Q43P - Fundamentals Of Physics

Question

An electron with kinetic energy ${\mathbf{K}}_{\mathbf{e}}$ travels in a circular path that is perpendicular to a uniform magnetic field, which is in the positive direction of a z axis. The electron’s motion is subject only to the force due to the field.

(a) Show that the magnetic dipole moment of the electron due to its orbital motion has magnitude $\mathbf{\mu }\mathbf{=}\frac{{\mathbf{K}}_{\mathbf{e}}}{\mathbf{B}}\mathbf{ }$ B and that it is in the direction opposite that of $\overrightarrow{\mathbf{B}}$.(b)What is the magnitude of the magnetic dipole moment of a positive ion with kinetic energy  $\mathbf{ }{\mathbf{K}}_{\mathbf{i}}$ under the same circumstances?  (c) What is the direction of the magnetic dipole moment of a positive ion with kinetic energy $\mathbf{ }{\mathbf{K}}_{\mathbf{i}}$ under the same circumstances? (d) An ionized gas consists of  $\mathbf{5.3}\mathbf{ }\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}$$\mathbf{electron}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$ and the same number density of ions. Take the average electron kinetic energy to be $\mathbf{6}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{20}}\mathbf{ }\mathbf{J}$ and the average ion kinetic energy to be $\mathbf{7}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{21}}\mathbf{J}$. Calculate the magnetization of the gas when it is in a magnetic field of $\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{ }\mathbf{T}$.

Step-by-Step Solution

Answer

1. The magnitude of magnetic dipole moment of electron is ${\mathrm{\mu }}_{\mathrm{e}}=\frac{{\mathrm{K}}_{\mathrm{e}}}{\mathrm{B}}$, and its direction is along negative $\mathrm{Z} \mathrm{axis}$.
2. The magnitude of magnetic dipole moment of positive ion is,${\mathrm{\mu }}_{+}=\frac{{\mathrm{K}}_{+}}{\mathrm{B}}$.
3. The direction of magnetic dipole moment of positive ion is along the negative $\text{Z axis}$.
4. Magnetization of ionized gas,$\mathrm{M}=3.1\times {10}^2\mathrm{A}/\mathrm{m}$

Step 1: Listing the given quantities

Positive ion concentration, ${\mathrm{n}}_{+}=5.3\times {10}^{21} \mathrm{ions}/{\mathrm{m}}^3$

Electron concentration, ${\mathrm{n}}_{\mathrm{e}}=5.3\times {10}^{21} \mathrm{electrons}/{\mathrm{m}}^3$

Average kinetic energy of electron, ${\mathrm{K}}_{\mathrm{e}}=6.2\times {10}^{-20}\mathrm{J}$

Average kinetic energy of positive ion, ${\mathrm{K}}_{+ }=7.6\times {10}^{-21}\mathrm{J}$

Applied magnetic field, $\overrightarrow{\mathrm{B}}=(1.2 \mathrm{T})\widehat{\mathrm{z}}$

 

Step 2: Understanding the concepts of magnetic moment

We will use the definition of magnetic moment and expression of current due to circular motion of electron and positive ion to find the magnetic moment of electron and positive ion. By using the necessary condition for circular motion, we can deduce the required expression.

Formula:

Force acting on charged particle moving in magnetic field is $\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}})$

Centripetal force, $\overrightarrow{\mathrm{F}}=-\frac{{\mathrm{mv}}^2}{\mathrm{r}}\widehat{\mathrm{r}}$

Magnetic moment,$\overrightarrow{\mathrm{\mu }}=\mathrm{i}\overrightarrow{\mathrm{A}}$

$\mathrm{K}.\mathrm{E}$ of mass $\mathrm{m}$ moving with velocity , $\mathrm{K}=\frac{1}{2}\cdot {\mathrm{mv}}^2$

Step 3: (a) Calculations of the magnetic dipole moment of electron

Magnetic moment of electron 

$\begin{array}{c}\overrightarrow{{\mathrm{\mu }}_{\mathrm{e}}}=\mathrm{i}\overrightarrow{\mathrm{A}}\\ =\frac{-\mathrm{e}}{\mathrm{T}}\overrightarrow{\mathrm{A}.}\end{array}$

The electron is moving in a circular orbit of radius $r.$ Hence the area vector is perpendicular to the plane of the orbit. We assume that the electron is moving in counter clockwise direction; then we get the direction of  $A$ as given below:

$\overrightarrow{\mathrm{A}}=\mathrm{\pi }{\mathrm{r}}^2\widehat{\mathrm{z}}$

Using this, we get 

Magnetic moment of electron

$\begin{array}{c}\overrightarrow{{\mathrm{\mu }}_{\mathrm{e}}}=\frac{-\mathrm{e}}{\mathrm{T}}\overrightarrow{\mathrm{A}}\\ =\frac{-\mathrm{e}}{\mathrm{T}}\cdot \mathrm{\pi }{\mathrm{r}}^2\widehat{\mathrm{z}}\\ =\frac{\mathrm{e}}{\mathrm{T}}\cdot {\mathrm{\pi r}}^2(-\widehat{\mathrm{z}})\end{array}$

Let’s assume that the electron is completing one revolution in time $\mathrm{T}=\frac{2\mathrm{\pi r}}{\mathrm{v}}$

Where, $\mathrm{v}$ is the velocity of electron.

Therefore,

$\begin{array}{c}\overrightarrow{{\mathrm{\mu }}_{\mathrm{e}}}=\frac{\mathrm{e}}{\mathrm{T}}\cdot {\mathrm{\pi r}}^2(-\widehat{\mathrm{z}})\\ =\frac{{\mathrm{e\pi r}}^2\mathrm{v}}{2\mathrm{\pi r}}\cdot (-\widehat{\mathrm{z}})\\ =\frac{\mathrm{evr}}{2}\cdot (-\widehat{\mathrm{z}})\end{array}$

Now the force acing on an electron moving in a magnetic field is $\overrightarrow{\mathrm{F}}=-\mathrm{e}(\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}})$

$\overrightarrow{\mathrm{F}}=\mathrm{evB}(-\widehat{\mathrm{r}})$

As the electron is moving in circular orbit 

$\begin{array}{c}\overrightarrow{\mathrm{F}}=\mathrm{evB}(-\widehat{\mathrm{r}})\\ =\frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2}{\mathrm{r}}(-\widehat{\mathrm{r}})\end{array}$

This gives the magnitude of force as given below:

$\mathrm{evB}=\frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2}{\mathrm{r}}$

Therefore,

$\mathrm{r}=\frac{{\mathrm{m}}_{\mathrm{e}}\mathrm{v}}{\mathrm{eB}}$

Substituting  $\mathrm{r}$ in  $\overrightarrow{{\mathrm{\mu }}_{\mathrm{e}}}$, we get 

$\begin{array}{c}\overrightarrow{{\mathrm{\mu }}_{\mathrm{e}}}=\frac{\mathrm{ev}}{2}\cdot \frac{{\mathrm{m}}_{\mathrm{e}}\mathrm{v}}{\mathrm{eB}}(-\widehat{\mathrm{z}})\\ =\frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2}{2\mathrm{B}}(-\widehat{\mathrm{z}})\\ =\frac{{\mathrm{K}}_{\mathrm{e}}}{\mathrm{B}}(-\widehat{\mathrm{z}})\end{array}$

Hence, the magnitude of magnetic dipole moment of electron is ${\mathrm{\mu }}_{\mathrm{e}}=\frac{{\mathrm{K}}_{\mathrm{e}}}{\mathrm{B}}$, and its direction is along the negative $\mathrm{Z} \mathrm{axis}$.

 

Step 4: (b) Calculations of the magnitude of magnetic dipole moment of a positive ion

Following similar steps for the positive ion, we will assume a charge on positive ion to be $\mathrm{q}=+\mathrm{e}.$

Magnetic moment of $(+)\mathrm{ion}$ is $\overrightarrow{{\mathrm{\mu }}_{+ }}=\mathrm{i}\overrightarrow{\mathrm{A}}=\frac{+\mathrm{e}}{\mathrm{T}}\overrightarrow{\mathrm{A}}=\frac{+\mathrm{e}}{\mathrm{T}}\cdot {\mathrm{\pi r}}^2(-\widehat{\mathrm{z}})$.

Let’s assume $(+)\mathrm{ion}$  is complete one revolution in time $\mathrm{T}=\frac{2\mathrm{\pi r}}{\mathrm{v}}$

where $\mathrm{v}$ is the velocity of  $(+)\mathrm{ion}.$

Therefore,

$\begin{array}{c}\overrightarrow{{\mathrm{\mu }}_{+ }}=\frac{+\mathrm{e}}{\mathrm{T}}\cdot {\mathrm{\pi r}}^2(-\widehat{\mathrm{z}})\\ =\frac{{\mathrm{e\pi r}}^2\mathrm{v}}{2\mathrm{\pi r}}\cdot (-\widehat{\mathrm{z}})\\ =\frac{\mathrm{evr}}{2}\cdot (-\widehat{\mathrm{z}})\end{array}$

Let  $\overrightarrow{\mathrm{F}}=\mathrm{evB}\left(\widehat{\mathrm{r}}\right)$

As the $(+) \mathrm{ion}$ is moving in circular orbit 

$\mathrm{r}=\frac{{\mathrm{m}}_{+}\mathrm{v}}{\mathrm{eB}}$

Substituting $r$ in $\overrightarrow{{\mathrm{\mu }}_{+}}$.  we get 

$\begin{array}{c}\overrightarrow{{\mathrm{\mu }}_{+}}=\frac{\mathrm{ev}}{2}\cdot \frac{{\mathrm{m}}_{+}\mathrm{v}}{\mathrm{eB}}(-\widehat{\mathrm{z}})\\ =\frac{{\mathrm{m}}_{+}{\mathrm{v}}^2}{2\mathrm{B}}(-\widehat{\mathrm{z}})\\ =\frac{{\mathrm{K}}_{+}}{\mathrm{B}}(-\widehat{\mathrm{z}})\end{array}$

Hence, the magnitude of magnetic dipole moment of a positive ion is ${\mathrm{\mu }}_{+}=\frac{{\mathrm{K}}_{+}}{\mathrm{B}}$.

Step 5: (c) Calculations of the direction of magnetic dipole moment of positive ion

We have 

$\overrightarrow{{\mathrm{\mu }}_{+}}=\frac{{\mathrm{K}}_{+}}{\mathrm{B}}(-\widehat{\mathrm{z}})$

Therefore, the direction of magnetic dipole moment of the positive ion is along the negative $\mathrm{Z}\mathrm{axis}$.

Step 6: (d) Calculations of magnetization of ionized gas

Magnetization of ionized gas:

Since the ionized gas is a mixture of positive ions and electrons, magnetization is   

$\begin{array}{c}\mathrm{M}={\mathrm{n}}_{\mathrm{e}}{\mathrm{\mu }}_{\mathrm{e}}+{\mathrm{n}}_{+}{\mathrm{\mu }}_{+}\\ ={\mathrm{n}}_{\mathrm{e}}\frac{{\mathrm{K}}_{\mathrm{e}}}{\mathrm{B}}+{\mathrm{n}}_{+}\frac{{\mathrm{K}}_{+}}{\mathrm{B}}\end{array}$

Also  ${\mathrm{n}}_{\mathrm{e}}={\mathrm{n}}_{+}$

Using this, we get 

$\begin{array}{c}\mathrm{M}=\frac{{\mathrm{n}}_{\mathrm{e}}}{\mathrm{B}}\cdot ({\mathrm{K}}_{\mathrm{e}}+{\mathrm{K}}_{+})\\ =\frac{5.3\times {10}^{21}}{1.2}\cdot (6.2\times {10}^{-20}+7.6\times {10}^{-21})\\ =\frac{5.3\times {10}^{21}\times {10}^{-21}\times (62+7.6)}{1.2}\\ =307.4\\ \approx 3.1\times {10}^2\mathrm{A}/\mathrm{m}\end{array}$

Magnetization of ionized gas,$\mathrm{M}=3.1\times {10}^2\mathrm{A}/\mathrm{m}$