Chapter 30, Problem 30.16P

In a long, .straight, vertical lightning stroke, electrons move downward and positive ions move upward and constitute a current of magnitude 20.0 kA. At a location 50.0 m east of the middle of the stroke, a free electron drifts through the air toward the west with a speed of 300 m/s. (a) Make a sketch showing the various vectors involved. Ignore the effect of the Earth's magnetic field. (b) Find the vector force the lightning stroke exerts on the electron. (c) Find the radius of the electron’s path. (d) Is it a good approximation to model the electron as moving in a uniform field? Explain your answer. (e) If it does not collide with any obstacles, how many revolutions will the electron complete during the 60.0-µs duration of the lightning stroke?

To determine

To draw: The various vectors involved to represent the lightning stroke of the electron and the positive ions.

Answer to Problem 30.16P

The various vectors involved to represent the lightning stroke of the electron and the positive ions as shown below,

Explanation of Solution

Given info: The electrons move downward and the positive ions move upwards. The magnitude of the uniform current is $20.0\text{kA}$ at the point of the $50.0\text{m}$ east of the middle of the stroke and the free electrons drift through the air towards the west with the speed of $300\text{m}/\text{s}$.

According to the Ampere’s right hand thumb rule, the index finger represents the direction of the velocity vector $\text{V}$, middle finger represents the direction of the magnetic field $\text{B}$ vector and the direction of thumb represent the direction of the cross product of force $\text{F}$.

Write the expression for the magnetic field.

$B=\frac{{\mu }_{0}i}{2\pi x}$

Here,

${\mu }_0$ is the permeability of the free space.

$i$ is the current in the wire.

$x$ is the location of the magnetic field.

Substitute $4\pi \times {10}^{-7}\text{T-m}/\text{A}$ for ${\mu }_0$, $20.0\text{kA}$ for $i$ and $50.0\text{m}$ for $x$ in the above expression for the value of the magnetic field.

$\begin{array}{c}B=\frac{\left(4\pi \times {10}^{-7}\text{T-m}/\text{A}\right)\left(20.0\text{kA}\right)}{2\pi \left(50.0\text{m}\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T-m}/\text{A}\right)\left(20.0\text{kA}\times \frac{{10}^3\text{A}}{1\text{kA}}\right)}{2\pi \left(50.0\text{m}\right)}\\ =8\times {10}^{-5}\text{T}\end{array}$

Write the expression for the direction of the magnetic field according to the ampere’s law of the magnetic field.

$\overrightarrow{\text{B}}=\left(8\times {10}^{-5}\text{T}\right)\widehat{\text{j}}$

Write the expression for the velocity vector pointed towards the west.

$\overrightarrow{\text{v}}=\left(300\text{m}/\text{s}\right)\left(-\widehat{\text{i}}\right)$

Write the expression for the force vector on the electron,

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

Here,

$e$ is the charge on the electron.

$\overrightarrow{v}$ is the velocity vector.

$\overrightarrow{B}$ is the magnetic field vector.

Substitute $\left(8\times {10}^{-5}\text{T}\right)\widehat{\text{j}}$ for $\overrightarrow{\text{B}}$, $\left(300\text{m}/\text{s}\right)\left(-\widehat{\text{i}}\right)$ for $\overrightarrow{\text{v}}$ and $-1.60\times {10}^{-19}\text{C}$ for $e$ in the above expression for the force vector on the electron.

$\begin{array}{c}\overrightarrow{\text{F}}=-1.60\times {10}^{-19}\text{C}\times \left(\left(300\text{m}/\text{s}\right)\left(-\widehat{\text{i}}\right)\times \left(8\times {10}^{-5}\text{T}\right)\widehat{\text{j}}\right)\\ =\left(1.60\times {10}^{-19}\text{C}\right)\times \left(2.4\times {10}^{-2}\text{T-m}/\text{s}\right)\widehat{\text{k}}\\ =\left(3.84\times {10}^{-21}\text{N}\right)\widehat{\text{k}}\end{array}$

From the result of the force vector, field vector and velocity vector the as shown below,

Figure (1)

To determine

The vector force lightning stroke exert on the electron.

Answer to Problem 30.16P

The vector force lightning stroke exert on the electron is $\left(3.84\times {10}^{-21}\text{N}\right)\widehat{\text{k}}$.

Explanation of Solution

Given info: The electrons move downward and the positive ions move upwards. The magnitude of the uniform current is $20.0\text{kA}$ at the point of the $50.0\text{m}$ east of the middle of the stroke and the free electrons drift through the air towards the west with the speed of $300\text{m}/\text{s}$.

From the part (a), the vector force on the electron.

$\overrightarrow{F}=\left(3.84\times {10}^{-21}\text{N}\right)\widehat{\text{k}}$

Conclusion:

Therefore, the vector force lightning stroke exert on the electron is $\left(3.84\times {10}^{-21}\text{N}\right)\widehat{\text{k}}$.

To determine

The radius of the electron path.

Answer to Problem 30.16P

The radius of the electron path is $2.135\times {10}^{-5}\text{m}$.

Explanation of Solution

Given info: The electrons move downward and the positive ions move upwards. The magnitude of the uniform current is $20.0\text{kA}$ at the point of the $50.0\text{m}$ east of the middle of the stroke and the free electrons drift through the air towards the west with the speed of $300\text{m}/\text{s}$.

Write the expression for the radius of the electron path.

$\begin{array}{c}Bev=\frac{m_{\text{e}}{v}^2}{r}\\ r=\frac{m_{\text{e}}v}{Be}\end{array}$

Here,

$v$ is the electron drift velocity.

$B$ is the magnetic field.

$e$ is the charge on the electron.

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

Substitute $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{e}}$, $300\text{m}/\text{s}$ for $v$, $8\times {10}^{-5}\text{T}$ for $B$ and $1.6\times {10}^{-19}\text{C}$ for $e$ in the above expression for the value of the radius of the electron path.

$\begin{array}{c}r=\frac{\left(9.11\times {10}^{-31}\text{kg}\right)\left(300\text{m}/\text{s}\right)}{\left(8\times {10}^{-5}\text{T}\right)\left(1.6\times {10}^{-19}\text{C}\right)}\\ =2.135\times {10}^{-5}\text{m}\end{array}$

Conclusion:

Therefore, the radius of the electron path is $2.135\times {10}^{-5}\text{m}$.

To determine

Whether it is a good approximation to model the electron as moving in a uniform field.

Answer to Problem 30.16P

The electron was not moving in a uniform field cause of the magnetic field is varies from the location of the lightning stroke.

Explanation of Solution

Given info: The electrons move downward and the positive ions move upwards. The magnitude of the uniform current is $20.0\text{kA}$ at the point of the $50.0\text{m}$ east of the middle of the stroke and the free electrons drift through the air towards the west with the speed of $300\text{m}/\text{s}$.

From the figure (1) of the part (a), the magnitude of the magnetic field is varies with the distance of the light stroke towards the positive $y$ axis and the radius of the electron is very small and it is negligible in comparison to the distance of $50.0\text{m}$. Thus, the electron was not moving in a uniform field.

Conclusion:

Therefore, the electron was not moving in a uniform field cause of the magnetic field is varies from the location of the lightning stroke.

To determine

The number of the revolutions will the electron complete during the $60.0\text{μs}$.

Answer to Problem 30.16P

The number of the revolutions will the electron complete during $60.0\text{μs}$ is $134$ revolutions.

Explanation of Solution

Given info: The electrons move downward and the positive ions move upwards. The magnitude of the uniform current is $20.0\text{kA}$ at the point of the $50.0\text{m}$ east of the middle of the stroke and the free electrons drift through the air towards the west with the speed of $300\text{m}/\text{s}$. The time of the complete revolution is $60.0\text{μs}$.

From the part (c) the radius of the electron path,

$r=2.135\times {10}^{-5}\text{m}$

Write the expression for the number of the revolution complete by the electron.

$\begin{array}{c}T=n\frac{2\pi r}{v}\\ n=\frac{vT}{2\pi r}\end{array}$

Here,

$T$ is the total time.

$r$ is the radius for the electron path.

$v$ is the speed of the electron.

$n$ is the total number of revolution of the electron.

Substitute $2.135\times {10}^{-5}\text{m}$ for $r$, $300\text{m}/\text{s}$ for $v$ and $60.0\text{μs}$ for $T$ in the above expression for the number of the revolution of the electron.

$\begin{array}{c}n=\frac{\left(300\text{m}/\text{s}\right)\left(60.0\text{μs}\right)}{2\pi \left(2.135\times {10}^{-5}\text{m}\right)}\\ =\frac{\left(300\text{m}/\text{s}\right)\left(60.0\text{μs}\times \frac{{10}^{-6}\text{s}}{1\text{μs}}\right)}{2\pi \left(2.135\times {10}^{-5}\text{m}\right)}\\ =134\end{array}$

Conclusion:

Therefore, the number of the revolutions will the electron complete during $60.0\text{μs}$ is $134$ revolutions.