Tusaale su'aal dood ah oo ku saabsan baaxadda EMF-da la kiciyey
Pendahuluan
Xoogga korantada ee la kiciyay (EMF) waa dhacdo jireed oo uu ogaaday Michael Faraday sanadkii 1831. Faraday wuxuu ogaaday in goobta birlabta ee isbeddelaysa ee ku jirta wareegga ama gariiradda gudbiyaha ay kicin karto koronto. Dhacdadan waxaa loo yaqaan kicinta elektaroonigga ah waana saldhigga tignoolajiyada casriga ah ee badan, sida matoorada, transformers-ka, iyo matoorada korontada. Maqaalkani wuxuu sharxi doonaa dhowr dhibaato oo tusaale ah wuxuuna ka hadli doonaa baaxadda EMF-ga la kiciyay, kaas oo aad waxtar u yeelan doona gaar ahaan ardayda baranaysa fiisigiska korontada iyo birlabta.
Fikradda Aasaasiga ah ee EMF-ga la kiciyay
Kahor inta aynaan guda gelin dhibaatada tusaalaha ah, waa fikrad wanaagsan in dib loo eego fikradda aasaasiga ah ee emf-ka la kiciyay. Sharciga Faraday wuxuu sheegayaa in emf-ka la kiciyay ee wareegga wadista uu si toos ah ugu dhigmayo heerka isbeddelka qulqulka birlabta ee wareegga. Xisaab ahaan, sharciga Faraday waxaa loo samayn karaa sidan soo socota:
\[ \mathcal{E} = – \frac{d\Phi}{dt} \]
Halkee:
– \( \mathcal{E} \) waa EMF-ga la soo saaray (volts)
– \( \Phi \) waa qulqulka birlabta (Weber, Wb)
– \( t \) waa waqti (labaad, s)
Calaamadda taban ee isle'egta waxay ka tarjumaysaa Sharciga Lenz kaas oo sheegaya in jihada hadda la kiciyay ay abuuri doonto goob birlab ah oo ka soo horjeeda isbeddelka qulqulka birlabka ee sababay.
Su'aalo iyo Doodo Tusaale ah
Su'aal 1aad: Hal Xidid
Su'aal: Hal gariirad oo leh bedka dusha sare ee 0.02 m² ayaa la dhigayaa goob birlab oo isku mid ah oo cabbirkeedu yahay 0.5 T, oo ku toosan dusha sare ee gariiradda. Goobta birlabtu si lama filaan ah ayay eber ugu noqotaa 0.1 ilbiriqsi gudahood. Xisaabi baaxadda EMF ee la kiciyay ee ka soo baxda gariiradda.
Dood:
Waa la ogyahay:
– Bedka dusha sare ee gariiradda \( A = 0.02 \, \text{m}^2 \)
– Cabbirka goobta birlabta \( B = 0.5 \, \text{T} \)
– Waqtiga isbeddelka goobta \( \Delta t = 0.1 \, \text{s} \)
Isbeddelka qulqulka birlabta:
\[ \Delta \Phi = B \times A \]
\[ \Delta \Phi = 0.5 \, \text{T} \times 0.02 \, \text{m}^2 \]
\[ \Delta \Phi = 0.01 \, \text{Wb} \]
Baaxadda EMF-ga la soo saaray:
\[ \mathcal{E} = – \frac{\Delta \Phi}{\Delta t} \]
\[ \mathcal{E} = – \frac{0.01 \, \text{Wb}}{0.1 \, \text{s}} \]
\[ \mathcal{E} = -0.1 \, \text{V} \]
Sababtoo ah waxa la weydiinayo waa baaxadda EMF (iyada oo aan la fiirin calaamadda):
\[ |\mathcal{E}| = 0.1 \, \ qoraal {V} \]
Su'aal 2: Coil oo leh N Turns
Su'aal: Gariirku wuxuu ka kooban yahay 100 wareeg, bedka dusha sarena waa 0.03 m². Gariirkani wuxuu ku jiraa goob birlabeed oo isku mid ah oo ah 0.4 T oo ku toosan gariiradda. Haddii goobta birlabeedku ay kordho 1.2 T 0.4 ilbiriqsi gudahood, xisaabi celceliska emf-ka la soo saaray ee lagu soo saaray gariiradda.
Dood:
Waa la ogyahay:
– Tirada wareegyada \( N = 100 \)
– Bedka dusha sare ee gariiradda \( A = 0.03 \, \text{m}^2 \)
– Isbeddel ku yimid goobta birlabta \( \Delta B = 1.2 \, \text{T} – 0.4 \, \text{T} \)
– Waqtiga isbeddelka goobta \( \Delta t = 0.4 \, \text{s} \)
Isbeddelka qulqulka birlabta halkii wareeg:
\[ \Delta \Phi = A \times \Delta B \]
\[ \Delta \Phi = 0.03 \, \text{m}^2 \times (1.2 \, \text{T} – 0.4 \, \text{T}) \]
\[ \Delta \Phi = 0.03 \, \qoraal{m}^2 \times 0.8 \, \qoraal{T} \]
\[ \Delta \Phi = 0.024 \, \text{Wb} \]
Isbeddelka guud ee qulqulka wareegyada N:
\[ \Delta \Phi_{\text{total}} = N \times \Delta \Phi \]
\[ \Delta \Phi_{\text{total}} = 100 \times 0.024 \, \text{Wb} \]
\[ \Delta \Phi_{\text{total}} = 2.4 \, \text{Wb} \]
Baaxadda EMF-ga la soo saaray:
\[ \mathcal{E} = – \frac{\Delta \Phi_{\text{wadarta}}}{\Delta t} \]
\[ \mathcal{E} = – \frac{2.4 \, \text{Wb}}{0.4 \, \text{s}} \]
\[ \mathcal{E} = -6 \, \text{V} \]
Sababtoo ah waxa la weydiinayo waa baaxadda EMF (iyada oo aan la fiirin calaamadda):
\[ |\mathcal{E}| = 6 \, \ qoraal {V} \]
Su'aal 3: Ku dhaqaajinta gariiradda goob birlab ah
Su'aal: Gariir leydi ah oo leh 50 wareeg, 4 cm dherer ah, iyo 2 cm ballac ah, ayaa lagu dhejiyaa goob birlabeed oo isku mid ah oo 0.3 T ah oo barbar socota dhererka gariiradda. Haddii gariiraddu ka baxdo goobta birlabeed xawaare joogto ah oo ah 5 cm/s, waa maxay EMF-ga la kiciyay ee gariiradda?
Dood:
Waa la ogyahay:
– Tirada wareegyada \( N = 50 \)
– Dhererka \( l = 0.04 \, \qoraalka{m} \)
– Ballaca \( w = 0.02 \, \qoraalka{m} \)
– Cabbirka goobta birlabta \( B = 0.3 \, \text{T} \)
– Xawaaraha bixitaanka ee goobta birlabta \( v = 0.05 \, \text{m/s} \)
Xoogga elektaroonigga ah ee ku jira gariiradda ku dhex socota goob birlab ah waxaa lagu bixiyaa:
\[ \mathcal{E} = B lv \]
Maadaama gariiraddu ay leedahay wareegyo N ah, wadarta guud ee EMF ee la kiciyay waa:
\[ \mathcal{E}_{\text{total}} = NB lv \]
\[ \mathcal{E}_{\text{total}} = 50 \times 0.3 \, \text{T} \times 0.04 \, \text{m} \times 0.05 \, \text{m/s} \]
\[ \mathcal{E}_{\text{total}} = 50 \jeer 0.0006 \, \text{V} \]
\[ \mathcal{E}_{\text{ total}} = 0.03 \, \text{V} \]
Gabagabo
Doodda ku saabsan dhibaatooyinka tusaalaha ah ee kor ku xusan waxay muujinaysaa sida emf-ka la kiciyay looga xisaabin karo isbeddellada ku yimaada qulqulka magnetic-ka ama dhaqdhaqaaqa gariiradda ee goobta birlabta. Fikraddani waa aasaas u ah electromagnetism-ka waxayna leedahay codsiyo badan oo ku saabsan tiknoolajiyada. Fahmidda sida saxda ah ee loo xisaabiyo emf-ka la kiciyay iyada oo loo marayo dhibaatooyinka wax ku oolka ah waxay si weyn uga caawin doontaa ardayda inay bartaan agabkan oo ay ku dabaqaan dhinacyo kala duwan.
