Moving Charges and Magnetism
चालित आवेश एवं चुंबकत्व | Class 12 Physics Chapter 4 | NCERT RBSE 2026 GOLD STANDARD
| Subject: | Physics | Chapter: | 4 |
| Class: | 12 (RBSE + NCERT) | Year: | 2025-26 |
| Topic: | Moving Charges and Magnetism (Complete with Examples) | ||
• Permeability of free space: μ₀ = 4π × 10⁻⁷ T·m/A
• Elementary charge: e = 1.6 × 10⁻¹⁹ C
• Electron mass: me = 9.1 × 10⁻³¹ kg
• Proton mass: mp = 1.67 × 10⁻²⁷ kg
• Conversion: 1 Tesla (T) = 10⁴ Gauss (G)
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1. Introduction
The connection between electricity and magnetism represents one of the most profound discoveries in physics. For centuries, electricity and magnetism were studied as separate phenomena. The breakthrough came in 1820 when Danish physicist Hans Christian Ørsted observed that a compass needle deflects when placed near a current-carrying wire. This simple experiment revealed that electric currents produce magnetic fields, forever linking these two fundamental forces.
In the previous chapters on electrostatics and current electricity, we studied charges at rest and charges in steady motion without considering their magnetic effects. This chapter introduces the magnetic effects produced by moving charges and the forces experienced by charges moving in magnetic fields. These principles form the foundation for understanding motors, generators, transformers, and countless modern technologies.
Figure 4.1: Historical Development of Electromagnetism
1820: Ørsted discovers current produces magnetic field↓
1820: Biot and Savart quantify magnetic field from current
↓
1826: Ampère develops mathematical theory
↓
1831: Faraday discovers electromagnetic induction
↓
1864: Maxwell unifies electricity and magnetism
2. Magnetic Field and Field Lines
A magnetic field is a region of space where a magnetic force can be detected. We represent the magnetic field by the vector B⃗, called the magnetic flux density or magnetic induction. The direction of the magnetic field at any point is defined as the direction in which the north pole of a small compass needle would point if placed at that location.
The SI unit of magnetic field is the tesla (T), named after Nikola Tesla. One tesla is defined as the magnetic field strength that exerts a force of one newton on a one-ampere current flowing through a one-meter length of conductor perpendicular to the field: 1 T = 1 N/(A·m) = 1 Wb/m².
Physical Interpretation: The tesla is a very large unit. Earth's magnetic field at the surface is approximately 5 × 10⁻⁵ T (0.5 G), while a small bar magnet produces a field of about 10⁻² T (100 G) near its poles.
- Magnetic field lines form continuous closed loops, unlike electric field lines
- The tangent to a field line at any point gives the direction of B⃗ at that point
- Field lines never intersect
- Field lines are more concentrated where the field is stronger
- Outside a magnet: N→S; inside: S→N, forming closed loops
Physical Interpretation: The closed-loop nature of magnetic field lines reflects that magnetic monopoles do not exist. Every magnetic field is produced by electric currents or intrinsic magnetic moments of particles.
3. Lorentz Force
When a charged particle moves through a region containing both electric and magnetic fields, it experiences the Lorentz force:
where q = charge (C), E⃗ = electric field (V/m), v⃗ = velocity (m/s), B⃗ = magnetic field (T)
The magnetic force component F⃗B = qv⃗ × B⃗ has magnitude F = qvB sin θ, where θ is the angle between v⃗ and B⃗. The force is maximum when perpendicular (θ = 90°) and zero when parallel (θ = 0°).
Physical Interpretation: The magnetic force is always perpendicular to velocity, so it cannot do work (W = ∫F⃗·ds⃗ = 0). It can only change the direction of motion, not the speed. This makes it act as a centripetal force.
4. Motion of Charged Particle in Magnetic Field
4.1 Circular Motion (v ⊥ B)
When a charged particle (charge q, mass m) enters a uniform magnetic field B⃗ with velocity perpendicular to the field, the magnetic force provides centripetal acceleration, causing circular motion.
Step 1: Magnetic force F = qvB provides centripetal force Fc = mv²/r:
Step 2: Solve for radius:
Step 3: Time period T = 2πr/v:
Step 4: Frequency f = 1/T:
Note: Frequency is independent of velocity!
Physical Interpretation: All particles with the same q/m ratio gyrate at the same frequency regardless of energy. This is the fundamental principle of the cyclotron accelerator.
4.2 Helical Motion (v at angle θ)
When velocity makes angle θ with field: v∥ = v cos θ (unchanged), v⊥ = v sin θ (causes circular motion). Result: helical path with radius r = mv⊥/(qB) and pitch p = v∥T = 2πmv cos θ/(qB).
5. The Cyclotron
The cyclotron accelerates charged particles to high energies using circular motion in a magnetic field combined with alternating electric field acceleration.
Principle: Particles move in semicircular paths inside D-shaped electrodes. Each time they cross the gap, an AC voltage (frequency = cyclotron frequency) accelerates them. Radius increases with energy, creating a spiral path until extraction.
Limitations: Relativistic effects at high speeds change particle mass, breaking resonance. Electrons reach relativistic speeds too quickly, making cyclotrons unsuitable for them.
6. Force on Current-Carrying Conductor
A current-carrying conductor in a magnetic field experiences a force because each moving charge carrier experiences a magnetic force.
Step 1: Each charge carrier q moving with drift velocity vd experiences force F₁ = qvdB.
Step 2: Total charges in length L, area A: N = nAL (n = number density).
Step 3: Total force: F = NF₁ = (nAL)qvdB.
Step 4: Since I = nAqvd:
7. Biot-Savart Law
The Biot-Savart law gives the magnetic field produced by a current element:
7.1 Application: Infinite Straight Wire
Step 1: For element at distance z, field point at perpendicular distance a: r = √(a² + z²), sin θ = a/r.
Step 2: dB = (μ₀Ia dz)/[4π(a² + z²)^(3/2)].
Step 3: Integrate z = −∞ to +∞ using substitution z = a tan φ:
Direction: Circular field lines (right-hand rule)
7.2 Application: Circular Loop (Axis)
At center (x = 0): B₀ = μ₀I/(2R)
Far from loop (x >> R): B ≈ μ₀IR²/(2x³) = μ₀m/(2πx³)
8. Ampère's Circuital Law
8.1 Application: Solenoid
Step 1: Choose rectangular loop: one side inside (length L), one outside, two perpendicular.
Step 2: ∮ B⃗ · dl⃗ = BL (only inside segment contributes).
Step 3: Enclosed current = nLI (n = turns per length).
Step 4: BL = μ₀nLI → B = μ₀nI.
8.2 Application: Toroid
Outside toroid: B = 0
9. Force Between Parallel Wires
Step 1: Wire 1 (current I₁) creates field B₁ = μ₀I₁/(2πd) at wire 2.
Step 2: Wire 2 (current I₂, length L) experiences force F = I₂LB₁.
Step 3: F/L = μ₀I₁I₂/(2πd).
Same direction currents: Attract | Opposite: Repel
One ampere is that current which, when flowing in two parallel wires 1 m apart, produces a force of 2 × 10⁻⁷ N/m.
10. Torque on Current Loop
where m = NIA (magnetic dipole moment)
11. The Moving Coil Galvanometer
A galvanometer detects and measures small currents using the torque on a current-carrying coil in a magnetic field.
11.1 Construction
Components: (1) Rectangular coil (N turns, area A) on non-magnetic frame, (2) Permanent horseshoe magnet with cylindrical pole pieces, (3) Soft iron cylindrical core, (4) Phosphor bronze spring (restoring torque + current lead), (5) Pointer and graduated scale.
11.2 Theory
Step 1: Magnetic torque on coil: τm = NIAB (radial field → θ = 90° always).
Step 2: Spring restoring torque: τs = kθ (k = torsion constant).
Step 3: At equilibrium: NIAB = kθ.
Step 4: Deflection: θ = (NAB/k)I.
11.3 Conversion to Ammeter
Connect low resistance S (shunt) in parallel with galvanometer to measure larger currents.
where Ig = full-scale galvanometer current, I = desired ammeter range
11.4 Conversion to Voltmeter
Connect high resistance R in series with galvanometer to measure larger voltages.
where V = desired voltmeter range
12. Solved Examples (15 Complete Examples)
Problem: An electron (q = −1.6 × 10⁻¹⁹ C) moves at 2 × 10⁶ m/s in +x direction in a 0.5 T field in +z direction. Find force.
Solution:
v⃗ × B⃗ = (2 × 10⁶ x̂) × (0.5 ẑ) = −10⁶ ŷ m·T/s
F⃗ = q(v⃗ × B⃗) = (−1.6 × 10⁻¹⁹)(−10⁶ ŷ) = 1.6 × 10⁻¹³ ŷ N
Answer: 1.6 × 10⁻¹³ N in +y direction
Problem: Proton (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C) at 5 × 10⁶ m/s in 2.0 T field. Find max and min force.
Solution:
Fmax = qvB (when v ⊥ B) = (1.6 × 10⁻¹⁹)(5 × 10⁶)(2.0) = 1.6 × 10⁻¹² N
Fmin = 0 (when v ∥ B)
Answer: Max: 1.6 × 10⁻¹² N, Min: 0
Problem: Electron at 2 × 10⁷ m/s perpendicular to 0.5 T field. Find (a) radius, (b) period, (c) frequency.
Solution:
(a) r = mv/(qB) = (9.1 × 10⁻³¹)(2 × 10⁷)/[(1.6 × 10⁻¹⁹)(0.5)] = 2.28 × 10⁻⁴ m = 0.228 mm
(b) T = 2πm/(qB) = 2π(9.1 × 10⁻³¹)/[(1.6 × 10⁻¹⁹)(0.5)] = 7.14 × 10⁻¹¹ s ≈ 71.4 ps
(c) f = 1/T = 1.4 × 10¹⁰ Hz = 14 GHz
Answer: (a) 0.228 mm, (b) 71.4 ps, (c) 14 GHz
Problem: Proton enters 0.3 T field at 60° with speed 4 × 10⁶ m/s. Find (a) radius, (b) pitch.
Solution:
v⊥ = v sin 60° = (4 × 10⁶)(0.866) = 3.46 × 10⁶ m/s
v∥ = v cos 60° = (4 × 10⁶)(0.5) = 2 × 10⁶ m/s
(a) r = mv⊥/(qB) = (1.67 × 10⁻²⁷)(3.46 × 10⁶)/[(1.6 × 10⁻¹⁹)(0.3)] = 0.12 m = 12 cm
T = 2πm/(qB) = 2.18 × 10⁻⁷ s
(b) p = v∥T = (2 × 10⁶)(2.18 × 10⁻⁷) = 0.436 m = 43.6 cm
Answer: (a) 12 cm, (b) 43.6 cm
Problem: Cyclotron (B = 1.5 T, rmax = 0.5 m) accelerates protons. Find maximum kinetic energy.
Solution:
Kmax = q²B²r²max/(2m)
= (1.6 × 10⁻¹⁹)²(1.5)²(0.5)²/[2(1.67 × 10⁻²⁷)]
= 1.44 × 10⁻¹² J = 9 MeV
Answer: 9 MeV
Problem: 50 cm wire carrying 10 A perpendicular to 0.8 T field. Find force.
Solution:
F = ILB sin 90° = (10)(0.5)(0.8)(1) = 4.0 N
Answer: 4.0 N
Problem: 2 m conductor with 5 A at 30° to 0.6 T field. Find force.
Solution:
F = ILB sin 30° = (5)(2)(0.6)(0.5) = 3.0 N
Answer: 3.0 N
Problem: Semicircular wire (radius R = 10 cm, current I = 5 A) in uniform perpendicular field B = 0.2 T. Find force on curved portion.
Solution:
For curved conductor, effective length = straight line distance = 2R
F = I(2R)B = (5)(2 × 0.1)(0.2) = 0.2 N
Answer: 0.2 N along diameter
Problem: Long wire carries 5 A. Find field at 10 cm distance.
Solution:
B = μ₀I/(2πr) = (4π × 10⁻⁷)(5)/(2π × 0.1) = 10⁻⁵ T = 10 μT
Answer: 10 μT
Problem: Circular coil (100 turns, radius 5 cm, current 0.5 A). Find field at center.
Solution:
B = μ₀NI/(2R) = (4π × 10⁻⁷)(100)(0.5)/(2 × 0.05) = 6.28 × 10⁻⁴ T = 0.628 mT
Answer: 0.628 mT
Problem: Circular loop (radius 10 cm, current 2 A). Find field at 10 cm on axis.
Solution:
x = R = 0.1 m
B = μ₀IR²/[2(R² + x²)^(3/2)] = (4π × 10⁻⁷)(2)(0.01)/[2(0.02)^(3/2)]
= 4.44 × 10⁻⁶ T ≈ 4.44 μT
Answer: 4.44 μT
Problem: 50 cm solenoid, 1000 turns, 5 A current. Find internal field.
Solution:
n = N/L = 1000/0.5 = 2000 turns/m
B = μ₀nI = (4π × 10⁻⁷)(2000)(5) = 0.0126 T = 12.6 mT
Answer: 12.6 mT
Problem: Toroid (3000 turns, mean radius 20 cm, current 2 A). Find field.
Solution:
B = μ₀NI/(2πr) = (4π × 10⁻⁷)(3000)(2)/(2π × 0.2) = 6 × 10⁻³ T = 6 mT
Answer: 6 mT
Problem: Two parallel wires 10 cm apart carry 5 A and 8 A in same direction. Find force per length.
Solution:
F/L = μ₀I₁I₂/(2πd) = (4π × 10⁻⁷)(5)(8)/(2π × 0.1) = 8 × 10⁻⁵ N/m = 80 μN/m
Answer: 80 μN/m (attractive)
Problem: Galvanometer (Rg = 50 Ω, Ig = 5 mA) to ammeter (0-5 A). Find shunt.
Solution:
S = IgRg/(I − Ig) = (0.005)(50)/(5 − 0.005) = 0.25/4.995 ≈ 0.050 Ω
Answer: 0.050 Ω (50 mΩ)
13. Multiple Choice Questions (15 Questions)
Q1. SI unit of magnetic field:
(a) Weber (b) Tesla (c) Gauss (d) Henry
1 T = 1 Wb/m² = 1 N/(A·m). Weber is flux unit, Gauss is CGS unit, Henry is inductance unit.
Q2. Magnetic field lines:
(a) Start at N, end at S (b) Form closed loops (c) Can intersect (d) Are equipotential
No magnetic monopoles exist. (a) incomplete—lines go S→N inside. (c) wrong—never intersect. (d) wrong—electric property.
Q3. Charged particle parallel to B experiences:
(a) Max force (b) Perpendicular force (c) No force (d) Centripetal force
F = qvB sin θ. When θ = 0°, sin θ = 0, so F = 0. (a),(d) wrong—max when perpendicular.
Q4. Cyclotron frequency depends on:
(a) Speed (b) Radius (c) q/m ratio (d) Kinetic energy
f = qB/(2πm) depends only on q, m, B. Independent of v, r, K.
Q5. Magnetic force does no work because:
(a) Force is zero (b) F ⊥ v (c) B is conservative (d) Speed constant
W = F⃗·ds⃗ = 0 when F⊥v. (a) wrong—F ≠ 0. (c) wrong—B not conservative. (d) is consequence.
Q6. Field at distance r from long wire:
(a) ∝ 1/r² (b) ∝ 1/r (c) ∝ 1/r³ (d) constant
B = μ₀I/(2πr) ∝ 1/r. Not 1/r² because infinite linear source.
Q7. Biot-Savart law gives:
(a) E from charge (b) B from current element (c) Force between currents (d) Torque on loop
dB⃗ = (μ₀/4π)(Idl⃗×r̂)/r². (a) is Coulomb's law. (c) needs two wires. (d) is τ = m×B.
Q8. At center of circular loop, B =
(a) μ₀I/R (b) μ₀I/(2R) (c) μ₀IR² (d) μ₀I/(2πR)
Standard formula for field at center. (a) missing factor 2. (c) has wrong dimensions. (d) is for straight wire.
Q9. Ampère's law useful when:
(a) High symmetry (b) No symmetry (c) B = 0 (d) I = 0
Like Gauss's law for E, requires symmetry to extract B from integral.
Q10. Field inside ideal solenoid:
(a) Zero (b) Varies with position (c) Uniform (d) Depends on radius
B = μ₀nI everywhere inside. (a) wrong—that's outside. (b),(d) wrong—independent of position/radius.
Q11. Parallel currents in same direction:
(a) Attract (b) Repel (c) No force (d) Torque only
F/L = μ₀I₁I₂/(2πd) attractive. Opposite currents repel.
Q12. One ampere produces between parallel wires 1 m apart:
(a) 2 × 10⁻⁷ N/m (b) 4π × 10⁻⁷ N/m (c) 10⁻⁶ N/m (d) 10⁻⁷ N/m
Definition of ampere: F/L = μ₀(1)(1)/(2π×1) = 2 × 10⁻⁷ N/m.
Q13. Magnetic dipole moment m =
(a) NI (b) IA (c) NIA (d) NIL
m = (number of turns)(current)(area). (a) missing area. (b) missing turns. (d) has length not area.
Q14. Galvanometer measures:
(a) Voltage (b) Small current (c) Resistance (d) Magnetic field
Detects μA-mA currents via torque on coil. Can be converted to ammeter/voltmeter.
Q15. To increase galvanometer current sensitivity:
(a) Decrease N (b) Increase k (c) Decrease B (d) Increase N
SI = NAB/k. Increase N or B, decrease k. (a),(c) decrease sensitivity. (b) decreases sensitivity.
14. Practice Questions
Short Answer Questions (2-3 marks)
Q1. State the SI unit of magnetic field and define it.
Answer: Tesla (T). 1 T is the field that exerts 1 N force on 1 A current in 1 m conductor perpendicular to field. 1 T = 1 N/(A·m) = 1 Wb/m².
Q2. Why don't magnetic field lines intersect?
Answer: If they intersect, there would be two directions of B⃗ at that point, which is impossible. Magnetic field has unique direction at each point.
Q3. State Biot-Savart law.
Answer: dB⃗ = (μ₀/4π)(Idl⃗ × r̂)/r². The magnetic field produced by a current element Idl⃗ at distance r is proportional to current, element length, sin θ, and inversely proportional to r².
Q4. Define magnetic dipole moment.
Answer: m⃗ = NIA⃗, where N = number of turns, I = current, A = area vector (perpendicular to loop, direction by right-hand rule). Unit: A·m² or J/T.
Q5. Why do parallel currents attract?
Answer: Wire 1 creates circular magnetic field. At wire 2's location, this field is perpendicular to wire 2. Force F⃗ = IL⃗ × B⃗ on wire 2 points toward wire 1 (right-hand rule), hence attraction. By Newton's third law, wire 1 also attracts wire 2.
Q6. State Ampère's circuital law.
Answer: ∮ B⃗·dl⃗ = μ₀Ienclosed. The line integral of magnetic field around any closed path equals μ₀ times the total current passing through the surface bounded by that path.
Long Answer Questions (5 marks)
Q1. Derive expression for radius and time period of charged particle moving perpendicular to uniform magnetic field.
Answer: [Complete derivation as shown in Section 4.1: equate qvB = mv²/r to get r = mv/(qB), then T = 2πr/v = 2πm/(qB)]
Q2. Derive magnetic field on axis of circular current loop using Biot-Savart law.
Answer: [Complete derivation as shown in Section 7.2: dB = (μ₀I/4π)(dl/r²), axial component dBx = dB(R/r), integrate to get B = μ₀IR²/[2(R²+x²)^(3/2)]]
Q3. Derive magnetic field inside solenoid using Ampère's law.
Answer: [Complete derivation as shown in Section 8.1: rectangular Amperian loop, ∮B⃗·dl⃗ = BL, enclosed current = nLI, result B = μ₀nI]
Q4. Describe construction and working of moving coil galvanometer. Derive expression for deflection.
Answer: [Construction from Section 11.1, working and derivation from 11.2: balance NIAB = kθ to get θ = (NAB/k)I]
Q5. Explain principle of cyclotron. Why can't it accelerate electrons to very high energies?
Answer: [Principle from Section 5: resonance condition fAC = qB/(2πm). Limitation: electrons reach relativistic speeds quickly, mass increases (m → γm), changing cyclotron frequency and breaking resonance]
15. Complete Formula Sheet
| Concept | Formula | Units |
|---|---|---|
| Lorentz Force | F⃗ = q(E⃗ + v⃗ × B⃗) | N |
| Magnetic Force | F = qvB sin θ | N |
| Cyclotron Radius | r = mv/(qB) | m |
| Cyclotron Frequency | f = qB/(2πm) | Hz |
| Cyclotron Period | T = 2πm/(qB) | s |
| Pitch of Helix | p = 2πmv cos θ/(qB) | m |
| Cyclotron Max Energy | Kmax = q²B²r²max/(2m) | J |
| Force on Conductor | F⃗ = IL⃗ × B⃗, F = ILB sin θ | N |
| Biot-Savart Law | dB⃗ = (μ₀/4π)(Idl⃗×r̂)/r² | T |
| Straight Wire | B = μ₀I/(2πr) | T |
| Circular Loop (center) | B = μ₀I/(2R) | T |
| Circular Loop (axis) | B = μ₀IR²/[2(R²+x²)^(3/2)] | T |
| N-turn Coil (center) | B = μ₀NI/(2R) | T |
| Ampère's Law | ∮ B⃗·dl⃗ = μ₀Ienc | — |
| Solenoid | B = μ₀nI = μ₀NI/L | T |
| Toroid | B = μ₀NI/(2πr) | T |
| Force Between Wires | F/L = μ₀I₁I₂/(2πd) | N/m |
| Torque on Loop | τ = NIAB sin θ | N·m |
| Torque (vector) | τ⃗ = m⃗ × B⃗ | N·m |
| Magnetic Dipole Moment | m = NIA | A·m² |
| Galvanometer Deflection | θ = (NAB/k)I | rad |
| Current Sensitivity | SI = NAB/k | rad/A |
| Voltage Sensitivity | SV = NAB/(kRg) | rad/V |
| Ammeter Shunt | S = IgRg/(I−Ig) | Ω |
| Voltmeter Series R | R = V/Ig − Rg | Ω |
16. Key Concepts Summary
- Magnetic field lines always form closed loops (no monopoles)
- Magnetic force is always perpendicular to velocity (F⊥v)
- Magnetic force does no work (cannot change kinetic energy)
- Cyclotron frequency independent of particle velocity
- Field from straight wire ∝ 1/r (not 1/r²)
- Field inside solenoid is uniform: B = μ₀nI
- Parallel currents attract, antiparallel currents repel
- Torque on loop: τ = NIAB sin θ (max when θ = 90°)
- Galvanometer: deflection ∝ current (linear device)
- Ammeter needs low resistance (shunt in parallel)
- Voltmeter needs high resistance (series resistance)
Complete Study Material | Class 12 Physics Chapter 4
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Chapter Summary & Exam Preparation
| Topic | Formula | Key Point |
|---|---|---|
| Lorentz Force | F⃗ = q(E⃗ + v⃗ × B⃗) | Magnetic part perpendicular to v⃗ |
| Circular Radius | r = mv/(qB) | Proportional to momentum |
| Cyclotron Frequency | f = qB/(2πm) | Independent of v, r, K |
| Cyclotron Energy | Kmax = q²B²r²/(2m) | Depends on dee radius |
| Force on Wire | F = ILB sin θ | Max when perpendicular |
| Biot-Savart | dB = (μ₀/4π)(Idl sin θ)/r² | For any shape |
| Straight Wire | B = μ₀I/(2πr) | Circular field lines |
| Circular Loop (center) | B = μ₀I/(2R) | Maximum at center |
| Circular Loop (axis) | B = μ₀IR²/[2(R²+x²)^(3/2)] | Decreases with distance |
| Solenoid | B = μ₀nI = μ₀NI/L | Uniform inside, zero outside |
| Toroid | B = μ₀NI/(2πr) | Completely confined |
| Parallel Wires | F/L = μ₀I₁I₂/(2πd) | Same I → attract |
| Torque on Loop | τ = NIAB sin θ | Zero when aligned |
| Magnetic Moment | m = NIA | Vector perpendicular to loop |
| Galvanometer | α = (NAB/k)I | Linear deflection |
| Current Sensitivity | SI = NAB/k | Deflection per ampere |
| Ammeter Shunt | S = IgRg/(I-Ig) | Low resistance parallel |
| Voltmeter Series R | R = V/Ig - Rg | High resistance series |
Very Short Answer (1 mark each):
- State SI unit of magnetic field
- Why do magnetic field lines form closed loops?
- Define magnetic dipole moment
- State Ampère's circuital law
Short Answer (2-3 marks each):
- Derive expression for cyclotron frequency
- Why do parallel currents attract?
- Explain principle of moving coil galvanometer
- How is galvanometer converted to ammeter?
Long Answer (5 marks each):
- Derive Lorentz force; explain circular motion of charged particle
- Cyclotron: Principle, construction, working, limitations
- Derive magnetic field from straight wire using Biot-Savart law
- Derive field inside solenoid using Ampère's law
- Moving coil galvanometer: Construction, theory, ammeter/voltmeter conversion
Numerical Problems (3-5 marks):
- Calculate radius, period, frequency for circular motion
- Force on current-carrying conductor
- Magnetic field from wire, loop, solenoid, toroid
- Galvanometer to ammeter/voltmeter conversion calculations
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