Positive point charges q= +8.00 mC and q_ = +3.00 mC are moving relative to an observer at point P, as shown in Fig. The distance dis 0.120 m, v= 4.50 * 106 m>s, and v_ = 9.00 * 106 m>s

(a) When the two charges are at the locations shown in the figure, what are the magnitude and direction of the net magnetic field they produce at point P?

(b) What are the magnitude and direction of the electric and magnetic forces that each charge exerts on the other, and what is the ratio of the magnitude of the electric force to the magnitude of the magnetic force?

(c) If the direction of vS_ is reversed, so both charges are moving in the same direction, what are the magnitude and direction of the magnetic forces that the two charges exert on each other?

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, $v=4.50\times {10}^{6}m/s$and$v\text{'}=9.00\times {10}^{6}m/s$ . The magnetic field due to this moving electron at the nucleus is given by,

$\overrightarrow{B}=\frac{{\mu }_o}{4\pi }\frac{q\overrightarrow{v}\times \widehat{r}}{r^2}$ (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 $\varphi ={\varphi }^{\mathrm{\prime }}={90}^{\circ },$ so we can write,

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

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

$\begin{array}{r}{B}_{\text{tetai}}=\frac{{\mu }_o}{4\pi }\left(\frac{\left(8.00\times {10}^{-6}\right)\left(6.00\times {10}^6\right)}{(0.120{)}^2}+\frac{\left(3.00\times {10}^{-6}\right)\left(v^{\mathrm{\prime }}=9.00\times {10}^6\mathrm{m}/\mathrm{s}\right)}{(0.120{)}^2}\right)\\ B_{\text{totai}}=4.38\times {10}^{-4}\mathrm{T}\end{array}$

The magnetic force between the charges is given by

$F_B=\frac{{\mu }_o}{4\pi }\frac{q{q}^{\mathrm{\prime }}{v}^{\mathrm{\prime }}}{r^2}$

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

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

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,

$\begin{array}{c}{F}_C=k\frac{q{q}^{\mathrm{\prime }}}{r^2}\\ =\left(8.99\times {10}^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{\left(8.0\times {10}^{-6}\mathrm{C}\right)\left(3.0\times {10}^{-6}\mathrm{C}\right)}{(0.240\mathrm{m}{)}^2}\\ =3.75\mathrm{N}\end{array}$

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

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