(a) At what speed will a proton move in a circular path of the same radius as the electron in Exercise \({\rm{22}}{\rm{.12}}\)? (b) What would the radius of the path be if the proton had the same speed as the electron? (c) What would the radius be if the proton had the same kinetic energy as the electron? (d) The same momentum?

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 operates on other currents and magnetic moments.

In exercise\({\rm{22}}{\rm{.12}}\), the radius that was obtained was\({\rm{4}}{\rm{.27 m}}\). To get the velocity which is needed by a proton to follow the a path having the same radius.

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

Solving it for\({\rm{v}}\)where the value of\({\rm{m}}\)is the proton's mass and the value of\({\rm{B}}\)is the strength of the magnetic field of the earth.

\({\text{V = }}\frac{{{\text{rqB}}}}{{\text{m}}}\)

\({\text{ = }}\frac{{{\text{4}}{\text{.27 m \times 1}}{\text{.6 \times 1}}{{\text{0}}^{{\text{ - 19}}}}{\text{C \times 1}}{{\text{0}}^{{\text{ - 5}}}}{\text{ T}}}}{{{\text{ 1}}{\text{.67 \times 1}}{{\text{0}}^{{\text{ - 27}}}}{\text{ kg}}}}\)

\({\text{ = 4}}{\text{.1 \times 1}}{{\text{0}}^{\text{3}}}{\text{ m/s V}}\)

Hence, the velocity of the proton must have: \({\rm{4}}{\rm{.1 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{ m/s}}\).

The proton has a speed of 7.5 x 10° m/s.

The radius of its path is then calculated with the help of the equation\({\rm{r = }}\frac{{{\rm{mv}}}}{{{\rm{qB}}}}\).

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

\({\text{ = }}\frac{{{\text{1}}{\text{.67 \times 1}}{{\text{0}}^{{\text{ - 27}}}}{\text{ kg \times 7}}{\text{.5 \times 1}}{{\text{0}}^{\text{6}}}{\text{ m/s}}}}{{{\text{ 1}}{\text{.6 \times 1}}{{\text{0}}^{{\text{ - 19}}}}{\text{C \times 1}}{{\text{0}}^{{\text{ - 5}}}}{\text{ T}}}}\)

\({\text{ = 7}}{\text{.8 x 1}}{{\text{0}}^{\text{3}}}{\text{ m}}\)

Therefore, the radius of the path the proton will be made is: \({\rm{7}}{\rm{.8 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{ m}}\).

(c)

The kinetic energy for the electron is:\({{\rm{(K}}{\rm{.E)}}_{\rm{e}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{m}}}{{\rm{m}}_{\rm{e}}}{\rm{v}}_{\rm{e}}^{\rm{2}}\)

The value of \({{\rm{m}}_{\rm{e}}}\) is the electron's mass and the value of \({{\rm{v}}_{\rm{e}}}\) is the electron's velocity. Then, to equate this kinetic energy such that the proton get’s its velocity from the equation: \({{\rm{(K}}{\rm{.E)}}_{\rm{p}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{{\rm{m}}_{\rm{p}}}{\rm{v}}_{\rm{p}}^{\rm{2}}\).

When getting the velocity, we can then get the radius of the path the proton makes with the help of the equation:\({\rm{r = }}\frac{{{{\rm{m}}_{\rm{p}}}{{\rm{v}}_{\rm{p}}}}}{{{{\rm{m}}_{\rm{p}}}{\rm{v}}}}\).

\({{\text{(K}}{\text{.E)}}_{\text{e}}}{\text{ = (K}}{\text{.E}}{{\text{)}}_{\text{p}}}\)

\(\frac{{\text{1}}}{{\text{2}}}{{\text{m}}_{\text{e}}}{\text{v}}_{\text{e}}^{\text{2}}{\text{ = }}\frac{{\text{1}}}{{\text{2}}}{{\text{m}}_{\text{p}}}{\text{v}}_{\text{p}}^{\text{2}}\)

\({{\text{v}}_{\text{p}}}{\text{ = }}\sqrt {\frac{{{{\text{m}}_{\text{e}}}{\text{v}}_{\text{e}}^{\text{2}}}}{{{{\text{m}}_{\text{p}}}}}} \)

\({\text{ = }}\sqrt {\frac{{{\text{9}}{\text{.11 \times 1}}{{\text{0}}^{{\text{ - 31}}}}{\text{kg \times (7}}{\text{.5 \times 1}}{{\text{0}}^{\text{6}}}{\text{ m/s)}}}}{{{\text{ 1}}{\text{.67 \times 1}}{{\text{0}}^{{\text{ - 27}}}}{\text{ kg}}}}} \)

\({\text{ = 1}}{\text{.8 \times 1}}{{\text{0}}^{\text{5}}}{\text{m/s}}\)

\({\text{r = }}\frac{{{{\text{m}}_{\text{p}}}{{\text{v}}_{\text{p}}}}}{{{\text{qB}}}}\)

\({\text{ = }}\frac{{{\text{ 1}}{\text{.67 \times 1}}{{\text{0}}^{{\text{ - 27}}}}{\text{ kg \times 1}}{\text{.8 \times 1}}{{\text{0}}^{\text{5}}}{\text{ m/s}}}}{{{\text{ 1}}{\text{.6 \times 1}}{{\text{0}}^{{\text{ - 19}}}}{\text{C \times 1}}{{\text{0}}^{{\text{ - 5}}}}{\text{ T}}}}\)

\({\text{ = 188m}}\)

Hence, the radius of the path proton will make: \({\rm{188 m}}\).\({\rm{188 m}}\)

(d)

Calculating the momentum for the electron:\({{\rm{P}}_{\rm{e}}}{\rm{ = }}{{\rm{m}}_{\rm{e}}}{{\rm{v}}_{\rm{e}}}\).

The value of\({{\rm{m}}_{\rm{e}}}\)is the electron's mass and the value of\({{\rm{v}}_{\rm{e}}}\)is the electron's velocity. Then, equating this momentum such that the proton get’s its velocity with the help of the equation:\({{\rm{P}}_{\rm{p}}}{\rm{ = }}{{\rm{m}}_{\rm{p}}}{{\rm{v}}_{\rm{p}}}\).

When getting the velocity, we can then get the radius of the path the proton makes with the help of the equation:\({\rm{r = }}\frac{{{{\rm{m}}_{\rm{p}}}{{\rm{v}}_{\rm{p}}}}}{{{{\rm{m}}_{\rm{p}}}{\rm{v}}}}\).

\({{\text{P}}_{\text{e}}}{\text{ = }}{{\text{P}}_{\text{p}}}\)

\({{\text{m}}_{\text{e}}}{{\text{v}}_{\text{e}}}{\text{ = }}{{\text{m}}_{\text{p}}}{{\text{v}}_{\text{p}}}\)

\({{\text{v}}_{\text{p}}}{\text{ = }}\frac{{{{\text{m}}_{\text{e}}}{{\text{v}}_{\text{e}}}}}{{{{\text{m}}_{\text{p}}}}}\)

\({\text{ = }}\frac{{{\text{9}}{\text{.11 \times 1}}{{\text{0}}^{{\text{ - 31}}}}{\text{kg \times (7}}{\text{.5 \times 1}}{{\text{0}}^{\text{6}}}{\text{ m/s)}}}}{{{\text{1}}{\text{.67 \times 1}}{{\text{0}}^{{\text{ - 27}}}}{\text{ kg}}}}\)

\({\text{r = }}\frac{{{{\text{m}}_{\text{p}}}{{\text{v}}_{\text{p}}}}}{{{\text{qB}}}}\)

\({\text{ = }}\frac{{{\text{ 1}}{\text{.67 \times 1}}{{\text{0}}^{{\text{ - 27}}}}{\text{ kg \times 4}}{\text{.1 \times 1}}{{\text{0}}^{\text{3}}}{\text{ m/s}}}}{{{\text{ 1}}{\text{.6 \times 1}}{{\text{0}}^{{\text{ - 19}}}}{\text{C \times 1}}{{\text{0}}^{{\text{ - 5}}}}{\text{ T}}}}\)

\({\text{ = 4}}{\text{.27 m}}\)

Therefore, the radius of the proton will make: \({\rm{4}}{\rm{.27 m}}\).