Question: An electron has velocity $\overrightarrow{\mathbf{v}}\mathbf{=}\left(32\hat{i}+40\hat{j}\right)\mathbf{\text{km/s}}$ as it enters a uniform magnetic field$\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{60}\hat{\mathbf{i}}\mathbf{ }\mathit{\mu }\mathit{T}$ What are (a) the radius of the helical path taken by the electron and (b) the pitch of that path? (c) To an observer looking into the magnetic field region from the entrance point of the electron, does the electron spiral clockwise or counterclockwise as it moves?

The velocity of electron is,$\overrightarrow{v}=(32\hat{i}+40\hat{j})\mathrm{km}/S$.

The magnetic field is, $\overrightarrow{B}=60\hat{i} \mu T$.

By using the concept of magnetic field, we can find the magnetic force direction, i.e., the direction of the electron and also find the radius of the helical path. By using the concept of the period of revolution, we can find the pitch of the path and the direction of electron motion.

Formula:

$\overrightarrow{\mathbf{F}}\mathbf{=}\mathit{q}\left(\overrightarrow{v}\times \overrightarrow{B}\right)$

$\mathit{r}\mathbf{=}\frac{\mathbf{m}\mathbf{v}}{\left|q\right|\mathbf{B}}$

First we find the magnetic force by using the formula

$\overrightarrow{F}=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$

$\overrightarrow{F}=q\left(v_x\hat{i}+v_y\hat{j}+v_z\hat{k}\right)\times B_x\hat{i}+B_y\hat{j}+B_z\hat{k}$

So the direction of force from the given value is

$\overrightarrow{F}=-q{v}_y{B}_x\hat{k}$

Electron charge is,$q=-e$

$\overrightarrow{F}=+e{v}_y{B}_x\hat{k}$

So the force along$+\hat{k}$

From equation 28-16, we can write

$r=\frac{mv}{\left|q\right|B}$

Here velocity is alongdirection, so that

$r=\frac{m{v}_y}{e{B}_x}$

By substituting the value, we can calculate

$$\begin{gathered} r=\frac{\left(9.1\times {10}^{-31}kg\times 40\times {10}^{3}m/s\right)}{1.6\times {10}^{-19}c\times 60\times {10}^{-6}T} \\ r=3.79\times {10}^{-3}m \\ r=0.00379m \end{gathered}$$

Hence the radius of the taken by the electron is, $r=0.0038 \text{m}$.

For one revolution, the electron takes

$$\begin{gathered} T=\frac{2\pi r}{v_y} \\ =\frac{2\pi \times 0.0038 m}{40\times {10}^3\text{m/s}} \\ =0.60\times {10}^{-6} s \end{gathered}$$

During that time, the drift of the electron in the x-direction, which is pitch of that path, is

$\Delta x=V_{x}T$

Substitute all the value in the above equation.

By using the equation, we can write

$$\begin{gathered} \Delta x=32\times {10}^3 \text{m/s}\times 0.60\times {10}^{-6} \text{s} \\ \Delta x=19\times {10}^{-3} \text{m} \end{gathered}$$

Hence the pitch of that path is,$\Delta x=19\times {10}^{-3} \text{m}$

As per the condition given in the problem, the magnetic field is along the positive x direction. But the observer sees the electron along the negative x- direction. As we know from the force calculation, the electron moves towards the $+\hat{k}$so that the electron moves away from him. He sees it entering the region with positive $v_y$but being pushed in the $+z$direction so that the electron spiral is clockwise.