Question

Calculate The charge of a meson that experiences a force of 6.1 x 10*(-20)N while moving at 4012m/s through a magnetic field of strength 0.13mT at an angle of 47

Answer 1

Given that the force is

$F=\text{ 6.1}\times10^{-20}\text{ N}$

The strength of the magnetic field is

$\begin{gathered} B\text{ = 0.13 mT} \\ =\text{ 0.13 }\times10^{-3}\text{ T} \end{gathered}$

The speed is v = 4012 m/s.

The angle is

$\theta=47^{\circ}$

We need to calculate the charge of the meson.

The formula to calculate charge is

$\begin{gathered} F=\text{ BqvSin}\theta \\ q=\frac{F}{Bv\text{ sin }\theta} \end{gathered}$

Substituting the values, the charge will be

$\begin{gathered} q=\frac{6.1\times10^{-20}}{0.13\times10^{-3}\times4012\times\sin 47^{\circ}} \\ =1.599\text{ }\times10^{-19}\text{ C} \end{gathered}$