Chapter 20, Problem 71GP

To determine

The magnitude and direction for the velocity of an electron.

Answer to Problem 71GP

The velocity of an electron is $v=1.1\times {10}^{-6}\text{m}/\text{s}$ .

The electron moves towards west.

Explanation of Solution

Given:

The earth magnetic field strength is $B=0.50\times {10}^{-4}\text{T}$ .

The mass of electron is $m=9.109\times {10}^{-31}\text{kg}$

The charge of electron is $q=1.602\times {10}^{-19}\text{C}$

The acceleration due to gravity is $g=9.8{\text{m}/\text{s}}^2$

Calculation:

Sketch the direction as shown in Figure 1.

  

The equation of magnetic force must be equal in magnitude to the weight of the electron is,

  $\begin{array}{c}qvB=mg\\ v=\frac{mg}{eB}\\ =\frac{\left(9.1\times {10}^{-31}\right)\left(9.8\right)}{1.6\times {10}^{-19}\times 0.5\times {10}^{-4}}\\ =1.1\times {10}^{-6}\text{m}/\text{s}\end{array}$

The magnetic force must point upwards , and so by the right hand rule and the negative charge of the electron, the electron must be moving west.

Conclusion:

Therefore, the velocity of an electron is $v=1.1\times {10}^{-6}\text{m}/\text{s}$ . Also, the electron must be moving west.