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}*10^(-20)\text{ N}`

The strength of the magnetic field is

`\begin{gathered} B\text{ = 0.13 mT} \\ =\text{ 0.13 }*10^(-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=(6.1*10^(-20))/(0.13*10^(-3)*4012*\sin 47^(\circ)) \\ =1.599\text{ }*10^(-19)\text{ C} \end{gathered}`