Class 12 Physics · Chapter 6
Short answer:
Ye chapter NCERT ke 6 exercise questions cover karta hai, lekin exam ke liye sirf textbook questions kaafi nahi — is page pe hum 16 questions solve kar rahe hain jisme magnetic flux calculation, Faraday's law se EMF nikalna, Lenz's law se induced current ki direction, motional EMF (rod moving in magnetic field), self-inductance aur mutual inductance ke numericals, aur AC generator ka basic EMF equation shamil hai. Concept + numerical dono tarah ke questions is chapter me aate hain, isliye formula aur unit dono pe pakad zaroori hai.
Electromagnetic Induction wo phenomenon hai jisme changing magnetic flux ek closed circuit me EMF (aur current) induce karta hai. Ye chapter Faraday aur Lenz ke experiments se shuru hota hai — jab bhi coil ke through magnetic flux change hota hai (chahe magnet move ho, coil move ho, ya field khud change ho), ek EMF induce hoti hai. Lenz's law batata hai ye induced EMF/current kis direction me hogi — hamesha us change ka opposition karte hue, jisse energy conservation maintain rehta hai. Aage hum dekhte hain ki motion se (motional EMF) aur coil ke apne current change se (self-inductance) bhi EMF induce ho sakti hai, aur ye principle hi AC generator ka base hai.
Chapter 6 Summary — 5 Minute Revision
1. Magnetic Flux
Magnetic flux ek surface se guzarne wali magnetic field lines ka measure hai.
Φ = B·A·cosθ
Jahan B = magnetic field (tesla, T), A = area (m²), θ = angle field aur area-vector ke beech. SI unit weber (Wb). Flux scalar quantity hai lekin sign matter karta hai (direction convention ke through).
2. Faraday's Laws of Induction
- First law: Jab bhi circuit se linked magnetic flux change hoti hai, ek EMF induce hoti hai.
- Second law: Induced EMF ka magnitude flux-linkage ke time rate of change ke proportional hota hai.
ε = −N(dΦ/dt)
N = coil me turns ki number. Negative sign Lenz's law represent karta hai.
3. Lenz's Law
Induced current hamesha us change (cause) ka virodh karta hai jisne use paida kiya. Ye energy conservation ka natural result hai — agar induced current apne hi cause ko support karta, to system se free energy nikalti rehti, jo possible nahi hai. Isliye jab flux badh raha ho, induced current apna khud ka magnetic field ulti direction me banata hai (flux opose karne ke liye), aur jab flux ghat raha ho, induced current usi direction ka field banata hai jisse flux ko maintain kare.
4. Motional EMF
Jab ek conducting rod magnetic field ke andar move karti hai, uske free electrons pe magnetic force lagti hai jo EMF paida karti hai.
ε = Blv
B = field, l = rod ki length (field ke perpendicular), v = velocity (field aur rod dono ke perpendicular). Direction Fleming's right-hand rule ya F = qv×B se determine hoti hai.
5. Self-Inductance
Coil me current change hone se uske apne flux-linkage me change hota hai, jo coil me hi EMF induce karta hai — is property ko self-inductance (L) kehte hain.
ε = −L(dI/dt)
SI unit henry (H). Long solenoid ke liye:
L = μ₀n²Al = μ₀N²A/l
n = turns per unit length, A = cross-section area, l = solenoid ki length.
6. Mutual Inductance
Do coils jab paas-paas hon, ek coil ka changing current dusri coil me EMF induce karta hai — is property ko mutual inductance (M) kehte hain.
ε₂ = −M(dI₁/dt)
Do coaxial solenoids (inner solenoid coil 2, outer coil 1, common length l) ke liye mutual inductance:
M = μ₀N₁N₂A/l
7. AC Generator — Basic Principle
AC generator electromagnetic induction ke principle pe kaam karta hai — coil ko uniform magnetic field me constant angular speed se rotate karaya jata hai, jisse flux sinusoidally change hoti hai aur alternating EMF induce hoti hai.
ε = NBAω sin(ωt) = ε₀ sin(ωt)
Yahan ε₀ = NBAω peak EMF hai, ω = angular frequency of rotation.
| Concept | Kya represent karta hai |
|---|---|
| Magnetic flux Φ | Surface se guzarne wali field lines |
| Faraday's law | EMF ∝ rate of change of flux-linkage |
| Lenz's law | Induced EMF ki direction — opposition of change |
| Motional EMF | Moving conductor me induced EMF |
| Self-inductance L | Coil ka apna current change EMF induce kare |
| Mutual inductance M | Ek coil ka current dusri coil me EMF induce kare |
| AC generator | Rotating coil se sinusoidal EMF |
↔ Table ko side me swipe karein
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IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q16)
Q1. Ek circular coil ka radius 8 cm aur turns 20 hain, ek uniform magnetic field B = 0.3 T me rakhi hai jo coil ke plane se 60° ka angle banati hai. Coil se linked magnetic flux nikaliye.
Flux formula lagate hain, dhyan rahe θ area-vector se angle hai — agar field plane se 60° ka angle banati hai, to area-vector se angle 30° hoga.
A = πr² = π(0.08)² = 0.0201 m²
θ (from normal) = 90° − 60° = 30°
Φ (per turn) = B·A·cosθ = 0.3 × 0.0201 × cos30° = 0.3 × 0.0201 × 0.866 = 5.22×10⁻³ Wb
Total flux linkage NΦ = 20 × 5.22×10⁻³ = 0.104 Wb
Q2. Ek coil ka flux 0.5 s me 0.02 Wb se 0.08 Wb tak badhta hai. Coil me 50 turns hain. Induced EMF nikaliye.
Faraday's law directly lagta hai.
dΦ = 0.08 − 0.02 = 0.06 Wb
dt = 0.5 s
ε = −N(dΦ/dt) = −50 × (0.06/0.5) = −50 × 0.12 = −6 V
Magnitude = 6 V
Q3. Ek 1.5 m lambi conducting rod, 4 T ke uniform magnetic field ke perpendicular, 2 m/s ki speed se field ke bhi perpendicular direction me move kar rahi hai. Rod ke ends ke beech induced EMF nikaliye.
Motional EMF formula ε = Blv seedha lagta hai kyunki B, l, v teeno mutually perpendicular hain.
ε = Blv = 4 × 1.5 × 2 = 12 V
Q4. Ek bar magnet ka N-pole ek coil ki taraf paas laya ja raha hai. Coil me induced current ki direction Lenz's law se batayein (coil ko us side se dekha ja raha hai jis side se magnet aa raha hai).
Lenz's law: induced current us change ka virodh karega jisne use paida kiya — yahan flux (magnet ki taraf, coil ke andar) badh rahi hai, isliye induced current apna field is badhti hui flux ko oppose karne ke liye banayega.
Coil ki near-side face ka polarity N-pole jaisi honi chahiye (taaki magnet ko repel kare aur uska paas aana oppose ho)
Is polarity ke liye current anticlockwise direction me flow karega (jab observer magnet ki side se dekhe)
Q5. Same setup me agar bar magnet ka N-pole coil se door le jaya jaye, to induced current ki direction kya hogi?
Ab flux ghat rahi hai (magnet door ja raha hai), isliye Lenz's law ke mutabik induced current is decrease ko oppose karega — matlab flux ko maintain karne ki koshish karega.
Near-side face S-pole jaisi banegi (taaki magnet ko attract kare, uska door jaana oppose ho)
Current clockwise direction me flow karega (jab observer magnet ki side se dekhe) — pehle wale case se opposite
Q6. 500 turns wali ek coil, jiska area 0.01 m² hai, ek magnetic field me hai jo uniformly 0.2 T se 0.8 T tak 0.3 s me badhta hai (field coil ke plane ke perpendicular). Average induced EMF nikaliye.
Yahan B change ho raha hai, A constant hai, θ = 0° (field area-vector ke parallel).
Φ_initial = B₁A = 0.2 × 0.01 = 2×10⁻³ Wb
Φ_final = B₂A = 0.8 × 0.01 = 8×10⁻³ Wb
dΦ = 8×10⁻³ − 2×10⁻³ = 6×10⁻³ Wb
ε = N(dΦ/dt) = 500 × (6×10⁻³/0.3) = 500 × 0.02 = 10 V
Q7. Ek solenoid ki length 0.5 m, cross-section area 4×10⁻⁴ m², aur turns 1000 hain. Solenoid ka self-inductance nikaliye. (μ₀ = 4π×10⁻⁷ T·m/A)
Long solenoid formula L = μ₀N²A/l use karte hain.
L = μ₀N²A/l
= (4π×10⁻⁷ × 1000² × 4×10⁻⁴) / 0.5
= (4π×10⁻⁷ × 10⁶ × 4×10⁻⁴) / 0.5
= (4π×10⁻⁷ × 400) / 0.5
= (5.03×10⁻⁴) / 0.5
= 1.0×10⁻³ H = 1 mH (approx)
Q8. Ek 0.6 H self-inductance wali coil me current 0.2 s me 2 A se 5 A tak badhta hai. Induced EMF (magnitude) nikaliye.
Self-inductance formula seedha lagta hai.
dI = 5 − 2 = 3 A
dt = 0.2 s
ε = L(dI/dt) = 0.6 × (3/0.2) = 0.6 × 15 = 9 V
Q9. Do coaxial solenoids: inner solenoid me 800 turns, outer solenoid me 1200 turns, common length 0.4 m, aur inner solenoid ka cross-section area 3×10⁻⁴ m² hai. Mutual inductance nikaliye. (μ₀ = 4π×10⁻⁷ T·m/A)
Mutual inductance formula M = μ₀N₁N₂A/l lagta hai (A = common/inner area jispe dono ka flux link hota hai).
M = μ₀N₁N₂A/l
= (4π×10⁻⁷ × 800 × 1200 × 3×10⁻⁴) / 0.4
= (4π×10⁻⁷ × 9.6×10⁵ × 3×10⁻⁴) / 0.4
= (4π×10⁻⁷ × 288) / 0.4
= (3.62×10⁻⁴) / 0.4
= 9.05×10⁻⁴ H ≈ 0.91 mH
Q10. Ek outer coil me current 0.1 s me 0 se 4 A tak badhta hai, aur is wajah se inner coil me 0.02 V ki EMF induce hoti hai. Mutual inductance nikaliye.
Mutual EMF formula se M nikal lete hain.
ε₂ = M(dI₁/dt)
0.02 = M × (4/0.1)
0.02 = M × 40
M = 0.02/40 = 5×10⁻⁴ H = 0.5 mH
Q11. Ek AC generator me coil ke 100 turns hain, area 0.05 m², field 0.4 T hai, aur coil 50 Hz se rotate ho rahi hai. Peak EMF nikaliye.
AC generator ka basic formula lagta hai, pehle ω nikalna hoga.
ω = 2πf = 2π × 50 = 314.16 rad/s
ε₀ = NBAω = 100 × 0.4 × 0.05 × 314.16
= 100 × 0.4 × 0.05 = 2, phir 2 × 314.16 = 628.3 V
Q12. Above wale generator me EMF ka instantaneous value uस waqt nikaliye jab coil ne apni starting position (plane field ke parallel) se 30° rotate kar liya ho.
ε = ε₀ sin(ωt), yahan ωt hi wo angle hai jitna coil rotate hui hai.
ε = ε₀ sinθ = 628.3 × sin30° = 628.3 × 0.5 = 314.15 V
Q13. Ek rod 0.8 m lambi, 0.5 T field ke perpendicular rakhi, field ke bhi perpendicular direction me move kar rahi hai aur uske ends pe 2 V EMF induce ho rahi hai. Rod ki velocity nikaliye.
Motional EMF formula ko v ke liye rearrange karte hain.
ε = Blv
2 = 0.5 × 0.8 × v
2 = 0.4v
v = 2/0.4 = 5 m/s
Q14. Ek square loop (side 10 cm) ek uniform magnetic field B = 0.6 T me hai jo loop ke plane ke parallel hai (field aur area-vector ke beech 90°). Loop se linked flux kitni hai?
Jab field area-vector ke perpendicular ho (θ = 90°), flux zero hoti hai — ye conceptual trap question hai.
θ = 90° (field plane ke parallel matlab area-vector ke perpendicular)
Φ = BAcos90° = BA × 0 = 0 Wb
Q15. Ek coil me 200 turns hain, aur usse linked flux ek equation Φ = (3t² + 2t) mWb se diya gaya hai (t seconds me). t = 2 s pe induced EMF nikaliye.
Yahan dΦ/dt nikalne ke liye calculus (differentiation) use hota hai.
Φ = (3t² + 2t) × 10⁻³ Wb
dΦ/dt = (6t + 2) × 10⁻³ Wb/s
t = 2 pe: dΦ/dt = (6×2 + 2) × 10⁻³ = 14×10⁻³ Wb/s
ε = N(dΦ/dt) = 200 × 14×10⁻³ = 2.8 V
Q16. Ek solenoid (self-inductance 2 mH) me current sinusoidally I = I₀sin(ωt) se change hota hai jahan I₀ = 3 A aur f = 60 Hz. Maximum self-induced EMF nikaliye.
dI/dt ka maximum value I₀ω hota hai (jab cos term apne peak pe ho), isliye maximum EMF = LI₀ω.
ω = 2πf = 2π × 60 = 376.99 rad/s
ε_max = L × I₀ × ω = 2×10⁻³ × 3 × 376.99
= 2×10⁻³ × 1130.97
= 2.26 V
Important Equations — Ek Nazar Me
| Quantity | Formula | Unit |
|---|---|---|
| Magnetic flux | Φ = BAcosθ | Wb (weber) |
| Faraday's law (EMF) | ε = −N(dΦ/dt) | V (volt) |
| Motional EMF | ε = Blv | V (volt) |
| Self-inductance (defining eqn) | ε = −L(dI/dt) | H (henry) |
| Self-inductance of solenoid | L = μ₀n²Al = μ₀N²A/l | H (henry) |
| Mutual EMF | ε₂ = −M(dI₁/dt) | V (volt) |
| Mutual inductance (coaxial solenoids) | M = μ₀N₁N₂A/l | H (henry) |
| AC generator EMF | ε = NBAω sin(ωt) = ε₀ sin(ωt) | V (volt) |
| Angular frequency | ω = 2πf | rad/s |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Negative sign bhool jaana Faraday's law me. ε = −N(dΦ/dt) me negative sign sirf formality nahi hai — ye Lenz's law ko represent karta hai (induced EMF apne cause ka virodh karti hai). Numerical me magnitude poochi ho to sign drop kar sakte hain, lekin conceptual questions me sign ka matlab samajhna zaroori hai.
- θ ko field aur plane ke beech ka angle maan lena, area-vector ke beech ka nahi. Φ = BAcosθ me θ hamesha field aur area-vector (jo plane ke perpendicular hota hai) ke beech ka angle hota hai. Agar question me field aur plane ke beech ka angle diya ho, to pehle 90° minus karke area-vector wala angle nikalo.
- Self-inductance aur mutual inductance ke formula confuse karna. Self-inductance ek hi coil ka property hai (L = μ₀N²A/l), jabki mutual inductance do coils ke beech hoti hai (M = μ₀N₁N₂A/l). Numerical me galat N use karna (jaise dono jagah same N) common mistake hai.
- Flux-linkage me N (number of turns) forget karna. Single loop ka flux Φ = BAcosθ hota hai, lekin N-turn coil ka total flux-linkage NΦ hota hai, aur EMF formula me bhi N zaroori hai — ε = N(dΦ/dt), sirf dΦ/dt nahi.
- Motional EMF ki direction galat nikalna. Rod me induced EMF ki direction F = qv×B (ya Fleming's right-hand rule) se milti hai, na ki random guess se. Velocity, field, aur force teeno vectors ka relative orientation carefully check karo — galat direction se galat polarity ka answer aata hai.
- AC generator ke formula me ω aur f mix kar dena. Peak EMF ε₀ = NBAω hota hai, jahan ω = 2πf hai — seedha f use kar dena (ε₀ = NBAf) common galti hai jo answer ko 2π factor se galat kar deti hai.
Board-Style Important Questions
- 1 mark: Lenz's law kis conservation principle se related hai?
- 2 marks: Ek coil ka self-inductance define karo aur uska SI unit likho.
- 2 marks: Ek uniform magnetic field me rotate karti hui coil me EMF induce hone ka basic principle samjhaiye (AC generator).
- 3 marks: Faraday's laws of electromagnetic induction likho aur ε = −N(dΦ/dt) formula derive/state karo, negative sign ka matlab explain karte hue.
- 3 marks: Ek conducting rod, magnetic field me perpendicular direction me move kar rahi hai — motional EMF ka expression derive karo.
- 5 marks: Do coaxial solenoids ke beech mutual inductance ka expression derive karo, aur ek numerical solve karo jisme turns, area, aur length diye hon.
Aksar Poochhe Jaane Wale Sawaal
Electromagnetic induction kya hai?
Ye wo phenomenon hai jisme ek closed circuit se linked magnetic flux change hone par uss circuit me EMF (aur agar circuit complete hai to current) induce hoti hai.
Faraday's law aur Lenz's law me kya difference hai?
Faraday's law batata hai induced EMF ka magnitude kitna hoga (flux change ki rate ke proportional), jabki Lenz's law batata hai us EMF/current ki direction kya hogi — hamesha change ka opposition karte hue.
Motional EMF aur Faraday's law wali EMF me kya farak hai?
Motional EMF tab hoti hai jab conductor khud magnetic field me move karta hai (field constant reh sakti hai). Faraday's law general case cover karta hai — chahe field change ho, coil move ho, ya dono.
Self-inductance aur mutual inductance me kya difference hai?
Self-inductance ek coil ka apna current change hone se usi coil me EMF induce karne ka property hai. Mutual inductance do alag coils ke beech hoti hai — ek coil ka current change dusri coil me EMF induce karta hai.
Kya eddy currents is chapter me exam ke liye important hain?
Current rationalised syllabus me eddy currents ka topic fully drop kar diya gaya hai — isliye eddy current applications ab exam-testable content nahi hain.
AC generator ka basic working principle kya hai?
Coil ko uniform magnetic field me constant angular speed se rotate karaya jata hai, jisse coil se linked flux sinusoidally change hoti hai aur electromagnetic induction ke through alternating EMF (ε = ε₀sinωt) generate hoti hai.
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