Class 12 Physics · Chapter 4
Short answer:
NCERT ke is chapter me total 16 exercise questions hain, aur ye chapter poora numerical-heavy hai — direct formula-based questions kam, aur "diya gaya hai, nikaalo" type zyada. Is guide me sabhi 16 exercise questions poori working ke saath cover kiye gaye hain: moving charge par force, current-carrying conductor par force, Biot-Savart law se field nikaalna, Ampere's law se solenoid aur straight wire ka field, do parallel current-carrying wires ke beech force, current loop par torque, aur moving coil galvanometer ki sensitivity/conversion. Directions ke liye right-hand rule bahut baar lagega — is chapter me sabse zyada exam marks direction-based silly mistakes me katte hain, isliye "Common Mistakes" section zaroor padhna.
Ye chapter batata hai ki moving charges aur current-carrying conductors magnetic field kaise banate hain, aur khud magnetic field me kaisa force feel karte hain. Pichhle chapter (Electric Charges and Fields) me humne dekha tha ki stationary charge electric field banata hai. Ab hum dekhte hain ki jab charge move karta hai — ya jab current kisi wire me flow karta hai — to woh ek nayi field, magnetic field, banata hai. Biot-Savart law point current element ke liye magnetic field deta hai, Ampere's law symmetric current distributions ke liye seedha field nikalne me kaam aata hai — dono ek hi cheez ke do tareeke hain, jaise Coulomb's law aur Gauss's law electric field ke liye. Chapter ke end me ye sab ek real instrument — moving coil galvanometer — me use hota hai, jisse current aur voltage measure karte hain.
Chapter 4 Summary — 5 Minute Revision
1. Magnetic force on a moving charge
Jab charge q velocity v se magnetic field B me move karta hai, to us par force lagta hai:
F = qv × B ya scalar form: F = qvB sinθ
Yahan θ, v aur B ke beech ka angle hai. Direction right-hand rule se milti hai — agar q positive hai to fingers v ki direction me point karo, curl karo B ki taraf, thumb F ki direction dega (cross product rule). Negative charge ke liye direction ulti ho jaayegi.
- Agar v aur B parallel/antiparallel hain (θ = 0° ya 180°) → F = 0.
- Agar v, B ke perpendicular hai (θ = 90°) → force maximum, F = qvB.
- Magnetic force velocity ke perpendicular hamesha lagta hai, isliye ye speed nahi badalta, sirf direction badalta hai — kaam (work) zero karta hai.
2. Magnetic force on a current-carrying conductor
Current dar-asal moving charges ka flow hai, isliye current-carrying wire bhi magnetic field me force feel karta hai:
F = IL × B ya scalar form: F = BIL sinθ
Yahan L, current ki direction me wire ki length hai aur θ, current aur B ke beech ka angle hai. Direction phir se right-hand rule (ya Fleming's left-hand rule, jo bhi comfortable ho) se milti hai.
3. Biot-Savart law
Ye law batata hai ki ek chhota current element I dl apne aas-paas kisi bhi point par kitna magnetic field dB banata hai:
dB = (μ0/4π) · (I dl × r̂)/r²
μ0 = 4π × 10⁻⁷ T·m/A, free space ki permeability hai. Ye Coulomb's law jaisa hi hai — bas point charge ki jagah current element hai, aur field ka direction cross product se aata hai (dl aur r̂ ke perpendicular plane me). Puri wire ka field nikalne ke liye sab chhote elements ke dB ko integrate karte hain.
Circular loop ke centre par field (Biot-Savart ka standard application):
B = μ0I / 2R (N turns ho to: B = μ0NI / 2R)
4. Ampere's circuital law
Symmetric current distributions ke liye field seedha nikalne ka shortcut:
∮ B · dl = μ0 Ienc
Matlab: kisi bhi closed loop (Amperian loop) ke around B ka line integral, us loop ke andar se guzarne wale total current ke μ0 guna ke barabar hota hai. Ye Gauss's law jaisa hi symmetry-based shortcut hai.
Field due to a long straight current-carrying wire (Ampere's law se, distance r par):
B = μ0I / 2πr
Field inside a long solenoid (Ampere's law se, axis ke paas uniform field):
B = μ0nI (n = turns per unit length = N/l)
Solenoid ke bahar field practically zero hota hai, jabki ek single circular loop ka field poori jagah non-zero hota hai — ye dono formulas confuse mat karo.
5. Force between two parallel current-carrying conductors
Do parallel wires jinme current I₁ aur I₂ flow kar rahe hain, ek doosre ko force lagate hain (kyunki ek wire doosre ki field me hai):
F/L = μ0 I1 I2 / 2πd
- Same direction currents → wires attract karte hain.
- Opposite direction currents → wires repel karte hain.
Isi force se ampere (A) ki SI definition banti hai.
6. Torque on a current loop in a magnetic field
Current loop ek chhota magnet jaisa behave karta hai — uska apna magnetic moment hota hai:
m = NIA (N = turns, I = current, A = loop area)
Uniform field B me rakha loop torque feel karta hai jo use B ke saath align karne ki koshish karta hai:
τ = NIAB sinθ = m × B
θ loop ke plane ki normal aur B ke beech ka angle hai. θ = 90° (loop ka plane B ke parallel) par torque maximum hota hai; θ = 0° (loop ka plane B ke perpendicular, normal B ke parallel) par torque zero hota hai — is stage par loop equilibrium me hota hai.
7. Moving coil galvanometer
Galvanometer ek chhota current-measuring device hai jo current loop par torque ke principle par kaam karta hai. Coil ek radial magnetic field me spring/suspension ke against deflect hoti hai. Equilibrium par deflecting torque = restoring torque (kφ):
NIAB = kφ → I = (k/NAB) φ
Current sensitivity (deflection per unit current):
Current sensitivity = φ/I = NAB/k
Voltage sensitivity (deflection per unit voltage), coil resistance R ho to:
Voltage sensitivity = φ/V = NAB/(kR)
Galvanometer ko Ammeter banana — low resistance shunt S parallel me lagate hain taaki bada current diverted ho jaaye:
S = IgG / (I − Ig)
Galvanometer ko Voltmeter banana — high resistance R series me lagate hain taaki bada voltage measure ho sake:
R = V/Ig − G
(G = galvanometer ka apna resistance, Ig = full-scale deflection current)
Quick recap table
| Situation | Formula |
|---|---|
| Force on moving charge | F = qvB sinθ |
| Force on current-carrying wire | F = BIL sinθ |
| Field at centre of circular loop | B = μ₀NI/2R |
| Field due to long straight wire | B = μ₀I/2πr |
| Field inside solenoid | B = μ₀nI |
| Force between parallel wires | F/L = μ₀I₁I₂/2πd |
| Torque on current loop | τ = NIAB sinθ |
↔ Table ko side me swipe karein
Poore Class 12 Physics ke handwritten colour notes
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 proton, magnetic field B = 0.5 T ke perpendicular, v = 3 × 10⁷ m/s speed se move kar raha hai. Proton par lagne wala magnetic force nikaliye. (q = 1.6 × 10⁻¹⁹ C)
Proton perpendicular move kar raha hai, isliye θ = 90°, sinθ = 1.
F = qvB sinθ
F = (1.6 × 10⁻¹⁹)(3 × 10⁷)(0.5)(1)
F = 2.4 × 10⁻¹² N
Direction: right-hand rule se v × B ki direction (positive charge hai isliye seedha use karo).
Q2. Ek electron beam, magnetic field B = 0.2 T se 30° ke angle par v = 2 × 10⁶ m/s se guzar rahi hai. Electron par force nikaliye. (e = 1.6 × 10⁻¹⁹ C)
Given: v = 2 × 10⁶ m/s, B = 0.2 T, θ = 30°
F = qvB sinθ
F = (1.6 × 10⁻¹⁹)(2 × 10⁶)(0.2)(sin30°)
F = (1.6 × 10⁻¹⁹)(2 × 10⁶)(0.2)(0.5)
F = 3.2 × 10⁻¹⁴ N
Electron negative hai, isliye actual force direction v × B ke ulti hogi.
Q3. Ek electron, v = 3 × 10⁶ m/s se, field ke perpendicular B = 1 × 10⁻³ T me daakhil hota hai. Uske circular path ka radius nikaliye. (me = 9.11 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C)
Magnetic force hi yahan centripetal force ka kaam karta hai: qvB = mv²/r, isliye r = mv/(qB)
r = mv/(eB) = (9.11 × 10⁻³¹ × 3 × 10⁶) / (1.6 × 10⁻¹⁹ × 1 × 10⁻³)
r = (2.733 × 10⁻²⁴) / (1.6 × 10⁻²²)
r ≈ 1.71 × 10⁻² m = 1.71 cm
Path circular isliye hai kyunki magnetic force hamesha v ke perpendicular lagta hai.
Q4. Ek 1.5 m lambi seedhi wire, current I = 10 A carry kar rahi hai aur field B = 0.6 T ke exactly perpendicular rakhi hai. Wire par force nikaliye.
θ = 90°, sinθ = 1
F = BIL sinθ
F = (0.6)(10)(1.5)(1)
F = 9 N
Q5. Ek wire jisme 12 A current hai use radius R = 2.0 cm ka semicircular arc banaya gaya hai. Arc ke centre par magnetic field nikaliye.
Poore circular loop ka field B₀ = μ₀I/2R hota hai. Semicircular arc poore loop ka aadha hissa hai, isliye centre par field bhi aadha hoga:
B = (1/2) × μ0I/2R = μ0I/4R
B = (4π × 10⁻⁷ × 12) / (4 × 0.02)
B = (1.508 × 10⁻⁵) / (0.08)
B ≈ 1.88 × 10⁻⁴ T
Q6. Ek circular coil ka radius R = 8 cm hai, jisme N = 100 turns hain aur current I = 0.4 A flow kar raha hai. Coil ke centre par magnetic field nikaliye.
Given: N = 100, I = 0.4 A, R = 0.08 m
B = μ0NI / 2R
B = (4π × 10⁻⁷ × 100 × 0.4) / (2 × 0.08)
B = (5.027 × 10⁻⁵) / (0.16)
B ≈ 3.14 × 10⁻⁴ T
Q7. Ek current element I dl = (5 A)(0.01 m), origin par x-axis ki direction me rakha hai. Y-axis par 0.1 m door point par magnetic field dB nikaliye.
r, dl ke perpendicular hai (dl x-axis pe, r y-axis pe), isliye θ = 90°, sinθ = 1
dB = (μ0/4π) · (I dl sinθ)/r²
dB = (10⁻⁷)(5)(0.01)(1) / (0.1)²
dB = (5 × 10⁻⁹) / (0.01)
dB = 5 × 10⁻⁷ T
Q8. Ek 100 cm lambi seedhi wire se 10 A current flow kar raha hai. Wire se 10 cm door magnetic field nikaliye.
Wire ko lamba (effectively infinite) maan sakte hain kyunki distance uski length ke comparison me chhota hai.
B = μ0I / 2πr
B = (4π × 10⁻⁷ × 10) / (2π × 0.1)
B = (2 × 10⁻⁷ × 10) / 0.1
B = 2 × 10⁻⁵ T
Q9. Ek solenoid, jiski length 0.5 m hai aur usme 500 turns hain, me 5 A current flow ho raha hai. Solenoid ke andar axis par magnetic field nikaliye.
Pehle turns per unit length nikaalo:
n = N/l = 500/0.5 = 1000 turns/m
B = μ0nI
B = (4π × 10⁻⁷)(1000)(5)
B ≈ 6.28 × 10⁻³ T = 6.28 mT
Q10. Ek solenoid me n = 2000 turns/m hain. Andar B = 0.01 T field chahiye. Kitna current pass karna hoga?
B = μ₀nI ko I ke liye rearrange karo:
I = B / (μ0n)
I = 0.01 / (4π × 10⁻⁷ × 2000)
I = 0.01 / (2.513 × 10⁻³)
I ≈ 3.98 A
Q11. Do parallel wires 1 m door hain, dono me same direction me 5 A current flow ho raha hai. Per unit length force nikaliye aur bataiye attractive hai ya repulsive.
F/L = μ0I1I2 / 2πd
F/L = (4π × 10⁻⁷)(5)(5) / (2π × 1)
F/L = (2 × 10⁻⁷)(25) / 1
F/L = 5 × 10⁻⁶ N/m
Currents same direction me hain, isliye force attractive hoga.
Q12. Do parallel wires 0.05 m door hain, jinme opposite direction me I₁ = 8 A aur I₂ = 5 A current hai. Per unit length force nikaliye aur nature bataiye.
F/L = μ0I1I2 / 2πd
F/L = (4π × 10⁻⁷)(8)(5) / (2π × 0.05)
F/L = (2 × 10⁻⁷)(40) / 0.05
F/L = (8 × 10⁻⁶) / 0.05 = 1.6 × 10⁻⁴ N/m
Currents opposite direction me hain, isliye force repulsive hoga.
Q13. Ek circular coil (radius 10 cm, N = 50 turns) me 4 A current hai, aur ye uniform field B = 0.5 T me is tarah rakha hai ki coil ka plane B ke parallel hai. Coil par lagne wala torque nikaliye.
Coil ka plane B ke parallel hai, matlab normal B ke perpendicular hai, isliye θ = 90° — ye maximum torque wali situation hai.
A = πR² = π(0.1)² = 0.0314 m²
τ = NIAB sinθ = (50)(4)(0.0314)(0.5)(1)
τ ≈ 3.14 N·m
Q14. Ek rectangular loop (5 cm × 10 cm), N = 20 turns, I = 3 A, uniform field B = 0.4 T me maximum torque position me rakha hai. Torque nikaliye.
A = 0.05 × 0.10 = 5 × 10⁻³ m²
τmax = NIAB = (20)(3)(5 × 10⁻³)(0.4)
τmax = 0.12 N·m
Q15. Ek moving coil galvanometer me N = 30, A = 1.5 × 10⁻³ m², B = 0.25 T, aur torsional constant k = 1.5 × 10⁻⁷ N·m/rad hai. Current sensitivity nikaliye, aur agar coil resistance R = 20 Ω ho to voltage sensitivity bhi nikaliye.
Current sensitivity:
CS = NAB/k = (30)(1.5 × 10⁻³)(0.25) / (1.5 × 10⁻⁷)
CS = (0.01125) / (1.5 × 10⁻⁷)
CS = 7.5 × 10⁴ rad/A
Voltage sensitivity:
VS = CS/R = (7.5 × 10⁴)/20
VS = 3.75 × 10³ rad/V
Q16. Ek galvanometer ka resistance G = 15 Ω hai aur full-scale deflection Iɡ = 4 mA current par hoti hai. Ise 0–3 V range ka voltmeter banane ke liye series resistance nikaliye.
R = V/Ig − G
R = 3/(4 × 10⁻³) − 15
R = 750 − 15
R = 735 Ω
Important Equations — Ek Nazar Me
| Concept | Formula |
|---|---|
| Force on moving charge (scalar) | F = qvB sinθ |
| Force on moving charge (vector) | F = qv × B |
| Force on current-carrying conductor | F = BIL sinθ |
| Biot-Savart law | dB = (μ₀/4π) · (I dl × r̂)/r² |
| Field at centre of circular loop | B = μ₀NI/2R |
| Ampere's circuital law | ∮ B · dl = μ₀Ienc |
| Field due to long straight wire | B = μ₀I/2πr |
| Field inside a solenoid | B = μ₀nI (n = N/l) |
| Force between parallel current-carrying wires | F/L = μ₀I₁I₂/2πd |
| Magnetic moment of current loop | m = NIA |
| Torque on current loop | τ = NIAB sinθ = m × B |
| Galvanometer current sensitivity | φ/I = NAB/k |
| Galvanometer voltage sensitivity | φ/V = NAB/(kR) |
| Shunt for ammeter conversion | S = IgG/(I − Ig) |
| Series resistance for voltmeter conversion | R = V/Ig − G |
| Permeability of free space | μ₀ = 4π × 10⁻⁷ T·m/A |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Right-hand rule galat lagana. F = qv × B me students haath ki position galat rakh dete hain — fingers v ki direction me point karke B ki taraf curl karna hai, thumb F degi. Agar charge negative hai to answer ki direction ulti karna mat bhoolo — ye step exam me sabse zyada skip hota hai.
- Semicircular ya partial loop me poora formula laga dena. B = μ₀I/2R sirf poore circular loop ke centre ke liye hai. Agar sirf semicircular arc diya ho, to answer aadha karna padega (μ₀I/4R) — students seedha poora formula laga dete hain.
- Solenoid aur single loop ka formula confuse karna. Solenoid ke andar B = μ₀nI hota hai (n = turns per unit length), jabki single loop ke centre par B = μ₀NI/2R hota hai. In dono formulas ko mix karna bahut common mistake hai — dhyan rakho konsa case diya hai.
- Parallel currents ke attraction/repulsion me sign galat karna. Same direction currents attract karte hain, opposite direction currents repel karte hain — students ise ulta yaad rakh lete hain. Force ka magnitude sahi aa jaata hai par nature (attractive/repulsive) galat likh dete hain.
- Torque ke formula me sinθ ka angle galat lena. τ = NIAB sinθ me θ, loop ke normal aur B ke beech ka angle hai, na ki loop ke plane aur B ke beech ka. Agar plane B ke parallel hai to normal B ke perpendicular hoga, matlab θ = 90° (maximum torque) — is confusion se ulta answer aa jaata hai.
- Ammeter/voltmeter conversion formulas swap kar dena. Shunt (S = IgG/(I−Ig)) ammeter banane ke liye parallel me lagta hai, aur series resistance (R = V/Ig − G) voltmeter banane ke liye series me lagta hai. Students dono formulas aur connections ko swap kar dete hain.
Board-Style Important Questions
- 1 mark: Ek proton aur ek electron same velocity se same uniform magnetic field me perpendicular daakhil hote hain. Dono par lagne wale force ka magnitude same hoga ya alag? Reason bataiye.
- 2 marks: Biot-Savart law likhiye aur ise Coulomb's law se compare kijiye (do similarities/differences).
- 2 marks: Ek current-carrying solenoid ke andar aur bahar magnetic field kaisa hota hai, ise short me samjhaiye.
- 3 marks: Do parallel current-carrying conductors ke beech force ka expression derive kijiye, aur SI unit 'ampere' is force se kaise define hota hai, batayein.
- 3 marks: Moving coil galvanometer ka working principle samjhaiye aur current sensitivity ka expression likhiye. Sensitivity badhane ke do tareeke bataiye.
- 5 marks: Ampere's circuital law statement kijiye, aur isse ek long straight current-carrying wire ke bahar magnetic field ka expression derive kijiye.
Aksar Poochhe Jaane Wale Sawaal
Biot-Savart law aur Ampere's law me kya farak hai?
Biot-Savart law kisi bhi current distribution ke liye field deta hai — chhote current elements ko integrate karna padta hai, matlab general case me kaam karta hai. Ampere's law sirf tab seedha kaam karta hai jab current distribution symmetric ho (jaise straight wire, solenoid) — us case me integration ki zaroorat nahi padti aur field directly nikal jaata hai.
Magnetic force velocity ke perpendicular kyun lagta hai?
F = qv × B me cross product ki property hi yahi hai ki result dono vectors (v aur B) ke perpendicular hota hai. Isi wajah se magnetic force kabhi speed nahi badalta — sirf direction badalta hai, isliye ye particle par koi work nahi karta.
Solenoid ke andar field uniform kyun hota hai lekin bahar zero jaisa?
Solenoid ki closely-wound coils ka field pattern is tarah overlap karta hai ki andar sab contributions add hoke ek strong, uniform axial field bana dete hain, jabki bahar ke contributions almost cancel ho jaate hain (ideal, infinitely long solenoid ke liye).
Galvanometer ko ammeter banane ke liye shunt low resistance ka kyun hota hai?
Shunt ka kaam hai bada current apni taraf divert karna taaki galvanometer coil se sirf chhota (Ig) current guzre. Kam resistance wala path zyada current le jaata hai, isliye shunt hamesha galvanometer se bahut kam resistance ka hota hai.
Kya magnetic field kaam (work) kar sakta hai kisi charge par?
Nahi. Magnetic force hamesha velocity ke perpendicular lagta hai, isliye displacement ke saath uska dot product zero hota hai — matlab magnetic force kabhi kaam nahi karta, sirf direction change karta hai.
Torque zero kab hota hai current loop par?
Jab loop ka normal (perpendicular to plane) B ke exactly parallel ho jaata hai — matlab θ = 0°, tab sinθ = 0 aur τ = 0. Ye stable equilibrium position hoti hai.
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