Chapter 34, Problem 34.4P

An election moves through a uniform electric field $\stackrel{\to }{\text{E}}$ = (2.50 $\hat{\text{i}}$ + 5.00 $\hat{\text{j}}$) V/m and a uniform magnetic field $\stackrel{\to }{\text{B}}$ = 0.400 $\hat{k}$ T. Determine the acceleration of the electron when it has a velocity $\stackrel{\to }{\text{v}}$ = 10.0 $\hat{\text{i}}$ m/s.

To determine

The acceleration of electron.

Answer to Problem 34.4P

The acceleration of electron is $\left(-4.39\widehat{\text{i}}-1.76\widehat{\text{j}}\right)\times {10}^{11}\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Given info: The uniform electric field is $\overrightarrow{\text{E}}\text{=}\left(2.50\widehat{\text{i}}\text{+5}\text{.00}\widehat{\text{j}}\right)\text{V}/\text{m}$ and uniform magnetic field is $\overrightarrow{\text{B}}\text{=}\left(\text{0}\text{.400}\widehat{\text{k}}\right)\text{T}$, and the velocity of electron is $10.0\widehat{\text{i}}\text{m}/\text{s}$.

The formula for force acting on an electron in an electric field is,

${\overrightarrow{\text{F}}}_{\text{e}}\text{=q}\overrightarrow{\text{E}}$ . (1)

Here,

$q$ is the charge.

$\overrightarrow{E}$ is the electric field.

${\overrightarrow{F}}_{\text{e}}$ is force due to electric field.

Substitute $-1.6\times {10}^{-19}\text{C}$ for $q$ and $\left(2.50\widehat{\text{i}}\text{+5}\text{.00}\widehat{\text{j}}\right)\text{V}/\text{m}$ for $\overrightarrow{\text{E}}$ in equation (1).

$\begin{array}{c}{\overrightarrow{\text{F}}}_{\text{e}}\text{=}\left(-1.6\times {10}^{-19}\text{C}\right)\left(\left(2.50\widehat{\text{i}}\text{+5}\text{.00}\widehat{\text{j}}\right)\text{V}/\text{m}\right)\\ =\left(-1.6\times {10}^{-19}\text{C}\right)\left(\left(2.50\widehat{\text{i}}\text{+5}\text{.00}\widehat{\text{j}}\right)\right)\text{N}/\text{C}\\ =\left(-4\times {10}^{-19}\widehat{\text{i}}-\text{8}\times {10}^{-19}\widehat{\text{j}}\right)\text{N}\end{array}$

The formula for force acting on an electron in an magnetic field is,

${\overrightarrow{\text{F}}}_{\text{m}}\text{=q}\left(\overrightarrow{\text{v}}\times \overrightarrow{\text{B}}\right)$ (2)

Here,

$\overrightarrow{B}$ is magnetic field.

$\overrightarrow{v}$ is velocity.

${\overrightarrow{F}}_{\text{m}}$ is force due to magnetic field.

Substitute $-1.6\times {10}^{-19}\text{C}$ for $q$, $\text{10}\text{.0}\widehat{\text{i}}\text{m}/\text{s}$ for $\overrightarrow{\text{v}}$ and $\left(\text{0}\text{.400}\widehat{\text{k}}\right)\text{T}$ for $\overrightarrow{\text{B}}$ in equation (2).

${\overrightarrow{\text{F}}}_{\text{m}}\text{=}\left(-1.6\times {10}^{-19}\text{C}\right)\left(\left(\text{10}\text{.0}\widehat{\text{i}}\text{m}/\text{s}\right)\times \left(\text{0}\text{.400}\widehat{\text{k}}\right)\text{T}\right)$ (3)

Write the formula to calculate cross product between $\overrightarrow{\text{v}}$ and $\overrightarrow{\text{B}}$.

$\overrightarrow{\text{v}}\times \overrightarrow{\text{B}}\text{=}\left|\begin{array}{ccc}\widehat{\text{i}}& \widehat{\text{j}}& \widehat{\text{k}}\\ a_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\end{array}\right|$

Substitute $10.0$ for $a_{11}$, $0$ for $a_{12}$, $0$ for $a_{13}$, $0$ for $a_{21}$, $0$ for $a_{22}$ and $0.400$ for $a_{23}$ in the above expression.

$\begin{array}{c}\overrightarrow{\text{v}}\times \overrightarrow{\text{B}}\text{=}\left|\begin{array}{ccc}\widehat{\text{i}}& \widehat{\text{j}}& \widehat{\text{k}}\\ 10.0& 0& 0\\ 0& 0& 0.400\end{array}\right|\\ =\left(0.400\times 0-0\times 0\right)\widehat{\text{i}}-\left(10.0\times 0.400-0\times 0\right)\widehat{\text{j}}\text{+}\left(10.0\times 0-0\times 0\right)\widehat{\text{k}}\\ \text{=}-\left(4.00\text{T}\cdot \text{m}/\text{s}\right)\widehat{\text{j}}\end{array}$

Substitute $-\left(4.00\text{T}\cdot \text{m}/\text{s}\right)\widehat{\text{j}}$ for $\overrightarrow{\text{v}}\times \overrightarrow{\text{B}}$ in equation (3).

$\begin{array}{c}{\overrightarrow{\text{F}}}_{\text{m}}\text{=}\left(-1.6\times {10}^{-19}\text{C}\right)\left(-\left(4.00\text{T}\cdot \text{m}/\text{s}\right)\widehat{\text{j}}\right)\\ =6.4\times {10}^{-19}\widehat{\text{j}}\text{Tm}/\text{s}\end{array}$

The net force on the object where both fields are present is,

${\overrightarrow{\text{F}}}_{\text{net}}={\overrightarrow{\text{F}}}_{\text{e}}+{\overrightarrow{\text{F}}}_{\text{m}}$ (3)

Substitute $\text{q}\left(\overrightarrow{\text{v}}\times \overrightarrow{\text{B}}\right)$ for ${\overrightarrow{\text{F}}}_{\text{m}}$ and $\text{q}\overrightarrow{\text{E}}$ for ${\overrightarrow{\text{F}}}_{\text{e}}$ above expression.

${\overrightarrow{\text{F}}}_{\text{net}}=\text{q}\overrightarrow{\text{E}}+\text{q}\left(\overrightarrow{\text{v}}\times \overrightarrow{\text{B}}\right)$

Substitute $6.4\times {10}^{-19}\widehat{\text{j}}\text{Tm}/\text{s}$ for ${\overrightarrow{\text{F}}}_{\text{m}}$ and $\left(-4\times {10}^{-19}\widehat{\text{i}}-\text{8}\times {10}^{-19}\widehat{\text{j}}\right)\text{N}$ for ${\overrightarrow{\text{F}}}_{\text{e}}$ in equation (3).

$\begin{array}{c}{\overrightarrow{\text{F}}}_{\text{net}}=\left(-4\times {10}^{-19}\widehat{\text{i}}-\text{8}\times {10}^{-19}\widehat{\text{j}}\right)\text{N}+\left(6.4\times {10}^{-19}\widehat{\text{j}}\text{Tm}/\text{s}\right)\\ =\left(-4\times {10}^{-19}\widehat{\text{i}}-\text{1}\text{.6}\times {10}^{-19}\widehat{\text{j}}\right)\text{N}\end{array}$

The formula to calculate acceleration is,

$\overrightarrow{\text{a}}=\frac{{\overrightarrow{\text{F}}}_{\text{net}}}{m_{\text{e}}}$

Here,

$m_{\text{e}}$ is mass of electron.

Substitute $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{e}}$ and $\left(-4\times {10}^{-19}\widehat{\text{i}}-\text{1}\text{.6}\times {10}^{-19}\widehat{\text{j}}\right)\text{N}$ for ${\overrightarrow{\text{F}}}_{\text{net}}$ in the above expression.

$\begin{array}{c}\overrightarrow{\text{a}}=\frac{\left(-4\times {10}^{-19}\widehat{\text{i}}-\text{1}\text{.6}\times {10}^{-19}\widehat{\text{j}}\right)\text{N}}{9.11\times {10}^{-31}\text{kg}}\\ =\frac{\left(-4\widehat{\text{i}}-1.6\widehat{\text{j}}\right)\times {10}^{-19}\text{N}}{9.11\times {10}^{-31}\text{kg}}\\ =\frac{-4\widehat{\text{i}}\times {\text{10}}^{-19+31}}{9.11}+\frac{\left(-1.6\widehat{\text{j}}\times {\text{10}}^{-19+31}\right)}{9.11}\\ =-0.439\widehat{\text{i}}\times {\text{10}}^{12}-0.176{\widehat{\text{j}}}^{\text{10}}\end{array}$

Further solve the expression,

$\begin{array}{c}\overrightarrow{\text{a}}=\left(-4.39\widehat{\text{i}}\times {\text{10}}^{11}-1.76\widehat{\text{j}}\times {\text{10}}^{11}\right)\text{m}/{\text{s}}^{\text{2}}\\ =\left(-4.39\widehat{\text{i}}-1.76\widehat{\text{j}}\right)\times {\text{10}}^{11}\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the acceleration of electron is $\left(-4.39\widehat{\text{i}}-1.76\widehat{\text{j}}\right)\times {10}^{11}\text{m}/{\text{s}}^{\text{2}}$