Chapter 24, Problem 79A

To determine

The number of elementary charges in a particle.

Answer to Problem 79A

The number of elementary charges carried by the particle is $2$.

Explanation of Solution

Given:

A force of $5.78\times {10}^{-16}\text{ N}$ is experienced by a particle traveling with a velocity $5.65\times {10}^4\text{ m/s}$

perpendicular to a uniform magnetic field of strength $3.20\times {10}^{-2}\text{ T}$.

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 given and based on the remaining parameters, the charge contained in the particle is to be calculated. Having found the charge, the number of elementary charges can eb obtained as,

  $q=ne$

Here, n denotes the number of elementary charges and 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 $B=3.20\times {10}^{-2}\text{ T}$ , particle velocity $v=5.65\times {10}^4\text{ m/s}$ and force $F=5.78\times {10}^{-16}\text{ N}$ , the charge corresponding to angle $\theta =90°$ is given by,

  $\begin{array}{c}q=\frac{F}{vB\left(\sin \theta \right)}\\ =\frac{5.78\times {10}^{-16}}{5.65\times {10}^4\times 3.20\times {10}^{-2}\left(\sin 90°\right)}\\ =0.319\times {10}^{-18}\text{ C}\end{array}$

Now, the number of elementary charges can be calculated as,

  $\begin{array}{c}q=n\times 1.6\times {10}^{-19}\\ n=\frac{q}{1.6\times {10}^{-19}}\\ =\frac{0.319\times {10}^{-18}}{1.6\times {10}^{-19}}\\ =1.998\\ \simeq 2\end{array}$

The above result has been rounded to the nearest whole number since the number of elementary charges is to be a whole number.

Conclusion:

Therefore, the number of elementary charges of the unknown particle is 2.