Question

An electron moves at 0.100 cas shown in Fig. Find the magnitude and direction of the magnetic field this electron produces at the following points, each 2.00 mm from the electron:

(a) points Aand B;

(b) point C;

(c) point D.

Short answer

1. ${\mathrm{B}}_{\mathrm{A}}=6.00\times {10}^{-8}\mathrm{T}\mathrm{and}{\mathrm{B}}_{\mathrm{B}}=6.00\times {10}^{-8}\mathrm{T}$
   
2. ${\mathrm{B}}_{\mathrm{C}}=1.20\times {10}^{-7}\mathrm{T}$
3. role="math" $B_D=0$
   

Step-by-step solution

Solving part (a) or the problem.

Consider an electron that moves with a speed of v = 0.100 c as shown in the figure.

First, we need to find the magnitude and direction of the magnetic field that this electron produces at A where the distance from this point to the electron is r= 2.00 um. The magnetic field due to this moving electron at the nucleus is given by,

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

where v 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, so we can write,

$\overrightarrow{B}=\frac{{\mu }_o}{4\pi }\frac{qv\sin \left(\varphi \right)}{r^2}$

substitute with the givens we get,

$B_A=\frac{{\mu }_0}{4\pi }\frac{qv\sin \left(\varphi \right)}{r^2}$

$$\begin{gathered} =\frac{4\mathrm{\pi }\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(3.00\times {10}^7\mathrm{m}/\mathrm{s}\right)\sin \left(30°\right)}{{\left(2.00\times {10}^{-6}\mathrm{m}\right)}^2} \\ =6.00\times {10}^{-8}\mathrm{T} \\ {\mathrm{B}}_{\mathrm{A}}=6.00\times {10}^{-8}\mathrm{T} \end{gathered}$$

according to the right-hand rule, the direction of the magnetic field$\overrightarrow{B}\infty q\overrightarrow{v}\times \hat{r}$ is out of the page. Now we need to find the magnitude and direction of the magnetic field that this

electron produces at B where the distance from this point to the electron is r= 2.00 $\mu m$. We get,

$$\begin{gathered} {\mathrm{B}}_{\mathrm{B}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{qvsin}\left(\mathrm{\varphi }\right)}{{\mathrm{r}}^2} \\ =\frac{4\mathrm{\pi }\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}}{4\mathrm{\pi }}\frac{\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(3.00\times {10}^7\mathrm{m}/\mathrm{s}\right)\sin \left(30°\right)}{{\left(2.00\times {10}^{-6}\mathrm{m}\right)}^2} \\ =6.00\times {10}^{-8}\mathrm{T} \\ {\mathrm{B}}_{\mathrm{A}}=6.00\times {10}^{-8}\mathrm{T} \end{gathered}$$

according to the right-hand rule, the direction of the magnetic field$\overrightarrow{\mathrm{B}}\infty q\overrightarrow{v}\times \hat{r}$ is out of the page.

Solving part (b) or the problem.

Now we need to find the magnitude and direction of the magnetic field that this electron produces at C where the distance from this point to the electron is r= 2.00, we can see that the angle between the velocity and the position vector is . We get,

$$\begin{gathered} {\mathrm{B}}_{\mathrm{C}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{qvsin}\left(\mathrm{\varphi }\right)}{{\mathrm{r}}^2} \\ =\frac{4\mathrm{\pi }\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}}{4\mathrm{\pi }}\frac{\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(3.00\times {10}^7\mathrm{m}/\mathrm{s}\right)\sin \left(30°\right)}{{\left(2.00\times {10}^{-6}\mathrm{m}\right)}^2} \\ =1.20\times {10}^{-7}\mathrm{T} \\ {\mathrm{B}}_{\mathrm{C}}=1.20\times {10}^{-7}\mathrm{T} \end{gathered}$$

according to the right-hand rule, the direction of the magnetic field$\overrightarrow{\mathrm{B}}\infty q\overrightarrow{v}\times \hat{r}$ is out of the page.

Solving part (c) or the problem.

Finally, we need to find the magnitude and direction of the magnetic field that this electron produces at D where the distance from this point to the electron is r= 2.00 $\mu m$, we can see that the angle between the velocity and the position vector is $\varphi =180°$So we get

$B_D=0$