Chapter 20, Problem 83GP

(a)

To determine

Value of magnetic field to make a beam go undeflected through an electric field.

Answer to Problem 83GP

The value of magnetic fieldis $2.08\times {10}^{-3} \text{T}$

Explanation of Solution

Given:

+x -axis is taken towards right side and +y -axis is taken vertically up.

The speed of electrons $\text{v = 4}{\text{.8×10}}^{\text{6}}\text{ m/s }\left(\text{along +x axis}\right)$

The electric field $\text{E=10000 V/m }\left(\text{along +y axis}\right)$$$

Let B be the magnetic field.

Formula used:

The electron is undeflected when the force due to magnetic field balances the force due to electric field.

  $\begin{array}{l}{\text{F}}_{\text{B}}{\text{=F}}_{\text{E}}\\ \text{qvB = qE}\\ \text{B =}\frac{\text{E}}{\text{v}}\end{array}$

Calculation:

Substituting the values,

  $\begin{array}{l}\text{B =}\frac{\text{10000 V/m}}{\text{4}{\text{.8×10}}^{\text{6}}\text{ m/s}}\\ \text{B = 2}{\text{.08×10}}^{\text{-3}}\text{ T}\end{array}$

  $$

Conclusion:

The value of magnetic field, so the t the electron is undeflected in the electric filed is $2.08\times {10}^{-3} \text{T}$ .

(b)

To determine

The direction of magnetic field.

Answer to Problem 83GP

The magnetic field is out of the plane (+z axis).

Explanation of Solution

Given:

+x -axis is taken towards right side and +y -axis is taken vertically up.

The speed of electrons $\text{v = 4}{\text{.8×10}}^{\text{6}}\text{ m/s }\left(\text{along +x axis}\right)$

The electric field $\text{E=10000 V/m }\left(\text{along +y axis}\right)$$$

Let B be the magnetic field.

Formula used:

The electron is undeflected when the force due to magnetic field balances the force due to electric field.

  $\begin{array}{l}{\text{F}}_{\text{B}}{\text{=F}}_{\text{E}}\\ \text{qvB = qE}\\ \text{B =}\frac{\text{E}}{\text{v}}\end{array}$

Calculation:

Due to the upward electric field, the electric force on the electron acts vertically downwards. Therefore, the magnetic force is in vertically upward direction. Using right hand rule with electron, the magnetic field is directed in the +z direction (out of the plane)

Conclusion:

The magnetic field is out of the plane.

(c)

To determine

The frequency of the circular orbit of the electrons when the electric field is turned off.

Answer to Problem 83GP

The frequency of the circular orbit of the electron when the electric field is $5.8\times {10}^7 \text{Hz}$

Explanation of Solution

Given:

+x -axis is taken towards right side and +y -axis is taken vertically up.

The speed of electrons $\text{v = 4}{\text{.8×10}}^{\text{6}}\text{ m/s }\left(\text{along +x axis}\right)$

The electric field $\text{E=10000 V/m }\left(\text{along +y axis}\right)$$$

The mass of electron $\text{m = 9}{\text{.1×10}}^{\text{-31}}\text{ kg}$

The charge of electron $\text{q }=1.6\times {10}^{-19} \text{C}$

Let f be the frequency of the circular orbit.

Formula used:

  $\text{f }=\frac{\text{qB}}{\text{2πm}}$

  $$

Calculation:

Substituting the values,

  $\begin{array}{l}\text{f =}\frac{\left(\text{1}{\text{.6×10}}^{\text{-19}}\text{C}\right)\left(\text{2}{\text{.08×10}}^{\text{-3}}\text{T}\right)}{\text{2}\left(3.14\right)\left(\text{9}{\text{.1×10}}^{\text{-31}}\text{Kg}\right)}\\ \text{f = 5}{\text{.8×10}}^{\text{7}}\text{ Hz}\end{array}$

Conclusion:

The frequency of the circular orbit of the electron when the electric field is turned off is 5.8×10⁷ Hz.