Chapter 24.2, Problem 29PP

To determine

Velocity of the ionized particle.

Answer to Problem 29PP

  $v=4.2\times {10}^6\text{m/s}$

Explanation of Solution

Given:

Charge on the singly ionized particle is, $q=1\times \left(1.6\times {10}^{-19}\right)\text{C}$

Force experienced is, $F=4.1\times {10}^{-13}\text{N}$

Strength of magnetic field, $B=0.61\text{T}$

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

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,

  $\begin{array}{c}4.1\times {10}^{-13}=\left(1\times 1.6\times {10}^{-19}\right)\times \left(v\right)\times 0.61\times \sin {90}^o\\ v=\frac{4.1\times {10}^{-13}}{\left(1.6\times {10}^{-19}\right)\times 0.61\times \sin {90}^o}\\ =4.2\times {10}^6\text{m/s}\end{array}$

Conclusion:

Therefore, the velocity of the ionized particle is $v=4.2\times {10}^6\text{m/s}$ .