Electromagnetic Induction: Class 12 Physics Chapter 6 Hinglish mein

Physics Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 28 Sep 2026 · Updated 28 Sep 2026

105 min
3 simulations
16 practice questions
29 MCQs

Is chapter mein aap kya seekhenge

Samjho ki Faraday aur Henry ke experiments kaise dikhate hain ki sirf changing magnetic flux hi emf induce karta hai
Kisi surface se guzarne wala magnetic flux calculate karo, aur Faraday's law se uske rate of change se bana emf nikaalo
Lenz's law se induced currents ki direction pata karo aur ise energy conservation se samjhao
Sliding rod ka aur ek end ke around ghoomti rod ka motional emf derive karo aur use karo
Mutual aur self-inductance define karo, long solenoids ke liye inhe derive karo aur inductor mein stored energy nikaalo
Samjhao ki ac generator sinusoidal emf kaise banata hai, aur uski peak value aur frequency calculate karo

NCERT Physics Class 12, Chapter 6, pp. 154–176 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

6.1 Introduction

Metro card ko gate reader par tap karo, phone ko wireless charger par rakho, ya dekho ki pedal tez chalane par cycle ke dynamo wala lamp kaise aur bright hota hai. Yeh sab isliye kaam karte hain kyunki ek changing magnetic field circuit mein charges ko dhakel sakta hai. 1820 tak Oersted electricity aur magnetism ke link ka ek hissa dikha chuke the: current paas rakhi compass needle ko ghuma deta hai. Agla obvious sawaal iska ulta tha (Fig. 6.1). Agar currents magnetic fields banate hain, to kya magnetic fields currents bana sakte hain?

3/3Magnet ko coil mein dhakelne se lamp jal jaata hai: 1831 mein mila ulta effect.

Fig. 6.1Electricity aur magnetism dono taraf se jude hain: current magnetic field banata hai, aur changing magnetic field current banata hai.

Do panels aamne-saamne. Left panel par 1820 likha hai; usmein ek copper wire compass ke upar north se south chal rahi hai. Right panel par 1831 likha hai; usmein ek coil chhote lamp se judi hai aur coil ke paas ek bar magnet rakha hai. Step 1: wire mein current nahi hai, isliye compass needle north dikhati hai; magnet ruka hai aur lamp band hai. Step 2: wire mein north ki taraf current behta hai, jise arrow dikhata hai, aur compass needle ghoomkar west ki taraf mud jaati hai. Step 3: needle mudi hui hi rehti hai, aur ab magnet ko coil mein dhakela jaata hai, jise arrow dikhata hai, aur lamp jal uthta hai.

Jawab 1831–32 mein aaya, London mein Michael Faraday aur Albany, New York mein Joseph Henry se, alag-alag, aur usmein ek twist tha. Steady field kuch nahi karta; closed coil se guzarne wala field jab change hota hai, tab woh coil mein current chalata hai. Is process ko electromagnetic induction (वैद्युतचुंबकीय प्रेरण) kehte hain. Aaj lagbhag har power station, transformer, induction cooktop aur electric-guitar pickup mein yahi kaam kar raha hai.

Is chapter mein hum is idea ko precise banayenge. Hum magnetic flux define karenge, induced emf ke size ke liye Faraday's law aur uski direction ke liye Lenz's law samjhenge, aur dekhenge ki dono energy conservation se kaise nikalte hain. Phir hum fields mein chalte conductors, coils ka inductance aur unmein stored energy padhenge, aur aakhir mein woh ac generator jo tumhare ghar ki mains supply deta hai.

Timeline

Hilti compass se power grid tak

1820
Oersted: current compass ko ghumata hai
Jab bhi paas ki wire mein current behta, compass needle ghoom jaati; isse pata chala ki currents magnetic fields banate hain.
1831
Faraday ki iron ring
Ek iron ring par do coils lapeti gayi; ek coil mein thoda sa current sirf usi pal aaya jab doosri coil ka current on ya off kiya gaya.
1831–32
Henry ki coils
Albany mein powerful electromagnets ke saath kaam karte hue Joseph Henry ne induction ko alag se dekha aur self-induction discover kiya.
1834
Lenz ka rule
Emil Lenz ne woh rule diya jo har induced current ki direction tay karta hai.
1880s
Kaam ki ac machines
Nikola Tesla aur doosron ne alternating-current generators aur motors banaye, jinse door tak power supply possible hua.
1902
Shivanasamudra mein hydroelectricity
Asia ke pehle bade hydroelectric stations mein se ek ne Karnataka ki Kaveri river se Kolar gold fields tak power bhejna shuru kiya.

Warm-up

2 mein se 1

1820 mein Oersted ne kya dekha?

Chalta magnet coil mein current chalata haiCoil ke paas ruka hua magnet usmein current behta rakhta haiCurrent-carrying wire paas ki compass needle ko ghuma deti hai

Answer dekho

Answer (C) Current-carrying wire paas ki compass needle ko ghuma deti hai

Oersted ne paaya ki currents magnetic fields banate hain, jo compass needle ko ghumate hain. Ulta effect, yaani chalta magnet current chalaye, Faraday aur Henry ne das saal baad discover kiya.

6.2 Faraday aur Henry ke experiments

Teen simple experiments mein poori kahani hai. Har ek mein coil ek sensitive centre-zero galvanometer se judi hai, jiski needle current ki direction ke hisaab se left ya right ghoomti hai. Dekho needle ko kya hilata hai — aur utna hi important, kya nahi hilata.Is chapter ki har coil insulated wire se lapeti gayi hai, taaki current turns ke saath-saath chale, unke across short-circuit na ho.

Experiment 6.1NCERT Experiment 6.1 · p. 155

Magnet aur coil

Tumhe chahiye

insulated copper wire ke 300–500 turns wali ek coil
ek strong bar magnet (neodymium magnet sabse achha rehta hai)
ek centre-zero galvanometer, ya lowest current range par digital multimeter
do connecting leads

Pehle guess karo

Tum magnet ka N pole coil mein dhakelte ho aur phir use andar hi rok kar rakhte ho. Jab magnet ruka hua hai, tab needle kya karti hai?

Jab tak magnet andar hai, needle ghoomi hui rehti haiNeedle doosri side ghoomkar wahin ruk jaati haiNeedle zero par wapas aakar wahin rehti hai

Coil ko galvanometer se jodo aur use bench par rakho, axis horizontal rakho.
Magnet ka N pole tezi se coil mein dhakelo aur needle dekho.
Magnet ko kuch seconds tak coil ke andar rok kar rakho.
Magnet ko bahar kheencho, pehle dheere aur phir tezi se.
Ab S pole aage rakh kar dohrao. Aakhir mein magnet ko fixed rakho aur coil ko uski taraf le jao.

Table 6.1Needle ne kya kiya?

Tum kya karte ho	Needle: left, right ya kuch nahi?	Swing bada ya chhota?
N pole andar dhakela
Magnet andar ruka hua
N pole dheere se bahar kheencha
N pole tezi se bahar kheencha
S pole andar dhakela
Coil ko ruke magnet par le gaye

Tumhari entries apne aap save hoti hain.

Abhi nahi kar sakte? Simulation try karo · Fig. 6.3

2/5N pole coil ki taraf chalta hai: needle ek side ghoomti hai.

Fig. 6.2Needle sirf tab hilti hai jab magnet coil ke relative chal raha ho.NCERT Fig. 6.1 · p. 155

Right side par ek horizontal coil hai jo neeche ke centre-zero galvanometer se judi hai, aur left mein usi axis par ek bar magnet hai jiska N pole coil ki taraf hai. Step 1: magnet door hai aur ruka hua hai, needle zero par hai. Step 2: magnet coil ki taraf chalta hai, jise arrow dikhata hai, aur needle right ghoomti hai. Step 3: magnet ka N end coil ke andar ruka hai aur needle wapas zero par hai. Step 4: magnet coil se door jaata hai aur needle left ghoomti hai. Step 5: magnet ko tezi se andar dhakela jaata hai, peeche speed lines hain, aur needle right mein kahin zyada ghoomti hai.

SimulationMagnet aur coil

Ek bar magnet jise tum centre-zero galvanometer se judi coil ki taraf, uske andar aur usse door drag kar sakte ho. Field lines aur flux readout dikhate hain ki coil se kitna field guzar raha hai, aur needle sirf tab ghoomti hai jab woh flux change ho raha ho.

Fig. 6.3Magnet ko drag karo aur flux readout aur galvanometer ko saath-saath dekho.

Try karo

Magnet ko coil ke andar rok kar rakho. Flux readout aur needle kya dikhate hain?
Magnet ko dheere se bahar kheencho, phir tezi se. Kisse bada swing milta hai, aur kya flux ka total change alag hai?
Turns ki sankhya badhao aur wahi push dohrao. Swing kaise badalta hai?

Experiment 6.2NCERT Experiment 6.2 · p. 155

Magnet ki jagah coil

Tumhe chahiye

Experiment 6.1 wali coil aur galvanometer
ek doosri, chhoti coil jo pehli coil ke paas ya andar fit ho jaaye
ek ya do 1.5 V cells

Pehle guess karo

Chhoti coil mein cell se steady current beh raha hai. Tum use badi coil ke paas rok kar rakhte ho. Galvanometer kya dikhata hai?

Bilkul deflection nahiEk steady deflectionDheere-dheere badhta deflection

Chhoti coil ko cell se jodo taaki usmein steady current bahe.
Chhoti coil ko badi coil ki taraf le jao, phir usse door, aur needle dekho.
Chhoti coil ko fixed rakho aur is baar badi coil ko hilao.

Experiment 6.3NCERT Experiment 6.3 · p. 156

Kuch nahi hilta, phir bhi needle jhatka khaati hai

Tumhe chahiye

ek common axis par end to end rakhi do coils
pehli coil ke liye ek cell aur tapping key (push switch)
doosri coil ke liye galvanometer
ek soft-iron rod jo dono coils se guzarne jitni lambi ho

Pehle guess karo

Is set-up mein kuch nahi hilta. Key dabane par doosri coil wali needle kya karegi?

Kuch nahi, kyunki kuch hil nahi rahaThoda jhatka khaakar zero par laut aayegiGhoomkar tab tak ghoomi rahegi jab tak key dabi hai

Coil P ko key ke through cell se jodo, aur coil S ko galvanometer se. Coils ek doosre ko touch nahi karni chahiye.
Key dabao aur needle dekho. Key ko kuch seconds dabaye rakho.
Key chhodo aur phir se dekho.
Soft-iron rod ko dono coils mein se guzaaro aur dohrao.

Table 6.2Needle ki readings

Pal	Needle kya karti hai
Key abhi dabayi
Key dabi hui
Key abhi chhodi
Iron rod lagakar key dabayi

Tumhari entries apne aap save hoti hain.

2/5P ka current zero se badhta hai, aur S ko ek chhota sa jhatka milta hai.

Fig. 6.4Coil S sirf tab respond karti hai jab coil P ka current change ho raha ho.NCERT Fig. 6.3 · p. 156

Left mein coil P cell aur key se judi hai; right mein coil S galvanometer se judi hai. Unke neeche ek blue graph P ka current time ke saath dikhata hai aur ek orange graph galvanometer ki reading dikhata hai. Step 1: key khuli hai, dono graphs zero par flat hain aur needle zero par hai. Step 2: key dabayi jaati hai, P ka current tezi se badhta hai aur G ki reading mein ek chhota upward spike aata hai; needle right ghoomti hai. Step 3: key dabi hai, P ka current flat hai aur G ki reading zero hai. Step 4: key chhodi jaati hai, P ka current zero par gir jaata hai aur G mein chhota downward spike aata hai; needle left ghoomti hai. Step 5: ek iron rod dono coils se guzarti hai aur G ki reading ke spikes kahin lambe ho jaate hain.

Teeno experiments mein common kya hai? Pehle do mein magnet ya current-carrying coil chalti hai, isliye source ke paas aane ya door jaane par coil par field change hota hai. Teesre mein source apni jagah rehta hai, lekin key dabane ya chhodne par uski strength change hoti hai. Needle kabhi us field par respond nahi karti jo bas wahan maujood hai, chahe kitna bhi strong ho; woh sirf coil se guzarne wale changing field par respond karti hai. Section 6.3 ‘coil se guzarne wale field’ ko ek measurable quantity banata hai.

6.3 Magnetic flux

Steady baarish mein ek hoop (chhalla) aage badha kar pakdo. Agar baarish seedhi neeche gir rahi hai aur hoop level hai, to woh har second sabse zyada boondein pakadta hai. Hoop ko tilt karo to kam boondein pakdega; seedha khada karo to lagbhag koi boond andar se nahi guzregi. Kitni boondein aayengi, yeh baarish kitni tez hai, hoop kitna bada hai aur kitna tilted hai, is par depend karta hai. Magnetic flux (चुंबकीय फ्लक्स) bilkul isi teen-part wale tareeke se naapta hai ki surface se kitna magnetic field guzar raha hai.

Definition

Magnetic flux

Uniform field $𝐁$ mein area $A$ wali flat surface ke liye magnetic flux $Φ_{B} = 𝐁 \cdot 𝐀 = B A \cos θ$ hai, jahan $θ$ , $𝐁$ aur surface ke normal ke beech ka angle hai.

Example $0.5 T$ field ke bilkul saamne rakhe $0.02 m^{2}$ area wale loop ka $Φ_{B} = 0.01 W b$ hota hai.

Yahan $𝐀$ surface ka area vector hai: iska magnitude area hai aur yeh normal ki direction mein point karta hai, bilkul waise hi jaise Chapter 1 mein electric flux ki definition mein tha. Fig. 6.5 teen cases dikhata hai. Jab loop field ke bilkul saamne ho ( $θ = 0$ ), to flux sabse zyada, $B A$ , hota hai. Itna tilt karo ki $θ = 60^{\circ}$ ho, to aadha hi pakadta hai. Edge-on ( $θ = 90^{\circ}$ ) koi field line andar se nahi guzarti, aur flux zero ho jaata hai.

2/3 $θ = 60^{\circ}$ : aadhi lines guzarti hain, $Φ_{B} = \frac{1}{2} B A$ .

Fig. 6.5Flux

𝐁

aur loop ke area vector ke beech ke angle par depend karta hai.NCERT Fig. 6.4 · p. 157

Das horizontal field lines left se right chal rahi hain. Beech mein ek flat circular loop hai, jo edge-on ek patle green ellipse jaisa dikhta hai, aur uske area vector ke liye ek red arrow hai. Step 1: loop field ke saamne hai, uska area vector field ki direction mein hai, aur aath highlighted lines usse guzarti hain. Step 2: loop itna tilt hai ki area vector field se 60 degree banata hai; ek angle arc bana hai aur sirf chaar lines guzarti hain. Step 3: loop field ke saath-saath pada hai, uska area vector 90 degree par seedha upar hai, aur koi field line usse nahi guzarti.

Magnetic flux (uniform field, flat surface)

Φ_{B} = 𝐁 \cdot 𝐀 = B A \cos θ

(6.1)

jahan

$θ$: $𝐁$ aur surface ke normal ke beech ka angle

Agar field jagah-jagah badalta hai, ya surface curved hai, to surface ko chhote patches $d 𝐀_{i}$ mein baant do, har patch itna chhota ki uspar field $𝐁_{i}$ practically uniform ho. Total flux saare patches ke contributions ka sum hai:

Φ_{B} = 𝐁_{1} \cdot d 𝐀_{1} + 𝐁_{2} \cdot d 𝐀_{2} + \dots = \sum_{all} 𝐁_{i} \cdot d 𝐀_{i}

(6.2)

Flux ek scalar hai, aur negative bhi ho sakta hai: jab field chune gaye normal ke against surface paar karta hai, to $\cos θ < 0$ . Iski SI unit weber (Wb) hai, jahan $1 W b = 1 T m^{2}$ .

Quick check

3 mein se 1

0.020 m^{2}

area wala ek flat loop

0.50 T

ke uniform field mein hai. Field loop ke normal se

60^{\circ}

ka angle banata hai. Loop se flux kitna hai, webers mein?

Answer dekho

Answer 0.005 Wb

Φ_{B} = 0.50 \times 0.020 \times \cos 60^{\circ} = 5.0 \times 10^{- 3} W b

. Galti se

\sin 60^{\circ}

lene par

8.7 \times 10^{- 3} W b

aata hai.

6.4 Faraday's law of induction

Faraday ka bada kadam yeh samajhna tha ki flux ki language mein describe karte hi Section 6.2 ke teeno experiments ek hi rule follow karte hain. Pehle do mein field ka source chalta hai, isliye meter se judi coil se flux change hota hai. Key wale experiment mein koi motion hi nahi chahiye. Coil P ka current zero se seedha apni final value par jump nahi kar sakta; key dabane ke baad second ke ek chhote hisse tak woh badhta hai, aur uske saath woh flux bhi jo P, S se hokar bhejta hai. Wahi chhota sa build-up jhatka hai. Current settle hote hi S se flux constant ho jaata hai aur meter zero dikhata hai. Key kholne par flux gir jaata hai, yeh ulti direction ka change hai, isliye jhatka bhi ulta hota hai.

LawFaraday's law of electromagnetic induction

Jab bhi kisi circuit se magnetic flux change hota hai, circuit mein emf induce hota hai. Emf ka magnitude flux ke change hone ki rate ke barabar hota hai.

Faraday's law

ε = - \frac{d Φ_{B}}{d t}

(6.3)

Minus sign emf ki direction batata hai, aur isliye closed circuit mein current ki bhi; Section 6.5 ise samjhata hai. $N$ turns wali tightly wound coil mein har turn same flux link karta hai, aur turns ke emf series mein lage cells ki tarah jud jaate hain:

N turns wali coil ke liye Faraday's law

ε = - N \frac{d Φ_{B}}{d t}

(6.4)

Product $N Φ_{B}$ ko coil ka flux linkage kehte hain. Isliye field ke same change se bada emf lene ka sabse aasaan tareeka hai zyada turns lapetna. Agar poore circuit ka resistance $R$ hai, to induced current $I = ε / R$ hota hai.

Kyunki $Φ_{B} = B A \cos θ$ hai, ise badalne ke teen independent tareeke hain, aur har ek emf induce karta hai. Tum field strength $B$ badal sakte ho. Tum circuit ka ghera hua area $A$ badal sakte ho, uske kisi hisse ko stretch, squeeze ya slide karke. Ya circuit ko ghuma kar angle $θ$ badal sakte ho. Fig. 6.6 ek-ek karke teeno dikhata hai. Faraday ke experiments pehla tareeka use karte hain, Section 6.6 mein motional emf doosra, aur Section 6.8 mein ac generator teesra.

2/4 $B$ double karne se flux double hokar $2 Φ_{0}$ ho jaata hai. Jab tak $B$ change ho raha hai, emf induce hota hai.

Fig. 6.6Flux badalne ke teen tareeke:

B

badlo,

A

badlo ya

θ

badlo. Emf sirf tab tak induce hota hai jab tak change ho raha ho.

Crosses ka ek panel page ke andar point karta magnetic field dikhata hai, usmein ek green rectangular loop hai, aur right mein ek vertical gauge loop se flux dikhata hai. Step 1: loop jaisa hai waisa; gauge aadha bhara hai, Phi nought mark ke saath. Step 2: aur crosses aa jaate hain, field double ho jaata hai aur gauge do Phi nought tak bhar jaata hai. Step 3: field normal par wapas hai aur loop sikudkar aadhe area ka ho gaya hai; gauge Phi nought ke aadhe par gir jaata hai. Step 4: loop ek vertical dashed axis ke around 60 degree ghoom jaata hai, isliye aadha chauda dikhta hai, aur gauge phir se Phi nought ka aadha dikhata hai.

Example 6.1NCERT Example 6.1 · p. 158

Jhatka bada kaise karein

Tum Experiment 6.2 dohraate ho, ek chhoti current-carrying coil ko galvanometer se judi badi coil ki taraf le jaate ho, lekin needle bas zara si hilti hai. (a) Teen changes suggest karo jo saaf-saaf bada deflection dein. (b) Galvanometer repair ke liye chala gaya. Tum phir bhi kaise dikhaoge ki current induce hota hai, aur yeh bhi ki woh kis taraf behta hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.2NCERT Example 6.2 · p. 158

Ghatte field mein ek loop

$20 c m \times 15 c m$ ke ek rectangular wire loop ka resistance $0.80 Ω$ hai. $0.25 T$ ka uniform field loop ke normal se $60^{\circ}$ ka angle banata hai. Field ko $0.50 s$ mein steadily zero tak kam kiya jaata hai. Induced emf aur current nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.3NCERT Example 6.3 · p. 159

Earth ke field mein coil palatna

600 turns aur $12 c m$ radius wali ek flat circular coil ka resistance $5.0 Ω$ hai. Woh seedhi khadi hai aur uska face north ki taraf hai, isliye Earth ke field ka horizontal component, $B_{H} = 3.6 \times 10^{- 5} T$ , usse seedha guzarta hai. Coil ko ek vertical diameter ke around $0.30 s$ mein $180^{\circ}$ ghumaya jaata hai. Usmein induced average emf aur current ka estimate lagao.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

50 turns wali ek coil ek aise field mein hai jo steadily badhta hai, jisse har turn ka flux

0.10 s

mein

2.0 m W b

8.0 m W b

ho jaata hai. Induced emf ka size kitna hai, volts mein?

Answer dekho

Answer 3 V

∣ ε ∣ = N Δ Φ_{B} / Δ t = 50 \times (6.0 \times 10^{- 3}) / 0.10 = 3.0 V

. Factor

N

chhod dene par ek turn ka emf,

0.060 V

, aata hai.

6.5 Lenz's law aur energy conservation

Faraday's law induced emf ka size deta hai. Uska sign — yaani induced current kis direction mein behta hai — ek rule se aata hai jo 1834 mein Baltic German physicist Emil Lenz ne diya.

LawLenz's law

Induced current hamesha us sense mein ghoomta hai jiska apna magnetic field flux ke us change ke against kaam kare jisne use paida kiya.

Dhyaan do ki Lenz's law change ko oppose karta hai, flux ko nahi. Agar loop se flux badh raha hai, to induced current badhte field ke against field banata hai, flux ko neeche rakhne ki koshish karta hai. Agar flux ghat raha hai, to induced current maujooda field ke saath field banata hai, flux ko sahara dene ki koshish karta hai. (6.3) ka minus sign isi rule ka mathematical form hai.

Ise apply karne ka ek aasaan tareeka current loop ke poles use karta hai. Loop ke ek face ko dekho: agar tumhe current anticlockwise ghoomta dikhe, to woh face north pole hai; clockwise dikhe to south pole. Fig. 6.7 mein magnet ka N pole ek ring ki taraf aata hai. Ring se flux badhta hai, isliye ring ke paas wale face ko north pole banna padega aur magnet ko peeche dhakelna padega: magnet ki taraf se dekhein to current anticlockwise hai. Magnet ko door kheencho to paas wala face south pole ban jaata hai aur magnet ko wapas kheenchta hai: ab current clockwise hai.

1/3Flux badhta hai, paas wala face N ban jaata hai, aur paas aata magnet peeche dhakela jaata hai.

Fig. 6.7Induced current hamesha us motion ko oppose karta hai jo use paida karta hai.NCERT Fig. 6.6 · p. 160

Left mein ek copper ring khadi hai aur right mein usi axis par ek bar magnet hai jiska N pole ring ki taraf hai; ring ke neeche ek face-on view magnet ki taraf se dikhne wali current direction dikhata hai. Step 1: magnet andar aata hai; magnet ke paas wale ring face par N aur door wale par S likha hai, ek green arrow dikhata hai ki magnet peeche dhakela gaya, aur face-on view anticlockwise current dikhata hai. Step 2: magnet bahar jaata hai; paas wala face ab S hai, green arrow dikhata hai ki magnet wapas kheencha gaya, aur current clockwise hai. Step 3: ek impossible case jahan magnet andar aata hai lekin paas wala face S hai; force arrow ring ki taraf point karta hai, magnet ki speed badhti hai, aur ek cross ise muft ki energy, yaani bina kisi source ke energy, mark karta hai.

Opposition kyun, help kyun nahi? Thought experiment ko ulta chala kar dekho. Agar paas aate N pole ko south face milta, to ring magnet ko andar kheenchti. Magnet aur tez pahunchta, flux aur tez change hota, induced current badhta, aur kheenchav aur strong ho jaata: ek runaway loop. Magnet ki kinetic energy aur ring ki heat dono badhti rehtin jabki koi energy de hi nahi raha. Aisi koi free-energy machine aaj tak kaam nahi kar paayi, kyunki energy conserved rehti hai. Us law ke saath match karne wali induced current ki sirf ek hi direction hai: woh jo change ko oppose kare.

Isliye asli ring mein magnet ko andar ya bahar le jaane mein hamesha mehnat lagti hai. Tumhari kharch ki hui energy pehle induced current ki energy ke roop mein aur phir ring ke resistance mein heat ke roop mein dikhti hai. Is nazar se dekho to Lenz's law bas induced currents ke liye likha gaya energy conservation hai.

SimulationLenz's law

Ek magnet coil ke axis par chalta hai, turns par arrows induced current ki direction dikhate hain aur labels batate hain ki coil par kaunse poles bante hain. Magnet ko andar ya bahar le jaane par dikhta hai ki induced current hamesha flux ke change ko oppose karta hai.

Fig. 6.8Arrows induced current dikhate hain; unhe magnet ke chalne ki direction se compare karo.

Try karo

N pole ko coil ki taraf le jao. Magnet ki side se dekhne par current clockwise hai ya anticlockwise, aur magnet ke saamne kaunsa pole hai?
Ab magnet ko door le jao. Dekhne se pehle arrows ki nayi direction predict karo.
Magnet ko bahut dheere aur phir bahut tezi se chalao. Kya direction badalti hai, ya sirf current ka size?

Try karke dekho

Ek magnet jo girna nahi chahta

Tumhe chahiye

ek strong neodymium magnet
lagbhag 50 cm lambi copper ya aluminium pipe, magnet se thodi chaudi
milte-julte size ka ek steel bolt ya iron nail

Pehle guess karo

Tum vertical pipe mein pehle nail aur phir magnet girate ho. Neeche pahunchne mein kise zyada time lagta hai?

NailDono ko same timeMagnet

Pipe ko cushion ya tah kiye kapde ke upar seedha pakdo.
Nail ko pipe mein girao aur uske girne ka time naapo.
Magnet ko pipe mein girao aur uske girne ka time naapo.
Ho sake to magnet girte waqt pipe ke andar jhaank kar dekho.

Magnet nail se kai guna dheere neeche utarta hai. Pipe rings ke stack ki tarah kaam karti hai. Magnet ke theek neeche wali rings apna flux badhta dekhti hain aur use repel karti hain; theek upar wali rings apna flux girta dekhti hain aur use wapas attract karti hain. Dono effects girne ko oppose karte hain, jaisa Lenz's law predict karta hai, aur magnet ki khoyi potential energy pipe ko zara sa garam kar deti hai.

Abhi nahi kar sakte? Simulation try karo · Fig. 6.8

Fig. 6.9 ek rectangular loop ko follow karta hai jo constant speed se uniform field ke region mein se slide karta hai. Loop se flux sirf tab change hota hai jab loop region ka koi edge paar kar raha ho, isliye emf bhi sirf tabhi induce hota hai. Andar jaate waqt flux badhta hai aur current anticlockwise hota hai. Poori tarah andar hone par flux steady hai aur kuch induce nahi hota. Bahar nikalte waqt flux girta hai aur current clockwise ho jaata hai. Kyunki ek seedha edge boundary ko steady rate se paar karta hai, emf jab tak rehta hai constant rehta hai, jise hum Section 6.6 mein derive karenge.

2/4Flux steady rate se badhta hai, isliye constant emf anticlockwise current chalata hai.

Fig. 6.9Emf sirf tab aata hai jab loop field region ka koi edge paar karta hai.NCERT Fig. 6.8 · p. 162

Crosses se bhara ek dashed square region page ke andar jaata field dikhata hai. Ek green rectangular loop usmein se left se right slide karta hai, aur neeche do graphs loop ki position ke against loop se flux aur induced emf plot karte hain; ek dashed guide line loop ke front edge ko dono graphs se jodti hai. Step 1: loop region ke left mein hai, aur dono graphs par markers zero par hain. Step 2: loop aadha andar hai; flux graph badh raha hai, emf graph axis ke neeche ek constant block dikhata hai jis par anticlockwise likha hai, aur loop mein ek curved arrow anticlockwise ghoomta hai. Step 3: loop poora andar hai; flux apni sabse badi value par flat hai aur emf zero hai. Step 4: loop right side se aadha bahar hai; flux gir raha hai, emf axis ke upar ek block hai jis par clockwise likha hai, aur loop ka arrow clockwise ghoomta hai.

Example 6.4NCERT Example 6.4 · p. 161

Field ka edge paar karte loops

Dashed square ke andar ek uniform magnetic field page se bahar point karta hai. Chaar conducting loops arrows ki direction mein chal rahe hain. Har loop ke liye tay karo ki current induce hota hai ya nahi, aur agar hota hai to tumhe clockwise dikhta hai ya anticlockwise.

Fig. 6.10NCERT Fig. 6.7 · p. 161

Dots se bhara ek dashed square page se bahar point karta field mark karta hai. Loop a, ek rectangle, left edge ke aadha andar hai aur right ki taraf field mein ja raha hai. Loop b, ek hexagon, top edge se bahar nikla hua hai aur upar ki taraf field se bahar ja raha hai. Loop c, ek circle, poori tarah field ke andar hai aur right ki taraf chal raha hai. Loop d, ek rectangle, right edge ke aadha andar hai aur left ki taraf field mein ja raha hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.5NCERT Example 6.5 · p. 161

Chaar quick faisle

(a) Ek closed wire loop hospital ke MRI magnet ke bahut strong, steady field ke andar ruka pada hai. Kya usmein current induce hota hai?
(b) Ek student kehti hai ki do charged plates ke beech ke steady electric field mein closed loop ko slide karne se usmein current ghoomega, ‘bilkul magnet ki tarah’. Jab loop poori tarah plates ke beech hai, tab kya woh sahi hai? Jab aadha bahar hai, tab?
(c) Do identical square loops same constant speed se uniform magnetic field ke region se bahar nikalte hain, field dono ke perpendicular hai. Loop 1 ki ek side field ke edge ke parallel hai; loop 2 ko $45^{\circ}$ ghumaya gaya hai, isliye woh corner se pehle nikalta hai. Bahar nikalte waqt kis loop ka induced emf constant rehta hai?
(d) Ek circular loop is page ke plane mein hai aur uski top side mein ek chhota capacitor laga hai: plate P left mein aur plate Q right mein. Page ke saamne pakde, page ki taraf point karte bar magnet ke N pole ko loop ki taraf dhakela jaata hai. Kaunsi plate positive banti hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

6.6 Motional electromotive force

Ab tak field change ho raha tha. Ab field ko uniform aur steady rakho, aur uski jagah circuit ke ek hisse ko field mein chalao. Fig. 6.11 mein $l$ length ki ek metal rod do parallel rails par bina friction slide karti hai; rails left end par ek resistor $R$ se judi hain. Ek uniform field $𝐁$ page ke andar point karta hai, rails ke plane ke perpendicular. Rod, rails aur resistor milkar ek closed loop banate hain jiska area rod ke $v$ speed se right chalne par badhta hai.

2/4Rod right chalti hai to ghera hua area badhta hai, aur current $I$ anticlockwise behta hai.

Fig. 6.11Rod slide karne se circuit ka area, aur isliye usse flux, change hota hai.NCERT Fig. 6.10 · p. 162

Do horizontal rails left end par resistor R se judi hain, aur unke upar l length ki copper rod rakhi hai, crosses se dikhaye gaye page ke andar jaate field mein. Circuit ka ghera hua area green shade kiya hai. Step 1: rod left end se x doori par hai aur flux B l x hai. Step 2: rod aur right slide kar gayi hai, shaded area bada hai, aur arrows anticlockwise current I dikhate hain jo rod mein upar behta hai. Step 3: rod ke top par plus signs aur bottom par minus signs jama hote hain, aur ek green arrow positive charges par force q v B ko rod ke upar ki taraf dikhata hai. Step 4: ek purple arrow rod par magnetic force I l B ko left ki taraf, motion ke against, dikhata hai, aur push likha ek black arrow tumhare dhakke ko right ki taraf dikhata hai.

DerivationFaraday's law se motional emf

Maan lo closed end se rod ki doori $x$ hai. Loop $l x$ area gherta hai, isliye usse flux hai

Φ_{B} = B l x

Sirf $x$ badalta hai, rate $d x / d t = v$ se. (6.3) se induced emf ka size hai

Motional emf

∣ ε ∣ = ∣ \frac{d Φ_{B}}{d t} ∣ = B l \frac{d x}{d t} = B l v

(6.5)

Lenz's law direction tay karta hai. Badhte into-page flux ko anticlockwise current oppose karta hai, jo rod mein upar behta hai. Rod ka top $B l v$ emf wale cell ke positive terminal ki tarah kaam karta hai.

Steady field mein conductor chalane se bane emf ko motional emf (गतिक विद्युत वाहक बल) kehte hain. Hum flux ka naam liye bina bhi yahi result paa sakte hain, chalti rod ke andar ke charges ko dekh kar.

DerivationLorentz force se wahi result

Rod ka har free charge $q$ rod ke saath $v$ speed se right jaata hai. Uspar magnetic force $q 𝐯 \times 𝐁$ ka magnitude $q v B$ hai aur yeh rod ke saath-saath point karta hai, positive charge ke liye top end ki taraf. Rod ke har point par force same hai.

Charge ko rod ke bottom se top tak le jaate hue yeh force work $W = q v B l$ karta hai. Kyunki emf per unit charge kiya gaya work hai,

ε = \frac{W}{q} = \frac{q v B l}{q} = B l v

Yeh (6.5) se match karta hai. Current behne lage to har charge rod ke saath drift bhi karta hai, aur yeh drift use ek doosra magnetic force deta hai jo rod ke across hota hai. Poori rod par jodne se yahi current par force $I l B$ banta hai, jise hum aage dekhenge.

Force wali picture energy ka hisaab bhi saaf kar deti hai. Induced current $I$ rod mein upar behta hai, aur rod field mein hai, isliye rod par $I l B$ size ka magnetic force lagta hai. Yeh force left ki taraf, motion ke against, point karta hai, bilkul jaisa Lenz's law chahta hai. Rod ko constant speed par chalaye rakhne ke liye tumhe barabar force se dhakelna padega, jo power $I l B v$ deta hai. Kyunki $I = B l v / R$ hai, yeh power $I^{2} R$ ke barabar hai, yaani wahi rate jis par resistor garam hota hai.

Example

Sliding-rod generator mein power

$0.50 m$ lambi ek rod rails ke perpendicular $0.30 T$ ke uniform field mein rails par steady $4.0 m / s$ se slide karti hai. Circuit ka total resistance $1.2 Ω$ hai. Nikaalo (a) induced emf, (b) current, (c) rod ko steadily chalane ke liye zaroori force, aur (d) us force se di gayi power. (d) ko heating ki rate se compare karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Ab situation ko ulta karo: conductor ko rok kar rakho aur field badlo, jaise Experiment 6.3 mein. Ruke hue charge par koi magnetic force nahi lagta, kyunki charge par poora Lorentz force hai

𝐅 = q (𝐄 + 𝐯 \times 𝐁) = q 𝐄 jab 𝐯 = 0

(6.6)

To stationary coil mein induced current jo bhi chalata hai, woh ek electric field hi hona chahiye: time ke saath change hota magnetic field apne aas-paas electric field banata hai. Yeh induced electric field Chapter 1 ke electrostatic field jaisa nahi hai. Iski field lines charges par shuru aur khatam hone ki jagah closed loops banati hain. Yeh loop par ek chakkar ghumaaye gaye charge par net work bhi karta hai, aur per unit charge yahi work induced emf hai. Oersted ne dikhaya tha ki chalte charges magnets ko dhakelte hain; Faraday ne ulta dikhaya, ki changing magnetic field ruke hue charges ko dhakelta hai.

Conductor ka seedhi line mein chalna zaroori nahi. Fig. 6.12 mein $R$ length ki metal rod ek end ke around angular speed $ω$ se ghoomti hai, ghadi ki sui ki tarah, rotation axis ke parallel uniform field mein. Iska free end ek circular metal ring par slide karta hai. Ab rod ke alag-alag hisse alag speeds se chalte hain: axis se $r$ doori wala point $v = ω r$ se chalta hai, isliye bahar wale hisse emf ka zyada hissa dete hain.

3/3Time $t$ mein rod $\frac{1}{2} R^{2} θ$ area ka sector sweep karti hai, jahan $θ = ω t$ , jisse $ε = \frac{1}{2} B ω R^{2}$ milta hai.

Fig. 6.12Uniform axial field mein ek end ke around ghoomti rod.NCERT Fig. 6.11 · p. 164

Ek metal ring dots ka grid gherti hai jo page se bahar aata field dikhata hai. Ek copper rod centre O se ring tak jaati hai aur anticlockwise ghoomti hai, jise omega likha curved arrow dikhata hai. Step 1: rod right ki taraf point karti hai aur uski length R likhi hai. Step 2: rod 40 degree ghoom chuki hai aur uske saath teen orange velocity arrows O se doori ke saath lambe hote jaate hain, jin par v equals omega r likha hai. Step 3: rod 70 degree ghoom chuki hai aur usne jo sector sweep kiya woh shaded hai, uska area one half R squared theta aur emf one half B omega R squared hai.

Example 6.6NCERT Example 6.6 · p. 163

Ghoomti rod

$0.80 m$ lambi ek metal rod $0.80 m$ radius wali metal ring ke centre par pivot hai aur ring ke plane mein 20 revolutions per second se ghoomti hai. $0.60 T$ ka uniform field rotation axis ke parallel lagta hai. Centre aur ring ke beech emf nikaalo (a) rod ke saath motional emfs jod kar aur (b) rod jis rate se area sweep karti hai usse.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.7NCERT Example 6.7 · p. 165

Cycle ke wheel ke spokes

Ek cyclist seedha east ki taraf ja raha hai, isliye front wheel ek vertical east–west plane mein ghoomta hai. Earth ke field ka horizontal component, $B_{H} = 3.5 \times 10^{- 5} T$ , north ki taraf hai, us plane ke right angle par. Wheel mein 32 metal spokes hain, har ek $0.30 m$ lamba, aur woh 180 revolutions per minute se ghoomta hai. Axle aur rim ke beech kitna emf aata hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Real life mein

Syllabus se aage: eddy currents

Induction ke liye wire loop zaroori nahi. Jab metal ka koi solid tukda magnetic field mein chalta hai, ya changing field mein rakha hota hai, to uske andar currents aise ghoomte hain jaise nadi mein bhanwar. Yeh eddy currents (भँवर धाराएँ) Lenz's law follow karte hain, isliye jo motion ya change inhe paida karta hai, use oppose karte hain.

1/2Solid plate mein bade eddy currents ghoomte hain aur use kuch hi swings mein rok dete hain.

Fig. 6.13Eddy currents solid plate ko zor se brake karte hain; slots unhe tod dete hain.

Ek metal plate pivot se pendulum ki tarah latki hai aur ek magnet pole face ke beech se jhoolti hai, jo plate par crosses ke dashed square se dikhaya gaya hai. Dashed outlines dikhati hain ki plate kitni door tak jhoolti hai. Step 1: plate solid hai, usmein do bade orange eddy-current loops ulti directions mein ghoomte hain, aur dashed outlines sirf chhota swing dikhati hain, jis par jaldi rukti likha hai. Step 2: plate mein vertical slots hain, eddy currents slots ke beech chhote patle loops mein toot jaate hain, aur dashed outlines chauda swing dikhati hain, jis par jhoolti rehti likha hai.

Inse bana drag kuch trains ke electromagnetic brakes mein kaam aata hai, aur us damping mein bhi jo galvanometer needle ko jaldi settle karati hai. Induction cooktops tezi se change hote field se chalne wale eddy currents se steel pan garam karte hain, aur metal detectors chhupe metal mein bane eddy currents ko sense karte hain. Jahan eddy currents sirf energy barbaad karte hain, jaise transformers aur motors ke iron cores mein, wahan core ko patli insulated sheets se banate hain jinhe laminations kehte hain; yeh current loops ko waise hi tod deti hain jaise pendulum mein slots.

Quick check

3 mein se 1

0.40 m

lambi ek seedhi rod

5.0 m / s

se apni length aur

0.25 T

ke uniform field, dono ke perpendicular chalti hai. Uske ends ke beech kitna emf induce hota hai, volts mein?

Answer dekho

Answer 0.5 V

ε = B l v = 0.25 \times 0.40 \times 5.0 = 0.50 V

6.7 Inductance

Current le jaane wali har coil apne turns se aur paas ki kisi bhi coil se magnetic flux bhejti hai. Agar coils apni shape aur position na badlein, to current double karne se har jagah field double hota hai, aur isliye har flux double hota hai: $Φ_{B} \propto I$ . $N$ turns wali tightly wound coil mein, jiske saare turns se same flux guzarta hai, Faraday's law mein aane wali quantity flux linkage $N Φ_{B}$ hai, isliye hum likh sakte hain

N Φ_{B} \propto I ⟹ N Φ_{B} = (constant) \times I

Proportionality constant ko inductance (प्रेरकत्व) kehte hain. Yeh sirf geometry aur materials se tay hota hai: coils ka size, shape, turns ki sankhya aur spacing, aur unke core mein jo bhi bhara ho. Capacitance bhi aise hi kaam karta hai, kyunki parallel-plate capacitor ki value uski plates, unke gap aur beech ke dielectric se tay hoti hai. Inductance ek scalar hai. Iski SI unit henry (H) hai, jahan $1 H = 1 W b / A = 1 V s / A$ , aur iske dimensions $[M L^{2} T^{- 2} A^{- 2}]$ hain. Flux paas ki coil se aa sakta hai ya coil ke apne current se; hum dono cases ek-ek karke dekhenge.

6.7.1 Mutual inductance

Same length $l$ ke do lambe coaxial solenoids lo, ek doosre ke andar (Fig. 6.14). Andar wale solenoid $S_{1}$ ka radius $r_{1}$ hai, per unit length $n_{1}$ turns aur kul $N_{1} = n_{1} l$ turns; bahar wale solenoid $S_{2}$ ka radius $r_{2}$ hai, per unit length $n_{2}$ turns aur $N_{2}$ turns. $S_{2}$ mein current $I_{2}$ , $S_{1}$ ke har turn se flux $Φ_{1}$ bhejta hai. Hum $S_{2}$ ke respect mein $S_{1}$ ka mutual inductance (अन्योन्य प्रेरकत्व) $M_{12}$ aise define karte hain

N_{1} Φ_{1} = M_{12} I_{2}

(6.7)

2/3 $S_{2}$ mein current poore $S_{2}$ ko uniform field $B_{2}$ se bhar deta hai, lekin $S_{1}$ se flux ke liye sirf area $π r_{1}^{2}$ count hota hai.

Fig. 6.14Current kisi bhi coil mein ho, doosri coil ko link karne wala flux area

π r_{1}^{2}

se hi guzarta hai.NCERT Fig. 6.12 · p. 166

Side view mein ek lamba bahar wala solenoid S2 hai, jo blue wire cross-sections ki do rows se bana hai, aur ek patla andar wala solenoid S1, jo orange wire cross-sections ki do rows se bana hai; dono ki length l hai. Right mein end view do concentric circles dikhata hai. Step 1: radii r1 aur r2 aur length l likhe hain aur koi current nahi behta. Step 2: bahar wali coil mein current behta hai, blue field arrows S2 ka poora andar ka hissa bhar dete hain, end view ke dono circles mein dots bhar jaate hain, aur pi r1 squared area wali andar ki disc shaded hai. Step 3: sirf andar wali coil mein current behta hai, orange field arrows sirf S1 ke andar aate hain, aur dots sirf end view ke andar wale circle mein aate hain.

DerivationDo lambe coaxial solenoids ka mutual inductance

Ends par edge effects ko ignore karo, jo $l ≫ r_{2}$ hone par achha approximation hai. $S_{2}$ mein current $I_{2}$ uske andar uniform field $B_{2} = μ_{0} n_{2} I_{2}$ banata hai. $S_{1}$ ke har turn ka area $π r_{1}^{2}$ hai, isliye $S_{1}$ ke saath flux linkage hai

N_{1} Φ_{1} = (n_{1} l) (π r_{1}^{2}) (μ_{0} n_{2} I_{2}) = μ_{0} n_{1} n_{2} π r_{1}^{2} l I_{2}

(6.8)

(6.7) se compare karne par,

M_{12} = μ_{0} n_{1} n_{2} π r_{1}^{2} l

(6.9)

Ab roles ulat do. $S_{1}$ mein current $I_{1}$ bahne do aur $M_{21}$ ko aise define karo

N_{2} Φ_{2} = M_{21} I_{1}

(6.10)

Andar wale solenoid ka field $B_{1} = μ_{0} n_{1} I_{1}$ sirf $S_{1}$ ke andar hota hai. $S_{2}$ ka har turn $S_{1}$ ko gherta hai, lekin field usse sirf area $π r_{1}^{2}$ par guzarta hai, isliye

N_{2} Φ_{2} = (n_{2} l) (π r_{1}^{2}) (μ_{0} n_{1} I_{1}) ⟹ M_{21} = μ_{0} n_{1} n_{2} π r_{1}^{2} l

(6.11)

Dono results ek jaise hain:

M_{12} = M_{21} = M

(6.12)

Humne $M_{12} = M_{21}$ sirf do lambe coaxial solenoids ke liye prove kiya, lekin yeh kisi bhi circuits ke pair ke liye sahi hai, unki shape kuch bhi ho, aur isse yeh ek powerful short cut ban jaata hai. Example 6.8 ise use karta hai. Badi coil ka field uske centre ke paas chhoti coil par likhna aasaan hai, jabki chhoti coil jo field badi coil mein failati hai woh har point par badalta hai aur use jodna bahut mushkil hota. Kyunki dono mutual inductances barabar hain, aasaan wala nikaalo aur mushkil wala free mein mil jaata hai.

Agar solenoids ke andar ki jagah relative permeability $μ_{r}$ wale material, jaise soft iron, se bhari ho, to field aur isliye mutual inductance $μ_{r}$ se multiply ho jaate hain: $M = μ_{r} μ_{0} n_{1} n_{2} π r_{1}^{2} l$ . Kisi bhi coil pair ke liye $M$ unki doori aur orientation par bhi depend karta hai. Common axis par paas-paas rakhi coils, door rakhi ya right angle par rakhi coils se kahin zyada flux link karti hain.

Ab Experiment 6.3 par wapas aao. $N_{1} Φ_{1} = M I_{2}$ aur time ke saath badalte current $I_{2}$ ke saath, Faraday's law coil 1 mein induced emf deta hai:

ε_{1} = - \frac{d (N_{1} Φ_{1})}{d t} = - M \frac{d I_{2}}{d t}

Ek coil mein changing current paas wali coil mein emf induce karta hai, mutual inductance ke aur current kitni tezi se badal raha hai uske proportion mein. Transformers, wireless phone chargers aur electric toothbrush ke charging stands, sab isi par chalte hain.

Example 6.8NCERT Example 6.8 · p. 167

Badi coil ke centre par chhoti coil

$r_{1} = 1.0 c m$ radius aur $N_{1} = 50$ turns wali ek chhoti circular coil, $r_{2} = 20 c m$ radius aur $N_{2} = 200$ turns wali badi circular coil ke centre par hai. Coils coaxial hain aur same plane mein hain. (a) Unke mutual inductance ka expression nikaalo, aur uski value bhi. (b) Badi coil ka current $50 A / s$ se badalne par chhoti coil mein induced emf nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check: mutual inductance

3 mein se 1

Coil A mein changing current paas ki coil B mein

2.0 m V

ka emf induce karta hai. Agar wahi changing current coil B mein bheja jaaye, to coil A mein induced emf hoga

2.0 m V

, pehle jitna hizyada, agar A mein B se zyada turns honkam, agar A andar wali coil hozero

Answer dekho

Answer (A) $2.0 m V$ , pehle jitna hi

Mutual inductance dono taraf same hai,

M_{12} = M_{21}

, isliye current ke change ki same rate same emf deti hai. Turns ki sankhya pehle se hi

M

mein shaamil hai aur dono directions mein fark nahi la sakti.

6.7.2 Self-inductance

Induction dikhane ke liye coil ko padosi ki zaroorat nahi. Coil ka current uske apne turns se flux bhejta hai, isliye jab woh current badalta hai, to coil ka apna flux linkage badalta hai aur coil mein hi emf aa jaata hai. Ise self-induction (स्वप्रेरण) kehte hain. Flux linkage coil ke apne current ke proportional hota hai:

N Φ_{B} = L I

(6.13)

Constant $L$ ko coil ka self-inductance, ya coefficient of self-induction, kehte hain. Phir Faraday's law deta hai

Self-induced emf

ε = - \frac{d (N Φ_{B})}{d t} = - L \frac{d I}{d t}

(6.14)

Minus sign phir se Lenz's law hai. Current badhane ki koshish karo to coil badhne ke against dhakelti hai; current kaatne ki koshish karo to coil aage dhakelti hai, use behta rakhne ki koshish karti hai (Fig. 6.15). Kyunki yeh hamesha change se ladta hai, self-induced emf ko back emf kehte hain.

1/3Jab $I$ badhta hai, self-induced emf uske against dhakelta hai.

Fig. 6.15Self-induced emf

- L d I / d t

current ke har change se ladta hai, upar ho ya neeche.

Loops ki row jaisi dikhne wali ek coil mein current I behta hai, jise blue arrow dikhata hai; uske neeche ek orange arrow self-induced emf dikhata hai. Coil ke neeche time ke against do graphs hain: current ramp up hota hai, level rehta hai, phir ramp down hota hai, aur emf pehle negative block, phir zero, phir positive block hai. Step 1: current ka badhta hissa highlight hai aur orange emf arrow current ke against point karta hai, jis par emf oppose karta likha hai. Step 2: level wala hissa highlight hai, emf arrow nahi hai, aur label par induced emf nahi likha hai. Step 3: girta hissa highlight hai aur orange emf arrow current ke saath point karta hai, jis par emf saath deta likha hai.

DerivationLambe solenoid ka self-inductance

$l$ length aur $A$ cross-sectional area wale solenoid mein per unit length $n$ turns hain, yaani kul $n l$ turns. End effects ignore karein to current $I$ uske andar uniform field $B = μ_{0} n I$ banata hai. Flux linkage hai

N Φ_{B} = (n l) (μ_{0} n I) (A) = μ_{0} n^{2} A l I

L = \frac{N Φ_{B}}{I} = μ_{0} n^{2} A l

(6.15)

Agar solenoid relative permeability $μ_{r}$ wale core par lapeta ho, to diye gaye current ke liye field $μ_{r}$ factor se badh jaata hai, aur $L$ bhi:

L = μ_{r} μ_{0} n^{2} A l

(6.16)

Mutual inductance ki tarah, $L$ coil ke size aur shape aur uske core mein kya bhara hai isse tay hota hai, us current se nahi jo coil mein behta hai. Kyunki $L \propto n^{2}$ , same former par do guna turns lapetne se $L$ chaar guna ho jaata hai, aur soft-iron core, jiska $μ_{r}$ sainkdon ya hazaaron mein hota hai, ise bahut zyada badha deta hai.

Example

Ek henry kitna bada hota hai?

$40 c m$ lambe solenoid mein 1000 turns hain, har turn ka cross-sectional area $5.0 c m^{2}$ hai, aur andar hawa hai. (a) Iska self-inductance nikaalo. (b) Isme $2.0 A$ ka current switch off kiya jaata hai aur $5.0 m s$ mein zero ho jaata hai. Kitna average emf induce hota hai? (c) $μ_{r} = 800$ wale soft-iron core ke saath isi solenoid ke liye (b) dohrao.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Back emf coil ko ek tarah ki electrical inertia deta hai. Bhari trolley ko chalana bhi mushkil hai aur rokna bhi; bade $L$ wali coil apne current ko shuru karna bhi mushkil banati hai aur rokna bhi. Current build karne ke liye source ko back emf ke against lagataar work karna padta hai. Yeh work bekaar nahi jaata. Yeh coil ke magnetic field mein store hota hai, bilkul waise hi jaise trolley ki speed badhane mein kiya gaya work uski kinetic energy ke roop mein store hota hai.

DerivationInductor mein stored energy

Maan lo kisi pal $L$ inductance wali coil mein current $I$ hai aur rate $d I / d t$ se badh raha hai. Resistance ignore karein to source ko current ko $L d I / d t$ size ke back emf ke against chalana padta hai, isliye woh is rate se work karta hai

\frac{d W}{d t} = ∣ ε ∣ I = L I \frac{d I}{d t}

Current ko zero se $I$ tak badhane ke liye kul work hai

Inductor mein stored energy

W = \int_{0}^{I} L I^{'} d I^{'} = \frac{1}{2} L I^{2}

(6.17)

Current zero tak girne par yeh energy wapas mil jaati hai. Yeh formula kinetic energy ke $\frac{1}{2} m v^{2}$ jaisa hai, jismein $L$ , $m$ ka role aur $I$ , $v$ ka role nibhata hai.

Compare

Mass aur inductance

Mechanics

Mass $m$ velocity ke changes ko resist karta hai
Velocity $v$
Force $F = m d v / d t$
Kinetic energy $\frac{1}{2} m v^{2}$

Circuits

Inductance $L$ current ke changes ko resist karta hai
Current $I$
$L d I / d t$ size ka back emf
Magnetic energy $\frac{1}{2} L I^{2}$

Jab paas-paas ki do coils dono mein changing currents hon, to coil 1 ke flux linkage ke do hisse hote hain, ek uske apne current se aur ek padosi ke current se: $N_{1} Φ_{1} = L_{1} I_{1} + M_{12} I_{2}$ . Tab coil 1 mein induced emf hai

ε_{1} = - L_{1} \frac{d I_{1}}{d t} - M_{12} \frac{d I_{2}}{d t}

Example 6.9NCERT Example 6.9 · p. 169

Magnetic field mein energy

(a) Lambe solenoid mein stored energy ko sirf uske field $B$ aur dimensions mein likho, aur field ki per cubic metre energy nikaalo. (b) Ek MRI magnet lagbhag $0.20 m^{3}$ volume mein $1.5 T$ ka field banata hai. Isme stored energy ka estimate lagao. (c) Kaunsa electric field isi density par energy store karega, aur kya hawa use sah sakti hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Magnetic field ki per unit volume energy

u_{B} = \frac{B^{2}}{2 μ_{0}}

(6.18)

Solenoid (6.18) derive karne ke liye bas ek convenient jagah tha. Jahan bhi magnetic field hai, magnet ke gap mein, power cable ke aas-paas ya light ki beam mein, woh har cubic metre mein $B^{2} / 2 μ_{0}$ joules store karta hai. Electric wala counterpart, $\frac{1}{2} ε_{0} E^{2}$ , utna hi general hai, aur dono field strength ke square ke saath badhte hain.

6.8 AC generator

Tum jo bhi electrical energy use karte ho, woh lagbhag saari ek generator mein shuru hoti hai, ek aisi machine jo electromagnetic induction se mechanical energy ko electrical energy mein badalti hai. Yeh flux badalne ke hamare teen tareekon mein se teesra use karti hai: coil magnetic field mein ghoomti hai, isliye field aur coil ke area vector ke beech ka angle lagataar badalta rehta hai. Practical alternating-current machines mein Serbian-American inventor Nikola Tesla ka bahut bada yogdaan hai, aur aaj ka ek hi generator sainkdon megawatts de sakta hai.

Fig. 6.16 ek simple ac generator (प्रत्यावर्ती धारा जनित्र) step by step banata hai. $N$ turns wali ek coil, jise armature kehte hain, magnet ke poles ke beech ek shaft par lagi hai, aur rotation ka axis field ke perpendicular hai. Ek turbine ya engine shaft ko ghumata hai. Coil ka har end apni alag metal slip ring se juda hai, jo shaft ke saath ghoomti hai. Ek pair stationary carbon brushes rings par dabe rehte hain aur current ko load tak bahar le jaate hain.

4/4Brushes se wires alternating current ko load, jaise lamp, tak le jaati hain.

Fig. 6.16Simple ac generator ke main parts.NCERT Fig. 6.13 · p. 170

Upar ek N pole block aur neeche ek S pole block hai, aur unke beech field lines neeche ki taraf chalti hain. Step 1: sirf magnet aur uska field dikhaya gaya hai. Step 2: poles ke beech ek horizontal shaft par ek rectangular armature coil aati hai, aur shaft ke end par ek curved arrow dikhata hai ki woh ghoomti hai. Step 3: coil ke dono ends shaft ke saath magnet ke right mein do slip rings tak jaate hain, paas wali ring ek end se aur door wali doosre end se judi hai, aur har ring par neeche se ek chhota brush dabta hai. Step 4: dono brushes se wires right mein ek lamp tak jaati hain, jo jal jaata hai.

DerivationGhoomti coil ka emf

Maan lo coil ka area $A$ hai aur woh constant angular speed $ω$ se ghoomti hai, aur $t = 0$ par uska area vector $𝐁$ ki direction mein hai. Time $t$ par unke beech angle $θ = ω t$ hai, isliye har turn se flux hai

Φ_{B} = B A \cos θ = B A \cos ω t

Faraday's law se, $N$ -turn coil ka emf hai

ε = - N \frac{d Φ_{B}}{d t} = - N B A \frac{d}{d t} (\cos ω t)

ε = N B A ω \sin ω t

(6.19)

Sabse badi value, jo jab bhi $\sin ω t = \pm 1$ ho tab aati hai, peak emf $ε_{0} = N B A ω$ hai, isliye

ε = ε_{0} \sin ω t, ε_{0} = N B A ω

(6.20)

$ω = 2 π ν$ ke saath, jahan $ν$ revolutions per second ki sankhya hai,

ε = ε_{0} \sin 2 π ν t

(6.21)

2/5Coil edge-on hai: flux zero hai lekin sabse tezi se change ho raha hai, isliye emf apne peak $+ ε_{0}$ par pahunchta hai. Round side mein current page ke andar behta hai.

Fig. 6.17Emf tab sabse zyada hota hai jab flux zero ho, aur zero tab jab flux sabse zyada ho.NCERT Fig. 6.14 · p. 171

Left mein ek end view coil dikhata hai, jo ek copper bar jaisi hai jiske ek end par round wire cross-section aur doosre par square wala hai; yeh upar N pole aur neeche S pole ke beech clockwise ghoomti hai, aur uske area vector ke liye green arrow hai. Right mein ek graph angle ke against, 0 se 360 degree tak, emf ko orange sine curve aur flux ko dashed blue cosine curve ke roop mein plot karta hai, aur dots current values mark karte hain. Step 1: 0 degree par coil level hai, field ke saamne; flux dot maximum par hai aur emf dot zero par. Step 2: 90 degree par coil seedhi khadi hai, round side upar; flux zero hai, emf dot positive peak par hai, aur round side mein cross aur square side mein dot dikhate hain ki current page ke andar aur bahar beh raha hai. Step 3: 180 degree par coil phir level hai, flux apne negative maximum par hai aur emf zero hai. Step 4: 270 degree par coil ulti taraf seedhi khadi hai, round side neeche hai aur ab uspar dot hai, yaani coil ka current ulta ho gaya hai, aur emf negative peak par hai. Step 5: 360 degree par coil wahin wapas aa gayi jahan se shuru hui thi aur emf ka ek poora cycle complete ho gaya.

Fig. 6.17 ek aisi baat saamne laata hai jo aksar students ko hairaan karti hai: emf tab sabse zyada hota hai jab coil se flux zero ho ( $θ = 90^{\circ}$ aur $270^{\circ}$ ), aur zero tab jab flux sabse zyada ho. Faraday's law flux curve ke slope ki parwaah karta hai, aur cosine wahan sabse steep hota hai jahan woh zero cross karta hai. Har half turn par emf ka sign badalta hai, isliye load mein uska chalaya current baar-baar ulta hota rehta hai. Aise current ko alternating current (ac) kehte hain; agla chapter ise handle karne ke tools, uski average values samet, develop karta hai.

SimulationAC generator

Ek rectangular coil do magnet poles ke beech ghoomti hai, aur uski speed tum badal sakte ho. Ek live graph induced emf ko time ke against coil se flux ke saath plot karta hai, taaki tum dekh sako ki emf kahan peak karta hai, kahan zero se guzarta hai aur kahan sign badalta hai.

Fig. 6.18Speed badlo aur emf peaks ki height aur unke beech ki doori, dono dekho.

Try karo

Us pal ko dekho jab coil poles ke bilkul saamne ho, jab flux sabse zyada ho. Us pal emf kya hai?
Rpm double karo. Peak emf ka kya hota hai, aur har second cycles ki sankhya ka?
Original speed par turns ki sankhya aadhi karo. Graph ka kaunsa feature badalta hai, aur kaunsa same rehta hai?

Power-station ke generators bhi isi principle par chalte hain, bas engineering ke sudhaaron ke saath. Badi machines mein electromagnets ghoomte hain, aur bhaari output coils ruki rehti hain; isse sliding contacts se bahut bade currents le jaane ki zaroorat nahi padti. Ek station se doosre mein fark bas yeh hai ki rotor kaise ghumaya jaata hai. Tamil Nadu ka Kudankulam nuclear fuel se steam banata hai, zyadatar Indian thermal stations coal jala kar, Bhakra Nangal girte paani se apni turbines ghumata hai, aur wind turbine chalti hawa use karti hai. India mein generators grid ke saath $50 H z$ par step mein chalte hain, jabki USA samet kuch desh $60 H z$ use karte hain.

Example 6.10NCERT Example 6.10 · p. 172

Ek micro-hydro generator

Pahadi naale ke paas ek micro-hydro plant ek generator chalata hai jiske armature mein 150 turns hain, har ek ka area $0.040 m^{2}$ . Armature $0.20 T$ ke uniform field mein 25 revolutions per second se ghoomta hai. (a) Peak emf nikaalo. (b) Is ac ki frequency kya hai, aur Indian mains frequency par ac dene ke liye coil ko kitni tezi se ghoomna hoga?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

200 turns wali generator coil, har turn ka area

0.030 m^{2}

0.15 T

ke field mein 50 revolutions per second se ghoomti hai. Iska peak emf kitna hai, volts mein?

Answer dekho

Answer 282.74 V

ε_{0} = N B A ω = 200 \times 0.15 \times 0.030 \times (2 π \times 50) \approx 283 V

ω = 2 π \times 50

ki jagah 50 lene par value

2 π

guna chhoti aati hai.

Table 6.3Is chapter ki quantitiesNCERT p. 174

Quantity	Symbol	SI unit	Dimensions	Key relation
Magnetic flux	$Φ_{B}$	weber (Wb)	$[M L^{2} T^{- 2} A^{- 1}]$	$Φ_{B} = 𝐁 \cdot 𝐀$
Induced emf	$ε$	volt (V)	$[M L^{2} T^{- 3} A^{- 1}]$	$ε = - N d Φ_{B} / d t$
Mutual inductance	$M$	henry (H)	$[M L^{2} T^{- 2} A^{- 2}]$	$ε_{1} = - M d I_{2} / d t$
Self-inductance	$L$	henry (H)	$[M L^{2} T^{- 2} A^{- 2}]$	$ε = - L d I / d t$
Magnetic energy density	$u_{B}$	$J / m^{3}$	$[M L^{- 1} T^{- 2}]$	$u_{B} = B^{2} / 2 μ_{0}$

Flash cards · 8

Quick recall

1 / 8

Recap

Summary aur key points

Faraday aur Henry ne paaya ki closed coil mein current sirf tab tak induce hota hai jab tak usse magnetic field badal raha ho, chahe motion se ya switching se (Section 6.2).
Flat surface se magnetic flux $Φ_{B} = 𝐁 \cdot 𝐀 = B A \cos θ$ hai, ek scalar jo webers mein naapa jaata hai (Section 6.3).
Faraday's law: $N$ -turn coil mein induced emf $ε = - N d Φ_{B} / d t$ hai, aur resistance $R$ wale closed circuit mein current $ε / R$ behta hai (Section 6.4).
Flux ko $B$ , area $A$ ya angle $θ$ badal kar badla ja sakta hai, aur har tareeka emf induce karta hai (Fig. 6.6).
Lenz's law: induced current flux ke us change ko oppose karta hai jo use banata hai, jaisa energy conservation maangta hai (Section 6.5).
Uniform field ke right angle par speed $v$ se chalti $l$ length ki rod ka motional emf $B l v$ hai; ek end ke around ghoomti $R$ length ki rod $\frac{1}{2} B ω R^{2}$ deti hai (Section 6.6).
Time ke saath badalta magnetic field ek electric field banata hai, jo stationary conductor mein induced current chalata hai ((6.6)).
Inductance per unit current flux linkage hai, $N Φ_{B} / I$ . Yeh sirf geometry aur core material par depend karta hai aur henries mein naapa jaata hai (Section 6.7).
Mutual inductance: $ε_{1} = - M d I_{2} / d t$ , jahan $M_{12} = M_{21}$ ; coaxial solenoids ke liye $M = μ_{0} n_{1} n_{2} π r_{1}^{2} l$ (Section 6.7.1).
Self-inductance: $ε = - L d I / d t$ current ke har change ko oppose karta hai; lambe solenoid ke liye $L = μ_{r} μ_{0} n^{2} A l$ (Section 6.7.2).
Current $I$ wali coil energy $\frac{1}{2} L I^{2}$ store karti hai; koi bhi magnetic field per unit volume $B^{2} / 2 μ_{0}$ store karta hai ((6.18)).
Ek ac generator magnetic field mein coil ghuma kar $ε = N B A ω \sin ω t$ deta hai, aur mechanical energy ko electrical energy mein badalta hai (Section 6.8).

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercises 6.1	Practice Q1
NCERT Exercises 6.2	Practice Q2
NCERT Exercises 6.3	Practice Q3
NCERT Exercises 6.4	Practice Q4
NCERT Exercises 6.5	Practice Q5
NCERT Exercises 6.6	Practice Q6
NCERT Exercises 6.7	Practice Q7
NCERT Exercises 6.8	Practice Q8

Quick check: MCQs

Warm-up2 questions
6.2 Faraday aur Henry ke experiments4 questions
6.3 Magnetic flux3 questions
6.4 Faraday's law of induction3 questions
6.5 Lenz's law aur energy conservation4 questions
6.6 Motional electromotive force3 questions
Quick check: mutual inductance3 questions
Quick check: self-inductance4 questions
6.8 AC generator3 questions

Glossary · 14

Key terms (English + Hindi)

ac generatorप्रत्यावर्ती धारा जनित्रpratyavarti dhara janitra: Ek machine jo magnetic field mein coil ghumaakar mechanical energy ko alternating emf mein badalti hai.Chapter mein pehli baar yahan aaya →

back emf: Coil mein self-induced emf, jo usmein current ke kisi bhi change ko oppose karta hai.Chapter mein pehli baar yahan aaya →

eddy currentsभँवर धाराएँbhanvar dharaen: Jab kisi solid metal ke tukde se flux change hota hai, to uske andar ghoomne wale induced current ke loops.Chapter mein pehli baar yahan aaya →
electromagnetic inductionवैद्युतचुंबकीय प्रेरणvaidyut-chumbakiya preran: Jab bhi kisi circuit se magnetic flux change hota hai, to usmein emf ka banna.Chapter mein pehli baar yahan aaya →

Faraday's lawफैराडे का नियमFaraday ka niyam: Circuit mein induced emf, usse linked flux ke rate of change ke minus ke barabar hota hai, $ε = - N d Φ_{B} / d t$ .Chapter mein pehli baar yahan aaya →
flux linkage: Coil ke saare turns se guzarne wala total flux; jab $N$ turns mein se har ek same flux $Φ_{B}$ link kare, to $N Φ_{B}$ .Chapter mein pehli baar yahan aaya →

henryहेनरीhenri: Inductance ki SI unit: $1 H = 1 W b / A = 1 V s / A$ .Chapter mein pehli baar yahan aaya →

inductanceप्रेरकत्वpreraktva: Per unit current flux linkage, jo coils ki geometry aur unke core ke material se tay hota hai.Chapter mein pehli baar yahan aaya →

Lenz's lawलेंज़ का नियमLenz ka niyam: Induced current aise ghoomta hai ki uska apna field, flux ke us change ke against kaam kare jisne use banaya.Chapter mein pehli baar yahan aaya →

magnetic fluxचुंबकीय फ्लक्सchumbakiya flux: Kisi surface se guzarne wale magnetic field ka measure; uniform field mein flat surface ke liye $Φ_{B} = B A \cos θ$ .Chapter mein pehli baar yahan aaya →
motional emfगतिक विद्युत वाहक बलgatik vidyut vahak bal: Magnetic field mein chalte conductor ka emf; $𝐁$ ke right angle par chalti rod ke liye $B l v$ .Chapter mein pehli baar yahan aaya →
mutual inductanceअन्योन्य प्रेरकत्वanyonya preraktva: Paas wali coil mein per unit current se ek coil ke saath flux linkage; isse $ε_{1} = - M d I_{2} / d t$ milta hai.Chapter mein pehli baar yahan aaya →

self-inductanceस्वप्रेरकत्वsvapreraktva: Usi coil mein per unit current se coil ke saath flux linkage; isse $ε = - L d I / d t$ milta hai.Chapter mein pehli baar yahan aaya →

weberवेबरvebar: Magnetic flux ki SI unit: $1 W b = 1 T m^{2}$ .Chapter mein pehli baar yahan aaya →

Electromagnetic Induction: Class 12 Physics Chapter 6 Hinglish mein

Is chapter mein aap kya seekhenge

6.1 Introduction

6.2 Faraday aur Henry ke experiments

6.3 Magnetic flux

6.4 Faraday's law of induction

6.5 Lenz's law aur energy conservation

6.6 Motional electromotive force

6.7 Inductance

6.7.1 Mutual inductance

6.7.2 Self-inductance

6.8 AC generator

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye