Chapter 3, Problem 1P

(a)

To determine

Verify the equation for the magnitude of electric field.

Answer to Problem 1P

It is proved that the magnitude of electric field is given by $\left(r/2\right)\left(dB/dt\right)$.

Explanation of Solution

Write the given equation for the relation between electric field and magnetic flux.

    ${\displaystyle \oint E}.ds=-\frac{d{\varphi }_B}{dt}$        (I)

Here, $E$ is the electric field induced by the electron over the surface $s$ with a circumference $2\pi r$, ${\varphi }_B$ is the magnetic flux which changes with time $t$.

Write the equation for the magnetic flux.

    $\begin{array}{c}{\varphi }_B=BA\\ =B\pi r^2\end{array}$        (II)

Here, ${\varphi }_B$ is the magnetic flux, $B$ is the magnetic flux, $A$ is the area and $r$ is the distance as in the given figure.

Conclusion:

Substitute equation (II) in equation (I) and also substitute $2\pi r$ for $s$ in equation (I).

    $\begin{array}{c}E2\pi r=\pi r^2\left(\frac{dB}{dt}\right)\\ E=\left(\frac{r}{2}\right)\left(\frac{dB}{dt}\right)\end{array}$        (III)

Therefore, it is proved that the magnitude of electric field is given by $\left(r/2\right)\left(dB/dt\right)$.

(b)

To determine

The change in speed of electron.

Answer to Problem 1P

The change in speed of the electron is $erB/2m_e$.                                                      

Explanation of Solution

Write the equation for the force acting on the electron.

    $F=Eq$        (IV)

Here, $F$ is the force acting on the electron, $E$ is the electric field induced by the electron and $e$ is the charge of electron.       

Write the equation for the force acting on the electron in term of the acceleration of electron.

     $Fdt=m_{e}dv$        (V)

Here, $F$ is the force acting on the electron, $m_e$ is the mass of electron and $dv$ is the acceleration of the electron.

Substitute equation (III) in equation (IV) and compare it with equation (V).

     $\begin{array}{c}Fdt=\left(\frac{r}{2}\right)\left(\frac{dB}{dt}\right)dt=m_{e}dv\\ dv=\left(\frac{re}{2m_e}\right)dB\end{array}$        (VI)

Conclusion:

Integrate equation (VI) by giving appropriate limits.

      $\begin{array}{c}{\displaystyle \underset{v}{\overset{v+\Delta v}{\int }}dv}=\left(\frac{re}{2m_e}\right){\displaystyle \underset{0}{\overset{B}{\int }}dB}\\ \Delta v=\frac{erB}{2m_e}\end{array}$        (VII)

Therefore, the change in speed is $erB/2m_e$.                                                       

(c)

To determine

The fractional change in frequency.

Answer to Problem 1P

The fractional change in frequency is $2.3\times {10}^{-5}$.

Explanation of Solution

Write the equation for the change in frequency of electron.

    $\Delta \omega =\frac{\Delta v}{r}$        (VIII)

Here, $\Delta \omega$ is the change in frequency, $\Delta v$ is the change in velocity and $r$ is the distance as in the given figure.

Substitute equation (VII) in equation (VIII).

    $\Delta \omega =\frac{eB}{2m_e}$        (IX)

Write the equation for the frequency of electron.

    $\omega =\frac{2\pi c}{\lambda }$        (X)

Here, $\omega$ is the frequency of electron, $c$ is the speed of light and $\lambda$ is the wavelength of emission line.       

Conclusion:

Substitute $1.6\times {10}^{-19}\text{C}$ for $e$, $1\text{T}$ for $B$ and $9.1\times {10}^{-31}\text{kg}$ for $m_e$ in equation (IX).

    $\begin{array}{c}\Delta \omega =\frac{\left(1.6\times {10}^{-19}\text{C}\right)\left(1\text{T}\right)}{2\left(9.1\times {10}^{-31}\text{kg}\right)}\\ =8.8\times {10}^{10}\text{rad/sec}\end{array}$               

Substitute $3\times {10}^8\text{m/s}$ for $c$ and $500\times {10}^{-9}\text{m}$ for $\lambda$ in equation (X).

     $\begin{array}{c}\omega =\frac{2\pi \left(3\times {10}^8\text{m/s}\right)}{500\times {10}^{-9}\text{m}}\\ =3.8\times {10}^{15}\text{rad/sec}\end{array}$

Find the ratio $\Delta \omega /\omega$.

    $\begin{array}{c}\frac{\Delta \omega }{\omega }=\frac{8.8\times {10}^{10}\text{rad/sec}}{3.8\times {10}^{15}\text{rad/sec}}\\ =2.3\times {10}^{-5}\end{array}$

Therefore, the fractional change in frequency is $2.3\times {10}^{-5}$.

(d)

To determine

The explanation for lines at ${\omega }_0$, ${\omega }_0+\Delta \omega$ and ${\omega }_0-\Delta \omega$. 

Answer to Problem 1P

The ${\omega }_0$ line and its components are justified. 

Explanation of Solution

The plane of electrons is parallel to $B$ for the ${\omega }_0$ line. The magnetic flux thus will always be zero. As a result, there will be no force on the electrons and there will be no $\Delta v$ for the electrons.

The ${\omega }_0+\Delta \omega$ line will have a $\Delta v$ given by equation (VII).

The ${\omega }_0-\Delta \omega$ line will have the same magnitude for $E$, $B$, and $\Delta v$ will be same in equation (VII) but in the opposite direction.

Conclusion:

Therefore, the ${\omega }_0$ line and its components are justified.