Chapter 20, Problem 76GP

(a)

To determine

Approximate deflection of the electron beam .

Answer to Problem 76GP

Deflection of the electron beam near the center of a TV screen due to the Earth’s magnetic field by $2.0\text{ kV}$ is $8\text{ mm}$ .

Explanation of Solution

Given:

Earth’s magnetic field is $5\times {10}^{-5}\text{ T}$ . CRT screen is $22\text{ cm}$ wide. Electron is moving with an electric potential of $2.0\text{ kV}$ .

Formula used:

According to the conservation of energy principle, total energy remains conserved that is potential energy converts in kinetic energy.

  $\begin{array}{l}{\text{PE}}_{\text{initial}}={\text{KE}}_{\text{final}}\\ \text{eV = }\frac{1}{2}{\text{mv}}_{\text{x}}^2\end{array}$

Calculation:

Since electron velocity is aligned horizontally, is perpendicular to the Earth’s magnetic field. Considering magnetic force of the electron is so small that electron follows the same path in the entire trajectory as shown in the figure below,

  

This results in the constant acceleration. Since electron’s voltage gained initially turns to be the kinetic energy of the electrons in the CRT. Thus, velocity of the electron is

  ${\text{v}}_{\text{x}}\text{=}\sqrt{\frac{2\text{eV}}{\text{m}}}$

Time in the field will be

  $\Delta {\text{x}}_{\text{field}}={\text{v}}_{\text{x}}{\text{t}}_{\text{field}}\Rightarrow {\text{t}}_{\text{field}}=\frac{\Delta {\text{x}}_{\text{field}}}{{\text{v}}_{\text{x}}}$

  $\begin{array}{l}{\text{F}}_{\text{y}}={\text{qv}}_{\text{x}}{\text{B}}_{\text{Earth}}={\text{ma}}_{\text{y}}\Rightarrow {\text{a}}_{\text{y}}=\frac{{\text{qv}}_{\text{x}}{\text{B}}_{\text{Earth}}}{\text{m}}\\ {\text{a}}_{\text{y}}=\frac{\text{e}\sqrt{\frac{2\text{eV}}{\text{m}}}{\text{B}}_{\text{Earth}}}{\text{m}}=\sqrt{\frac{2{\text{e}}^3\text{V}}{{\text{m}}^3}}{\text{B}}_{\text{Earth}}\end{array}$

Since electrons are starting at rest. Using second equation of motion,

  $\begin{array}{}$ \Delta \text{y = 0}\text{.t+}\frac{1}{2}{\text{a}}_{\text{y}}{\text{t}}^2=\frac{1}{2}{\text{a}}_{\text{y}}{\text{t}}^2$$ \Delta \text{y =}\frac{1}{2}\sqrt{\frac{2{\text{e}}^3\text{V}}{{\text{m}}^3}}{\text{B}}_{\text{Earth}}{\left(\frac{\Delta {\text{x}}_{\text{field}}}{{\text{v}}_{\text{x}}}\right)}^2$$ \Delta \text{y =}\frac{1}{2}\sqrt{\frac{2{\text{e}}^3\text{V}}{{\text{m}}^3}}{\text{B}}_{\text{Earth}}{(\Delta {\text{x}}_{\text{field}})}^2\frac{\text{m}}{2\text{eV}}$$ \Delta \text{y =}\sqrt{\frac{\text{e}}{\text{8mV}}}{\text{B}}_{\text{Earth}}{(\Delta {\text{x}}_{\text{field}})}^2$$ \Delta \text{y =}\sqrt{\frac{1.6\times {10}^{-19}\text{ C}}{\text{8(9}\text{.11}\times {10}^{-31}\text{ kg)(2000 V)}}}\text{(0}.5\times {10}^{-4}\text{ T)}{(0.22\text{ m})}^2$$ \Delta \text{y =}8.02\times {10}^{-3}\text{ m}\approx \text{8 mm}$\end{array}$

Conclusion:

Thus, deflection of the electron beam near the center of a TV screen due to the Earth’s magnetic field by $2.0\text{ kV}$ is $8\text{ mm}$ .

(b)

To determine

Approximate deflection of the electron beam .

Answer to Problem 76GP

Deflection of the electron beam near the center of a TV screen due to the Earth’s magnetic field by $30\text{ kV}$ is $\text{2 mm}$ .

Explanation of Solution

Given:

Earth’s magnetic field is $5\times {10}^{-5}\text{ T}$ . CRT screen is $22\text{ cm}$ wide. Electron is moving with an electric potential of $30\text{ kV}$ .

Formula used:

According to the conservation of energy principle, total energy remains conserved that is potential energy converts in kinetic energy.

  $\begin{array}{l}{\text{PE}}_{\text{initial}}={\text{KE}}_{\text{final}}\\ \text{eV = }\frac{1}{2}{\text{mv}}_{\text{x}}^2\end{array}$

Calculation:

Deflection of the electron beam is

  $\begin{array}{l}$ \Delta \text{y =}\sqrt{\frac{\text{e}}{\text{8mV}}}{\text{B}}_{\text{Earth}}{(\Delta {\text{x}}_{\text{field}})}^2$$ \Delta \text{y =}\sqrt{\frac{1.6\times {10}^{-19}\text{ C}}{\text{8(9}\text{.11}\times {10}^{-31}\text{ kg)(30000 V)}}}\text{(0}.5\times {10}^{-4}\text{ T)}{(0.22\text{ m})}^2$\Delta \text{y =}2.07\times {10}^{-3}\text{ m}\approx 2\text{ mm}\end{array}$

Since the deflection is so small that the horizontal distance traveled and the assumptions are verified.

Conclusion:

Thus, deflection of the electron beam near the center of a TV screen due to the Earth’s magnetic field by $30\text{ kV}$ is $\text{2 mm}$ .