An electron and a proton are each moving at 735 km>s in perpendicular paths as shown in Fig.. At the instant when they are at the positions shown, find the magnitude and direction of

(a) the total magnetic field they produce at the origin;

(b) the magnetic field the electron produces at the location of the proton;

(c) the total electric force and the total magnetic force that the electron exerts on the proton.

Consider an electron and a proton which are each moving at v = 735 km/s in perpendicular paths as shown in the textbook's figure. When the electron is at (x, y, z) = (0,5.00 nm, 0) and the proton is at (x, y, z) =(4.00 nm, 0, 0).

First we need to find the total magnetic field the they produce at the origin. The magnetic field due to a moving charge 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 unit vector that points from the charge to the point that we want to find the field at it, we can see that the angle between ö and f is 90°, and according to the right hand rule, both fields are into the page (note that we take the sign of the electron charge into account), so we can write,

$B=B_e+B_p$

where,

$B_e=\frac{{\mu }_o}{4\pi }\frac{ev}{r_e^2},B_p=\frac{{\mu }_o}{4\pi }\frac{ev}{r_p^2}$

Substitute with the givens (note that $r_{\theta }=5.00\mathrm{nm}$and $r_p=4.00\mathrm{nm}$) to get,

$\begin{array}{c}B=\frac{\left(4\pi \times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}\right)\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(7.35\times {10}^5\mathrm{m}/\mathrm{s}\right)}{4\pi }\\ \times \left(\frac{1}{{\left(5.00\times {10}^{-9}\mathrm{m}\right)}^2}+\frac{1}{{\left(4.00\times {10}^{-9}\mathrm{m}\right)}^2}\right)\\ =1.21\times {10}^{-3}\mathrm{T}\\ B=1.21\times {10}^{-3}\mathrm{T}\end{array}$

Now we need to find the magnetic field that the electron produces at the location of the proton, the distance between the electron and the proton can be found from the right triangle as,

$r=\sqrt{(4.00\mathrm{nm}{)}^2+(5.00\mathrm{nm}{)}^2}=\sqrt{41\mathrm{nm}}$

and the angle that the line between the electron and the proton makes relative to the vertical axis can be determined from the triangle, that is,

$\tan \theta =\frac{4.00\mathrm{nm}}{5.00\mathrm{nm}}$

Or

$\theta ={\tan }^{-1}\frac{4.00\mathrm{nm}}{5.00\mathrm{nm}}={38.7}^{\circ }$

but the angle between the vertical line and the velocity of the electron is 90", so the angle between the velocity and the position vector is$\varphi ={90}^{\circ }+{38.7}^{\circ }={128.7}^{\circ }$ , so the magnetic field that the electron produces at the location of the proton is,

$\begin{array}{c}B=\frac{{\mu }_0}{4\pi }\frac{\mathrm{evsin}\left(\varphi \right)}{r^2}\\ =\frac{\left(4\pi \times {10}^{-7}\mathrm{T}\cdot \frac{\mathrm{m}}{\mathrm{A}}\right)}{4\pi }\frac{\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(7.35\times \frac{{10}^5\mathrm{m}}{\mathrm{s}}\right)\sin \left({128.7}^{\circ }\right)}{{\left(\sqrt{41}\times {10}^{-9}\mathrm{nm}\right)}^2}\\ =2.24\times {10}^{-4}\mathrm{T}\\ B=2.24\times {10}^{-4}\mathrm{T}\end{array}$

according to the right hand rule, the direction of the field is into the page

Now we need to find the total electric force and the total magnetic force that the electron exerts on the proton. That is,

$F=F_B+F_C$

where

$F_B=qvB\sin \left({90}^{\circ }\right),F_C=\frac{e^2}{4\pi {\epsilon }_o{r}^2}$

So,

$\begin{array}{c}F=\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(7.35\times {10}^5\mathrm{m}/\mathrm{s}\right)\left(2.24\times {10}^{-4}\mathrm{T}\right)\sin \left({90.0}^{\circ }\right)\\ +\frac{\left(9.00\times {10}^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2\right){\left(1.60\times {10}^{-19}\mathrm{C}\right)}^2}{{\left(\sqrt{41}\times {10}^{-9}\mathrm{nm}\right)}^2}\\ =5.62\times {10}^{-12}\mathrm{N}\\ F=5.62\times {10}^{-12}\mathrm{N}\end{array}$

this force is at 128.7° counterclockwise from the positive horizontal axis.