Question

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.

Short answer

\({p_1} + {p_2} = p_1^\prime + p_2^\prime \)

Step-by-step solution

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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:

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

\(\begin{aligned}r &= \frac{{mv}}{{qB}}\\ &= \frac{{\left( {1.67 \times {{10}^{ - 27}}\;kg} \right) \times \left( {6.92 \times {{10}^7}\;m/s} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}\;C} \right) \times 1.20\;T}}\\ &= 0.60\;m\end{aligned}\)

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

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