Edits to “Statement”, “Solution”
en/10.2.12.md
+2 −2
| @@ -1,6 +1,6 @@ | |||
| ### Statement | |||
| − | $10.2.12.$ Find the drift velocity of the electron and proton in the gravity field and the Earth's magnetic field, the induction of which is equal to 0.7 | ||
| + | $10.2.12.$ Find the drift velocity of the electron and proton in the gravity field and the Earth's magnetic field, the induction of which is equal to $0.7 \cdot 10^{-4} \text{ T}$. The magnetic field is perpendicular to the gravity field. | ||
| ### Solution | |||
| Drift velocity is achieved when particle's path is unaffected respect to force fields that acting over it. So, absolute values of these forces are equal and their directions are opposed. Hence, | |||
| $$mg = q~v_d~B$$ | |||
| @@ -10,7 +10,7 @@Solution | |||
| $$v_d = \frac{mg}{qB}$$ | |||
| − | Now, considering the following data, calculations for each particle are possible: mass of electron $m_e = 9.1 \cdot 10^{-31}\text{ kg}$, mass of proton $m_p = 1.672 \cdot 10^{-27}\text{ kg}$, absolute value of charge $q = 1.6 \cdot 10^{-19}\text{ C}$ and acceleration due to gravity $g = 9.8 \text{m/s}^2$. Calculating, | ||
| + | Now, considering the following data, calculations for each particle are possible: mass of electron $m_e = 9.1 \cdot 10^{-31}\text{ kg}$, mass of proton $m_p = 1.672 \cdot 10^{-27}\text{ kg}$, absolute value of charge $q = 1.6 \cdot 10^{-19}\text{ C}$ and acceleration due to gravity $g = 9.8 \text{ m/s}^2$. Calculating, | ||
| $$v_e = 7.9625 \cdot 10^{-7} \text{ m/s}$$ | |||
| $$v_p = 1.423625 \cdot 10^{-3} \text{ m/s}$$ | |||
| #### Answer | |||
| $$v_e \simeq 8 \cdot 10^{-7} m/s, v_p \simeq 1.4 \cdot 10^{-3} \text{ m/s}$$ | |||
| unchanged lines 5 | |||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $10.2.12.$ Find the drift velocity of the electron and proton in the gravity field and the Earth's magnetic field, the induction of which is equal to 0.7 |
$10.2.12.$ Find the drift velocity of the electron and proton in the gravity field and the Earth's magnetic field, the induction of which is equal to $0.7 \cdot 10^{-4} \text{ T}$. The magnetic field is perpendicular to the gravity field. | ||
| ### Solution | ### Solution | ||
| Drift velocity is achieved when particle's path is unaffected respect to force fields that acting over it. So, absolute values of these forces are equal and their directions are opposed. Hence, | Drift velocity is achieved when particle's path is unaffected respect to force fields that acting over it. So, absolute values of these forces are equal and their directions are opposed. Hence, | ||
| $$mg = q~v_d~B$$ | $$mg = q~v_d~B$$ | ||
| @@ -10,7 +10,7 @@Solution | |||
| $$v_d = \frac{mg}{qB}$$ | $$v_d = \frac{mg}{qB}$$ | ||
| Now, considering the following data, calculations for each particle are possible: mass of electron $m_e = 9.1 \cdot 10^{-31}\text{ kg}$, mass of proton $m_p = 1.672 \cdot 10^{-27}\text{ kg}$, absolute value of charge $q = 1.6 \cdot 10^{-19}\text{ C}$ and acceleration due to gravity $g = 9.8 \text{m/s}^2$. Calculating, | Now, considering the following data, calculations for each particle are possible: mass of electron $m_e = 9.1 \cdot 10^{-31}\text{ kg}$, mass of proton $m_p = 1.672 \cdot 10^{-27}\text{ kg}$, absolute value of charge $q = 1.6 \cdot 10^{-19}\text{ C}$ and acceleration due to gravity $g = 9.8 \text{ m/s}^2$. Calculating, | ||
| $$v_e = 7.9625 \cdot 10^{-7} \text{ m/s}$$ | $$v_e = 7.9625 \cdot 10^{-7} \text{ m/s}$$ | ||
| $$v_p = 1.423625 \cdot 10^{-3} \text{ m/s}$$ | $$v_p = 1.423625 \cdot 10^{-3} \text{ m/s}$$ | ||
| #### Answer | #### Answer | ||
| $$v_e \simeq 8 \cdot 10^{-7} m/s, v_p \simeq 1.4 \cdot 10^{-3} \text{ m/s}$$ | $$v_e \simeq 8 \cdot 10^{-7} m/s, v_p \simeq 1.4 \cdot 10^{-3} \text{ m/s}$$ | ||
| unchanged lines 5 | |||