91Ó°ÊÓ

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: \({\rm{4}}{\rm{.27 m}}\).