−7}\mathrm{T}â‹\dots \mathrm{m}/\mathrm{A}\right)\left(−3.60×{10}^{−6}\mathrm{C}\right)\left(−1.27×{10}^4{\mathrm{m}}^2/\mathrm{s}\right)}{(0.3606\mathrm{m}{)}^3}\widehat{k}\\ =\left(9.75×{10}^{−8}\mathrm{T}\right)\widehat{k}\\ \stackrel{→}{B}=\left(9.75×{10}^{−8}\mathrm{T}\right)\widehat{k}\end{array}$