Question

A cosmic ray electron moves at\({\rm{7}}{\rm{.50 \times 1}}{{\rm{0}}^{\rm{6}}}{\rm{ m/s}}\)perpendicular to the Earth’s magnetic field at an altitude where field strength is\({\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ T}}\). What is the radius of the circular path the electron follows?

Short answer

The circular path the electron follows has obtained a path of measurement \({\rm{4}}{\rm{.27 m}}\).

Step-by-step solution

Effect of magnetic field on the charged particle

When a charged particle moves in the presence of a magnetic field, a force acts on the mong charge particle given by Fleming’s right-hand rule and which deviates the path of the moving charged particle.

Evaluating the radius of the circular path that the electron follows

To obtain the circular path that the electron follows.

The equation used is:\({\rm{r = }}\frac{{{\rm{mv}}}}{{{\rm{qB}}}}\).

Solving it for the value of\({\rm{r}}\)where the value of\({\rm{m}}\)is the mass of the electron.

\(\begin{array}{c}{\rm{r}} = \frac{{{\rm{mv}}}}{{{\rm{qB}}}}\\ = \frac{{\left( {{\rm{9}}{\rm{.1 \times 1}}{{\rm{0}}^{{\rm{ - 31}}}}\;{\rm{kg}}} \right){\rm{ \times }}\left( {{\rm{7}}{\rm{.50 \times 1}}{{\rm{0}}^{\rm{6}}}{\rm{ m/s}}} \right)}}{{\left( {{\rm{1}}{\rm{.6 \times 10}}{}^{{\rm{ - 19}}}\;{\rm{C}}} \right){\rm{ \times }}\left( {{\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ T}}} \right)}}\\ = {\rm{4}}{\rm{.27 m}}\end{array}\)

Therefore, the circular path the electron follows has obtained a path of measurement \({\rm{4}}{\rm{.27 m}}\).