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Q16 PE
Expert-verifiedWhat radius circular path does an electron travel if it moves at the same speed and in the same magnetic field as the proton in Exercise \({\rm{22}}{\rm{.13}}\)?
The radius of the circular path of the electron is obtained as: \({\rm{4}}{\rm{.36}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ m}}\).
In the presence of a magnetic field, a force acts on the moving charged particle. The direction of this force can be obtained using Fleming’s right-hand rule and can be numerically expressed as
\({\rm{F}} = {\rm{q}}\left( {\vec v \times \vec B} \right)\)
where\({\rm{q,}}\;{\rm{v,}}\;{\rm{and B}}\) are given as the amount of charge, the velocity of the charged particle, and the strength of the magnetic field.
To obtain the circular path of the electron.
The equation used is: \({\rm{r = }}\frac{{{\rm{mv}}}}{{{\rm{qB}}}}\).
The value of \({\rm{m}}\) is the mass of the electron.
The value of \({\rm{q}}\) is the charge of the electron.
The value of \({\rm{e}}\) and \({\rm{B}}\) is the strength of the magnetic field obtained in the exercise \({\rm{22}}{\rm{.13}}\), which is \({\rm{0}}{\rm{.98 T}}\).
Now, solve the previous equation for the value of \({\rm{r}}\) which will be the radius of the path.
\(\begin{aligned}{\rm{r}} &= \frac{{{\rm{mv}}}}{{{\rm{eB}}}}\\ &= \frac{{\left( {{\rm{9}}{\rm{.1 \times 1}}{{\rm{0}}^{{\rm{ - 31}}}}\;{\rm{kg}}} \right){\rm{ \times }}\left( {{\rm{7}}{\rm{.5 \times 1}}{{\rm{0}}^{\rm{7}}}{\rm{ m/s}}} \right)}}{{\left( {{\rm{1}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}\;{\rm{C}}} \right) \times \left( {{\rm{0}}{\rm{.98 T}}} \right)}}\\ &= {\rm{4}}{\rm{.36}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ m}}\end{aligned}\)
Therefore, the radius of the circular path of the electron is: \({\rm{4}}{\rm{.36}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ m}}\).
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