Chapter 24.2, Problem 28PP

To determine

Magnitude of force on the ionized particle.

Answer to Problem 28PP

  $F=1.728\times {10}^{-13}\text{N}$

Explanation of Solution

Given:

Charge on the triply ionized ion is, $q=3\times \left(1.6\times {10}^{-19}\right)\text{C}$

Velocity, $v=9\times {10}^6\text{m/s}$

Strength of magnetic field, $B=4\times {10}^{-2}\text{T}$

Angle between $v\text{and }B$ is $\theta =90°$

Formula Used:

The force on a moving charge (Lorentz force) is,

  $F=qvB\sin \theta$

Where, $q$ is the moving charge, $v$ is velocity, $B$ is magnetic field strength and $\theta$ is the minimum angle between the direction of $v\text{and }B$ .

Calculation:

On putting the given values in above relation,

  $\begin{array}{c}F=\left(3\times 1.6\times {10}^{-19}\right)\times \left(9\times {10}^6\right)\times 0.04\times \sin 90°\\ =1.728\times {10}^{-13}\text{N}\end{array}$

Conclusion:

Therefore, the force on the ionized particle is $F=1.728\times {10}^{-13}\text{N}$ .