Chapter 24, Problem 76A

To determine

The magnitude of force acting on an electron.

Answer to Problem 76A

The force on each electron is $24\times {10}^{-15}\text{ N}$

Explanation of Solution

Given:

A beam of electrons traveling with a velocity $2.5\times {10}^6\text{ m/s}$ perpendicular to a uniform magnetic field of strength $6.0\times {10}^{-2}\text{ T}$ .

Formula Used:

The force acting on a particle is dependent on the strength of magnetic field, velocity of motion, charge and the angle which the velocity vector makes with the magnetic field. It is mathematically expressed as,

  $F=qvB\left(\sin \theta \right)$

Here, the force is to be calculated for a single electron. Hence, the charge can be written as,

  $q=e$

Here, e denotes the charge of one electron which is $e=1.6\times {10}^{-19}\text{ C}$ .

Calculation:

For the given strength of magnetic field $B=6.0\times {10}^{-2}\text{ T}$ , particle velocity $v=2.5\times {10}^6\text{ m/s}$ , force corresponding to angle $\theta =90°$ is given by,

  $\begin{array}{c}F=evB\left(\sin \theta \right)\\ =1.6\times {10}^{-19}\times 2.5\times {10}^6\times 6.0\times {10}^{-2}\left(\sin 90°\right)\\ =24\times {10}^{-15}\text{ N}\end{array}$

Conclusion:

Therefore, the force on a single electron in the beam is $24\times {10}^{-15}\text{ N}$ .