91Ó°ÊÓ

Consider positive point charges, which have charges of q = +8.00 and q'= +3.00 . As shown in the textbook's figure, both charges are moving relative to an observer at point P, where d= 0.120, $v=4.50×{10}^{6}m/s$and$v\text{'}=9.00×{10}^{6}m/s$ . The magnetic field due to this moving electron at the nucleus is given by,

$\stackrel{→}{B}=\frac{{\mathrm{μ}}_o}{4\mathrm{π}}\frac{q\stackrel{→}{v}×\widehat{r}}{r^2}$ (1)

where $\hat{r}$is the direction of the position vector (from the charge to the point that we want to find the field at it) we can see that the angles between the velocity and the position vectors are $\mathrm{ϕ}={\mathrm{ϕ}}^{\mathrm{′}}={90}^{∘},$ so we can write,

$B_{\text{tota}}=B+B^{\mathrm{′}}=\frac{{\mathrm{μ}}_o}{4\mathrm{π}}\left(\frac{qv}{d^2}+\frac{q^{\mathrm{′}}{v}^{\mathrm{′}}}{d^2}\right)$

note that the fields are in the same direction (which is into the page according to the right-hand rule) substitute with the givens to get

$\begin{array}{r}{B}_{\text{tetai}}=\frac{{\mathrm{μ}}_o}{4\mathrm{π}}\left(\frac{\left(8.00×{10}^{−6}\right)\left(6.00×{10}^6\right)}{(0.120{)}^2}+\frac{\left(3.00×{10}^{−6}\right)\left(v^{\mathrm{′}}=9.00×{10}^6\mathrm{m}/\mathrm{s}\right)}{(0.120{)}^2}\right)\\ B_{\text{totai}}=4.38×{10}^{−4}\mathrm{T}\end{array}$

The magnetic force between the charges is given by

$F_B=\frac{{\mathrm{μ}}_o}{4\mathrm{π}}\frac{q{q}^{\mathrm{′}}{v}^{\mathrm{′}}}{r^2}$

where r is the distance between them, which is r=2d, substitute with the givens to get,

$\begin{array}{r}{F}_B=\frac{\left(8.00×{10}^{−6}\mathrm{C}\right)\left(3.00×{10}^{−6}\mathrm{C}\right)\left(4.50×{10}^6\mathrm{m}/\mathrm{s}\right)\left(9.00×{10}^6\mathrm{m}/\mathrm{s}\right)}{(0.240{)}^2}\\ F_B=1.69×{10}^{−3}\mathrm{N}\end{array}$

where the force on the upper charge points up and the force on the lower charge points down. The magnitude of the electric force is,

$\begin{array}{c}{F}_C=k\frac{q{q}^{\mathrm{′}}}{r^2}\\ =\left(8.99×{10}^9\mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{\left(8.0×{10}^{−6}\mathrm{C}\right)\left(3.0×{10}^{−6}\mathrm{C}\right)}{(0.240\mathrm{m}{)}^2}\\ =3.75\mathrm{N}\end{array}$

the ratio of the Coulomb force to the magnetic force is, therefore,

$\begin{array}{r}\frac{F_B}{F_C}=\frac{3.75\mathrm{N}}{1.69×{10}^{−3}\mathrm{N}}=2.22×{10}^3\\ \frac{F_B}{F_C}=2.22×{10}^3\end{array}$$\begin{array}{r}\frac{F_B}{F_C}=\frac{3.75\mathrm{N}}{1.69×{10}^{−3}\mathrm{N}}=2.22×{10}^3\\ \frac{F_B}{F_C}=2.22×{10}^3\end{array}$

If the direction of i is reversed, the magnetic forces are reversed but the magnitude of the force is unchanged.