Chapter 24, Problem 93A

To determine

Force on the beta particle.

Answer to Problem 93A

  $F=-2.4\times {10}^{-12}\text{N}$

Explanation of Solution

Given:

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

Velocity, $v=2.5\times {10}^7\text{m/s}$

Strength of magnetic field, $B=0.6\text{T}$

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,

  $\begin{array}{c}F=\left(-1.6\times {10}^{-19}\right)\times \left(2.5\times {10}^7\right)\times 0.6\times \sin 90°\\ =-2.4\times {10}^{-12}\text{N}\end{array}$

Conclusion:

Therefore, the force on the beta particle is $F=-2.4\times {10}^{-12}\text{N}$ .