Find the radius of curvature of the path of a \({\rm{25}}{\rm{.0 - MeV}}\) proton moving perpendicularly to the \({\rm{1}}{\rm{.20 - T}}\) field of a cyclotron.

Magnetic fields mediate a class of physical properties known as magnetism. A magnetic field is created by electric currents and the magnetic moments of elementary particles, which operate on other currents and magnetic moments.

The equation used to get the radius of the curvature is:

\({\rm{r = }}\frac{{{\rm{mv}}}}{{{\rm{qB}}}}\)

Firstly, getting the velocity of the proton using its kinetic energy with the help of the equation is:

\({\rm{K}}{\rm{.E = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}{{\rm{v}}^{\rm{2}}}\)

The radius is then obtained as:

\(\begin{aligned}{\rm{v }} &= \sqrt {\frac{{{\rm{2E}}}}{{\rm{m}}}} \\ &= \sqrt {\frac{{{\rm{2}} \times {\rm{25}}{\rm{.0}}\;{\rm{MeV}}\left( {\frac{{{\rm{1}}{{\rm{0}}^{\rm{6}}}\;{\rm{eV}}}}{{{\rm{MeV}}}}} \right) \times \left( {{\rm{1}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}\;{\rm{C}}} \right)}}{{{\rm{1}}{\rm{.67 \times 1}}{{\rm{0}}^{{\rm{ - 27}}}}\;{\rm{kg}}}}} \\ &= {\rm{6}}{\rm{.92}} \times {\rm{1}}{{\rm{0}}^{\rm{7}}}\;{\rm{m/s}}\end{aligned}\)

\({\rm{r }} = \frac{{{\rm{mv}}}}{{{\rm{qB}}}}\)

\({\text{ = }}\frac{{\left( {{\text{1}}{\rm{.67 \times 1}}{{\text{0}}^{{\text{ - 27}}}}\;{\text{kg}}} \right){\rm{ \times }}\left( {{\text{6}}{\rm{.92 \times 1}}{{\text{0}}^{\text{7}}}\;{\text{m/s}}} \right)}}{{\left( {{\text{1}}{\rm{.6 \times 1}}{{\text{0}}^{{\text{ - 19}}}}\;{\text{C}}} \right){\rm{ \times }}\left( {{\text{1}}{\text{.20T}}} \right)}}\)

\({\rm{ = 0}}{\rm{.60 m}}\)

Therefore, the radius of the curvature is \({\rm{0}}{\rm{.60 m}}\).