An alpha particle (charge +2e) and an electron move in opposite directions from the same point, each with a speed of 2.50 x 105 m/s (Fig.). Find the magnitude and direction of the total magnetic field these charges produce at point P, which is 8.65 nm from each charge.

Consider an alpha particle with q= +2e and an electron with q=-e, which move in opposite directions from the same point as shown in the textbook's figure, each with the speed of$\mathrm{v}=2.50\times {10}^5\mathrm{m}/\mathrm{s}$ . We need to find the magnitude and direction of the total magnetic field these charges produce at point P, where this point is at a distance of r = 8.65 nm from each of the charges. The magnetic field due to this moving electron at the nucleus is given by,

$\overrightarrow{B}=\frac{{\mu }_o}{4\pi }\frac{q\overrightarrow{v}\times \hat{r}}{r^2}$ (1)

where 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 angle between the velocity and the position vector is ${\varphi }_{\alpha }=40°$and ${\varphi }_e=140°$for the alpha particle and the electron respectively. So we can write,

$B=B_{\alpha }+B_e=\frac{{\mu }_{0}v}{4\pi r^2}\left[e\sin \left({\varphi }_{\alpha }\right)+2e\sin \left({\varphi }_e\right)\right]$

note that the electron is negatively charged and its velocity is negative, that is why the term of the electron is positive. Substitute with the givens we get,

$$\begin{gathered} \mathrm{B}=\frac{4\mathrm{\pi }\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}}{4\mathrm{\pi }}\frac{\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(2.50\times {10}^5\mathrm{m}/\mathrm{s}\right)}{{\left(8.65\times {10}^{-9}\mathrm{m}\right)}^2} \\ \times \left[\sin \left(40°\right)+2\sin \left(140°\right)\right] \\ =1.03\times {10}^{-4}\mathrm{T} \\ \mathrm{B}=1.03\times {10}^{-4}\mathrm{T} \end{gathered}$$