Tambayoyi Misali Game da Ƙarfin da ke Kan Kuɗin Canjawa
Pendahuluan
Ilimin kimiyyar lissafi shine nazarin abubuwan da suka faru na halitta, gami da ƙarfin da ke aiki akan abubuwa. Wani batu mai ban sha'awa da ake yawan tattaunawa akai shine ƙarfin da ke aiki akan cajin motsi, musamman a cikin mahallin filayen lantarki da maganadisu. Ƙarfin da ke aiki akan cajin motsi a cikin filin lantarki ko maganadisu ana kiransa da ƙarfin Lorentz. Wannan labarin zai tattauna misalai da yawa na matsaloli da tattaunawarsu game da ƙarfin da ke aiki akan cajin motsi.
Ƙarfin Lorentz
Ƙarfin Lorentz haɗuwa ne na ƙarfin lantarki da ƙarfin maganadisu da ke aiki akan caji da ke motsawa a cikin filin lantarki da filin maganadisu. A lissafi, ana iya bayyana ƙarfin Lorentz (F) ta hanyar lissafi:
\[ \mathbf{F} = q (\mathbf{E} + \mathbf{v} \times \mathbf{B}) \]
Ina:
– \ ( \mathbf {F} \) shine ƙarfin Lorentz
– \( q \) shine cajin
– \( \mathbf{E} \) shine filin lantarki
– \( \mathbf{v} \) shine saurin cajin
– \( \mathbf{B} \) shine filin maganadisu
Da wannan lissafi, za mu iya yin nazarin ƙarfin da ke aiki akan caji da ke motsawa a cikin filin lantarki da filin maganadisu.
Tambayoyi da Tattaunawa Samfura
Tambaya ta 1: Ƙarfin Caji a Filin Wutar Lantarki
Tambaya:
Cajin da ke da kyau \( q = 2 \sau 10^{-6} \, C \) yana cikin filin lantarki iri ɗaya \( E = 5 \sau 10^4 \, N/C \) wanda aka nuna zuwa dama. Lissafa ƙarfin da ke aiki akan cajin.
Tattaunawa:
Don caji a filin lantarki ba tare da kasancewar filin maganadisu ba, ƙarfin Lorentz ya ƙunshi ɓangaren ƙarfin lantarki kawai:
\[ \mathbf{F} = q \mathbf{E} \]
Tare da \( q = sau 2 10^{-6} \, C \) da \( \mathbf{E} = sau 5 10^4 \, N/C \):
\[ \mathbf{F} = (sau 2 10^{-6} \, C) \sau (sau 5 10^4 \, N/C) \]
\[ \mathbf{F} = 0.1 \, N \]
Alkiblar ƙarfin \( \mathbf{F} \) tana daidai da alkiblar filin lantarki saboda cajin yana da kyau. Don haka, ƙarfin da ke aiki akan cajin shine 0.1 N zuwa dama.
Tambaya ta 2: Ƙarfin caji a filin maganadisu
Tambaya:
Cajin mara kyau \( q = -3 \sau 10^{-6} \, C \) yana motsawa da gudu \( \mathbf{v} = 2 \sau 10^3 \, m/s \) tare da axis ɗin x a cikin filin maganadisu iri ɗaya \( \mathbf{B} = 0.5 \, T \) wanda aka nuna tare da axis ɗin z. Lissafa ƙarfin da ke aiki akan cajin.
Tattaunawa:
Ga wani caji da ke motsawa a cikin filin maganadisu ba tare da kasancewar filin lantarki ba, ƙarfin Lorentz ya ƙunshi kawai abubuwan da ke cikin ƙarfin maganadisu:
\[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \]
Tare da \( q = -3 \sau 10^{-6} \, C \), \( \mathbf{v} = 2 \sau 10^3 \, m/s \) a cikin alkiblar x, da kuma \( \mathbf{B} = 0.5 \, T \) a cikin alkiblar z:
Lissafi \( \mathbf{v} \times \mathbf{B} \):
\[ \mathbf{v} = sau 2 10^3 \, m/s \, \hat{i} \]
\[ \mathbf{B} = 0.5 \, T \, \hat{k} \]
\[ \mathbf{v} \times \mathbf{B} = (2 \sau 10^3 \, m/s \, \hat{i}) \sau (0.5 \, T \, \hat{k}) \]
\[ \mathbf{v} \times \mathbf{B} = 2 \times 10^3 \, m/s \times 0.5 \, T \, \hat{i} \times \hat{k} \]
\[ \hat{i} \times \hat{k} = -\hat{j} \]
\[ \mathbf{v} \times \mathbf{B} = – (1 \times 10^3 \, T \cdot m/s) \, \hat{j} \]
Saboda haka, ƙarfin maganadisu:
\[ \mathbf{F} = q \mathbf{v} \times \mathbf{B} \]
\[ \mathbf{F} = (-3 \sau 10^{-6} C) \sau ( -10^3 \, T \cdot m/s \, \hat{j}) \]
\[ \mathbf{F} = 3 \ lokuta 10 ^ {-3} \, N \, \ hat {j} \]
\[ \mathbf{F} = 0.003 \, N \, \hat{j} \]
Alkiblar ƙarfin \( \mathbf{F} \) tana zuwa ga axis mai kyau na y. Don haka, ƙarfin da ke aiki akan cajin shine 0.003 N sama (a cikin alkiblar y mai kyau).
Tambaya ta 3: Ƙarfin da ake caji a filayen lantarki da maganadisu
Tambaya:
Cajin mai kyau \( q = 1.5 \sau 10^{-6} \, C \) yana motsawa da sauri \( \mathbf{v} = 4 \sau 10^3 \, m/s \) a cikin y-direction a cikin filin lantarki \( \mathbf{E} = 3 \sau 10^4 \, N/C \) a cikin x-direction, da kuma filin maganadisu \( \mathbf{B} = 0.2 \, T \) a cikin z-direction. Lissafa ƙarfin da ke aiki akan cajin.
Tattaunawa:
Jimlar ƙarfin Lorentz:
\[ \mathbf{F} = q (\mathbf{E} + \mathbf{v} \times \mathbf{B}) \]
Da farko, lissafta \( \mathbf{v} \times \mathbf{B} \):
\[ \mathbf{v} = sau 4 10^3 \, m/s \, \hat{j} \]
\[ \mathbf{B} = 0.2 \, T \, \hat{k} \]
\[ \mathbf{v} \times \mathbf{B} = (4 \sau 10^3 \, m/s \, \hat{j}) \sau (0.2 \, T \, \hat{k}) \]
\[ \mathbf{v} \times \mathbf{B} = 4 \times 10^3 \, m/s \times 0.2 \, T \, \hat{j} \times \hat{k} \]
\[ \hat{j} \times \hat{k} = \hat{i} \]
\[ \mathbf{v} \times \mathbf{B} = (0.8 \times 10^3 \, T \cdot m/s) \, \hat{i} \]
\[ \mathbf{v} \times \mathbf{B} = 800 \, T \cdot m/s \, \hat{i} \]
Sannan, ƙarfin lantarki:
\[ q \mathbf{E} = (1.5 \sau 10^{-6} \, C) \sau (sau 3 \sau 10^4 \, N/C \, \hat{i}) \]
\[q \mathbf{E} = 0.045 \, N \, \hat{i} \]
Ƙarfin maganadisu:
\[ q (\mathbf{v} \times \mathbf{B}) = (1.5 \times 10^{-6} \, C) \times (800 \, T \cdot m/s \, \hat{i}) \]
\[q (\mathbf{v} \times \mathbf{B}) = 0.0012 \, N \, \hat{i} \]
Cikakken salo:
\[ \mathbf{F} = q \mathbf{E} + q (\mathbf{v} \times \mathbf{B}) \]
\[ \mathbf{F} = 0.045 \, N \, \hat{i} + 0.0012 \, N \, \ hat {i} \]
\[ \mathbf{F} = 0.0462 \, N \, \hat{i} \]
Don haka, jimlar ƙarfin da ke aiki akan cajin shine 0.0462 N zuwa dama (axis mai kyau na x).
Kammalawa
Ƙarfin da ke kan caji da ke motsawa a cikin filayen lantarki da maganadisu ya dogara sosai akan alkibla da girman kowane fili da kuma saurin da nau'in caji. Ta hanyar misalan tambayoyi da tattaunawa da ke sama, ana fatan masu karatu za su iya fahimtar yadda ake amfani da ƙa'idar ƙarfin Lorentz a cikin yanayi daban-daban. Wannan fahimta ba wai kawai tana da mahimmanci a ka'ida ba har ma a aikace-aikace a fannoni na fasaha da kimiyya kamar ƙira injinan lantarki, fahimtar abin da ke faruwa a cikin aurora, da aikin barbashi a cikin na'urorin haɓaka barbashi.
