Chapter 30, Problem 16P

(a)

To determine

The sketch showing the various vectors involved.

Explanation of Solution

The sketch showing the various vectors involved is shown in the figure below.

Figure-(1)

Here, $\overrightarrow{B}$ is the magnetic field vector, $\overrightarrow{F}$ is the force vector and $v$ is the velocity vector and $r$ is the distance between the middle of the vertical lightening stoke and the given point.

(b)

To determine

The vector force by lightening stoke on the electron.

Answer to Problem 16P

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

Explanation of Solution

Write the expression to obtain the magnetic field along the conductor.

    $B=\frac{{\mu }_{0}i}{2\pi r}$                                                                                                                (I)

Here, $B$ is the magnetic field, ${\mu }_0$ is the magnetic permittivity, $i$ is the current and $r$ is the distance between the conductor and the point where the magnetic field is to be determined.

Write the expression to obtain the vector force on the electron.

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

Here, $\overrightarrow{F}$ is the vector force, $q$ is the charge on the electron, $\overrightarrow{v}$ is the velocity of the electron and $\overrightarrow{B}$ is the magnetic field vector.

Conclusion:

Substitute $20.0\text{kA}$ for $i$, $4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}$ for ${\mu }_0$ and $50.0\text{m}$ for $r$ in equation (I) to calculate $B$.

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

The magnetic field in vector form.

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

Substitute $-1.60\times {10}^{-19}\text{C}$ for $q$, $\left(300\text{m}/\text{s}\right)\left(-\widehat{i}\right)$ for $\overrightarrow{v}$ and $\left(8\times {10}^{-5}\text{T}\right)\widehat{j}$ for $\overrightarrow{B}$ in equation (II) to calculate $\overrightarrow{F}$.

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

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

(c)

To determine

The radius of the electron’s path.

Answer to Problem 16P

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

Explanation of Solution

Write the expression to obtain the radius of the electron’s path.

    $r=\frac{mv}{Bq}$

Here, $r$ is the radius of the electron’s path, $m$ is the mass of the electron, $v$ is the velocity of the electron, $B$ is the magnetic field and $q$ is the charge on the electron.

Conclusion:

Substitute $1.6\times {10}^{-19}\text{C}$ for $q$, $9.11\times {10}^{-31}\text{kg}$ for $m$, $300\text{m}/\text{s}$ for $v$, $8\times {10}^{-5}\text{T}$ for $B$ in the above equation to calculate $r$.

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

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

(d)

To determine

Weather it is good approximation to model the electron is moving in the uniform magnetic field.

Explanation of Solution

It is not good approximation to model the electron is moving in the uniform magnetic field because the magnitude of the magnetic field is not same through-out that is the magnetic field varies with the distance from the lightening stoke.

(e)

To determine

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

Answer to Problem 16P

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

Explanation of Solution

Write the expression to obtain the number of revolutions the electron will complete during $60.0\mu \text{s}$ duration.

    $N=\frac{T_{\text{D}}}{T}$                                                                                                                (III)

Here, $N$ is the number of revolutions the electron will complete during $60.0\mu \text{s}$ duration, $T_{\text{D}}$ is the total duration and $T$ is the time to complete one revolution.

Write the expression to obtain the time to complete one revolution.

    $T=\frac{2\pi r}{v}$

Here $T$ is the time to complete one revolution, $r$ is the radius of the path of the electron and $v$ is the velocity of the electron.

Conclusion:

Substitute $2.135\times {10}^{-5}\text{m}$ for $r$ and $300\text{m}/\text{s}$ for $v$ in the above equation to calculate $T$.

    $\begin{array}{c}T=\frac{2\pi \left(2.135\times {10}^{-5}\text{m}\right)}{\left(300\text{m}/\text{s}\right)}\\ =4.47\times {10}^{-7}\text{s}\end{array}$

Substitute $4.47\times {10}^{-7}\text{s}$ for $T$ and $60.0\mu \text{s}$ for $T_{\text{D}}$ in equation (III) to calculate $N$.

    $\begin{array}{c}N=\frac{60.0\mu \text{s}}{\left(4.47\times {10}^{-7}\text{s}\right)}\\ =\frac{60.0\mu \text{s}\times \frac{1\text{s}}{{10}^6\mu \text{s}}}{\left(4.47\times {10}^{-7}\text{s}\right)}\\ =134\end{array}$

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