🧲 Moving Charges and Magnetism 🧲
📖 Welcome to Moving Charges and Magnetism Class 12 Notes
Discover the fascinating world of Moving Charges and Magnetism with our comprehensive Class 12 Physics Chapter 4 notes. This chapter explores how moving electric charges create magnetic fields and experience forces in magnetic fields. Master the Lorentz force derivation, understand the cyclotron working principle Class 12, and learn about Biot-Savart law applications.
Our complete guide covers all NCERT syllabus topics including magnetic force on current-carrying conductor, Ampere's circuital law Class 12, solenoid magnetic field formula, and galvanometer to ammeter conversion. Perfect for board exams and competitive preparation (JEE Main, JEE Advanced, NEET 2025).
📊 Chapter 4 Quick Summary
NCERT Reference: Class 12 Physics Part 1, Chapter 4 - Moving Charges and Magnetism
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📑 Complete Chapter Contents (Click to Jump)
1. Introduction to Moving Charges and Magnetism
The phenomenon of magnetism has fascinated humanity for over two thousand years. Ancient civilizations discovered that certain naturally occurring rocks, called lodestones (magnetite, Fe₃O₄), possessed the mysterious ability to attract iron objects. This discovery laid the foundation for one of the most fundamental forces in nature.
1.1 Historical Development
The scientific understanding of magnetism evolved through several key discoveries:
- 600 BCE: Thales of Miletus documented lodestone properties
- 1269: Petrus Peregrinus mapped magnetic field lines using iron filings
- 1600: William Gilbert published "De Magnete" - first systematic study of magnetism
- 1820: Hans Christian Ørsted discovered that electric current produces magnetic field
- 1820: André-Marie Ampère developed mathematical theory of electrodynamics
- 1831: Michael Faraday discovered electromagnetic induction
- 1865: James Clerk Maxwell unified electricity and magnetism through his equations
🔬 Physical Significance
Ørsted's discovery in 1820 was revolutionary because it established that electricity and magnetism are not separate phenomena, but are intimately connected. This discovery showed that:
- Moving electric charges (currents) produce magnetic fields
- Magnetism is fundamentally a relativistic effect of moving charges
- The universe contains only one fundamental force: the electromagnetic force
1.2 Relationship Between Electricity and Magnetism
Modern physics reveals that electricity and magnetism are two aspects of a single fundamental interaction - the electromagnetic force. The key insights are:
| Phenomenon | Fundamental Cause | Field Type |
|---|---|---|
| Static charges | Stationary charges | Electric field only |
| Steady current | Moving charges (constant velocity) | Electric + Magnetic fields |
| Accelerating charges | Charges with acceleration | Electric + Magnetic + Electromagnetic radiation |
📌 Key Concept
There is no such thing as an isolated "magnetic charge" (magnetic monopole). All magnetism arises from moving electric charges or from the intrinsic magnetic moments of elementary particles (which is a quantum mechanical property).
1.3 Sources of Magnetic Fields
Magnetic fields in nature arise from two fundamental sources:
1.3.1 Macroscopic Sources (Classical)
- Moving charges: Electric currents in wires, particle beams
- Current loops: Circular currents, solenoids, electromagnets
- Permanent magnets: Aligned atomic magnetic moments in ferromagnetic materials
1.3.2 Microscopic Sources (Quantum)
- Orbital motion of electrons: Electrons orbiting atomic nuclei
- Electron spin: Intrinsic angular momentum of electrons
- Nuclear spin: Intrinsic magnetic moments of protons and neutrons
1.4 Magnetic Field - The Fundamental Concept
A magnetic field is a region of space where a magnetic force can be detected. We denote the magnetic field by the vector 𝐁 (magnetic field vector or magnetic flux density).
SI Unit: Tesla (T) or Weber/meter² (Wb/m²)
Alternative unit: Gauss (G), where 1 T = 10⁴ G
Dimensions: [MT⁻²A⁻¹]
🎯 Physical Interpretation of Magnetic Field
The magnetic field 𝐁 at a point represents:
- Direction: The direction a north magnetic pole would move if placed at that point
- Magnitude: The strength of the magnetic influence at that point
- Field lines: Lines tangent to 𝐁 at every point, showing field direction
Unlike electric field lines (which begin and end on charges), magnetic field lines always form closed loops - they never begin or end. This reflects the absence of magnetic monopoles.
1.5 Applications of Magnetism
Understanding moving charges and magnetism is crucial for numerous modern technologies:
| Application | Principle Used | Examples |
|---|---|---|
| Electric Motors | Force on current-carrying conductor | Industrial machinery, electric vehicles, fans |
| Generators | Electromagnetic induction | Power plants, alternators |
| Transformers | Mutual induction | Power transmission, voltage conversion |
| Particle Accelerators | Lorentz force on moving charges | LHC, medical cyclotrons |
| MRI Scanners | Nuclear magnetic resonance | Medical imaging |
| Magnetic Storage | Magnetic domains | Hard disks, magnetic tapes |
| Magnetic Levitation | Repulsive magnetic force | Maglev trains |
1.6 Scope of This Chapter
In this chapter, we shall systematically develop the theory of magnetism, covering:
- The fundamental force laws (Lorentz force)
- Motion of charged particles in magnetic fields
- Laws governing magnetic fields produced by currents (Biot-Savart law, Ampère's law)
- Magnetic properties of current loops (magnetic dipoles)
- Practical applications (galvanometers, measuring instruments)
2. Magnetic Force: Sources and Fields
2.1 Experimental Observations
The existence of magnetic force can be demonstrated through simple experiments:
Experiment 1: Permanent Magnets
When two bar magnets are brought near each other:
- Like poles repel: North-North or South-South poles push each other away
- Unlike poles attract: North-South poles pull each other together
- The force acts at a distance (no physical contact needed)
- The force increases as magnets approach each other
Experiment 2: Magnet and Iron Filings
When iron filings are sprinkled around a bar magnet, they align themselves along curved lines connecting the poles. These lines represent the magnetic field lines.
📌 Properties of Magnetic Field Lines
- Field lines emerge from North pole and enter South pole (outside the magnet)
- Field lines are continuous closed loops (inside magnet: S→N, outside: N→S)
- Field lines never intersect each other
- Tangent to field line at any point gives direction of 𝐁 at that point
- Density of field lines indicates field strength
- Field lines always form closed loops (no magnetic monopoles)
2.2 Comparison: Electric vs Magnetic Fields
| Property | Electric Field (𝐄) | Magnetic Field (𝐁) |
|---|---|---|
| Source | Electric charges (stationary or moving) | Moving electric charges (currents) |
| Monopoles | Exist (positive and negative charges) | Do not exist (no isolated N or S poles) |
| Field Lines | Begin on + charges, end on − charges | Form closed loops (no beginning or end) |
| Force Direction | Along field direction (for + charge) | Perpendicular to both 𝐁 and velocity |
| Work Done | Can do work on charged particle | Cannot do work (F ⊥ v always) |
| Mathematical Law | Gauss's law: ∮𝐄·d𝐀 = Q/ε₀ | Gauss's law: ∮𝐁·d𝐀 = 0 |
🔬 Why Magnetic Monopoles Don't Exist
If you cut a bar magnet into two pieces, you do not get isolated North and South poles. Instead, you get two smaller magnets, each with its own North and South pole. This can be repeated indefinitely - you can never isolate a single magnetic pole.
This fundamental difference arises because magnetism is caused by circulating currents (or equivalent quantum effects). You cannot have a circulating current with only one end!
The non-existence of monopoles is expressed mathematically as: $$\nabla \cdot \vec{\mathbf{B}} = 0$$ or $$\oint \vec{\mathbf{B}} \cdot d\vec{\mathbf{A}} = 0$$ (magnetic flux through any closed surface is zero)
2.3 Magnitude of Magnetic Field
Typical magnetic field strengths in various situations:
| Source | Magnetic Field Strength |
|---|---|
| Earth's magnetic field at surface | ~5 × 10⁻⁵ T = 0.5 G |
| Small bar magnet near poles | ~10⁻² T = 100 G |
| Electromagnet (laboratory) | ~1 T = 10⁴ G |
| Superconducting magnet | ~10 T = 10⁵ G |
| Strongest continuous lab field | ~45 T |
| Neutron star surface | ~10⁸ T |
| Magnetar (magnetic neutron star) | ~10¹¹ T |
3. Lorentz Force - The Fundamental Force Law
3.1 Definition and Mathematical Form
When a charged particle moves through a region containing both electric and magnetic fields, it experiences a force called the Lorentz force. This is the most fundamental force law in classical electromagnetism.
Where:
- 𝐅 = total electromagnetic force on the particle (N)
- q = charge of the particle (C)
- 𝐄 = electric field vector (V/m or N/C)
- 𝐯 = velocity vector of the particle (m/s)
- 𝐁 = magnetic field vector (T)
🎯 Physical Interpretation of Each Term
Electric Force Term: q𝐄
- Acts on charged particle whether at rest or moving
- Direction: along 𝐄 (for positive charge), opposite for negative
- Can do work and change kinetic energy
Magnetic Force Term: q(𝐯 × 𝐁)
- Acts only when particle is moving (requires 𝐯 ≠ 0)
- Direction: perpendicular to both 𝐯 and 𝐁 (right-hand rule)
- Cannot do work (F ⊥ v, so F·v = 0)
- Changes direction of motion but not speed
3.2 Magnetic Force on a Moving Charge
In the absence of an electric field (𝐄 = 0), the Lorentz force reduces to the pure magnetic force:
Magnitude: $$F_B = qvB\sin\theta$$
Where θ is the angle between 𝐯 and 𝐁
3.2.1 Direction of Magnetic Force - Right-Hand Rule
📌 Right-Hand Rule for Magnetic Force
For a positive charge (+q):
- Point fingers of right hand along velocity direction (𝐯)
- Curl fingers toward magnetic field direction (𝐁)
- Thumb points in direction of force (𝐅)
For a negative charge (−q):
Force direction is opposite to that given by right-hand rule
3.2.2 Special Cases of Magnetic Force
| Condition | Angle θ | sin θ | Force F_B | Physical Situation |
|---|---|---|---|---|
| 𝐯 ∥ 𝐁 (parallel) | 0° | 0 | 0 | No force, particle moves straight |
| 𝐯 ⊥ 𝐁 (perpendicular) | 90° | 1 | qvB (maximum) | Maximum force, circular motion |
| 𝐯 antiparallel to 𝐁 | 180° | 0 | 0 | No force, particle moves straight |
| General angle | θ | sin θ | qvB sin θ | Helical motion |
⚠️ Critical Observation: Magnetic Force Does No Work
Since 𝐅 is always perpendicular to 𝐯, the work done by magnetic force is:
$$W = \int \vec{\mathbf{F}} \cdot d\vec{\mathbf{s}} = \int \vec{\mathbf{F}} \cdot \vec{\mathbf{v}}dt = 0$$
Consequences:
- Kinetic energy remains constant: ½mv² = constant
- Speed |v| remains constant
- Only the direction of velocity changes
- Magnetic force acts as a centripetal force
3.3 Solved Examples - Lorentz Force
Problem: An electron (charge = −1.6 × 10⁻¹⁹ C) moves with velocity 2 × 10⁶ m/s in the +x direction through a uniform magnetic field of 0.5 T in the +z direction. Find the magnetic force on the electron.
Given:
- q = −1.6 × 10⁻¹⁹ C
- 𝐯 = (2 × 10⁶ m/s) x̂
- 𝐁 = (0.5 T) ẑ
Solution:
Using 𝐅 = q(𝐯 × 𝐁)
First, calculate 𝐯 × 𝐁:
$$\vec{\mathbf{v}} \times \vec{\mathbf{B}} = (2 \times 10^6 \hat{x}) \times (0.5 \hat{z})$$
$$= (2 \times 10^6)(0.5)(\hat{x} \times \hat{z})$$
$$= 10^6 (-\hat{y})$$ [since x̂ × ẑ = −ŷ]
$$= -10^6 \hat{y} \text{ (m/s)(T)}$$
Now, 𝐅 = q(𝐯 × 𝐁):
$$\vec{\mathbf{F}} = (-1.6 \times 10^{-19})(-10^6 \hat{y})$$
$$= 1.6 \times 10^{-13} \hat{y} \text{ N}$$
Answer: The magnetic force is 1.6 × 10⁻¹³ N in the +y direction
Physical Interpretation: The electron experiences a force perpendicular to both its velocity (x-direction) and the magnetic field (z-direction), causing it to curve in the y-direction.
Problem: A proton (mass = 1.67 × 10⁻²⁷ kg, charge = +1.6 × 10⁻¹⁹ C) moves with speed 5 × 10⁶ m/s in a magnetic field of 2.0 T. Find: (a) maximum possible magnetic force, (b) minimum possible magnetic force.
Given:
- q = +1.6 × 10⁻¹⁹ C
- v = 5 × 10⁶ m/s
- B = 2.0 T
(a) Maximum force occurs when θ = 90° (v ⊥ B):
$$F_{max} = qvB = (1.6 \times 10^{-19})(5 \times 10^6)(2.0)$$
$$= 1.6 \times 10^{-12} \text{ N}$$
(b) Minimum force occurs when θ = 0° or 180° (v ∥ B):
$$F_{min} = 0$$
Answer: (a) F_max = 1.6 × 10⁻¹² N, (b) F_min = 0
❌ Common Student Mistakes
- Forgetting the vector nature: Students often calculate F = qvB without considering direction. Always use the vector cross product or right-hand rule.
- Sign confusion: For negative charges, force direction is opposite to right-hand rule prediction.
- Assuming force does work: Magnetic force is always perpendicular to velocity, so it can never do work on the particle.
- Units confusion: Remember 1 T = 1 N/(A·m) = 1 Wb/m² = 1 kg/(A·s²)
4. Magnetic Force on a Current-Carrying Conductor
4.1 Conceptual Foundation
We have seen that a single moving charge experiences a magnetic force. When many charges move together in a conductor (forming an electric current), each charge experiences a force, and the net effect is a force on the entire conductor.
Consider a straight conductor of length L carrying current I in a uniform magnetic field 𝐁. The current consists of charge carriers (typically electrons in metals) moving with drift velocity v_d.
4.2 Derivation of Force on Current-Carrying Conductor
📐 Complete Derivation: Force on Current Element
Setup: Consider a conductor of length L, cross-sectional area A, carrying current I in magnetic field 𝐁.
Step 1: Force on a single charge carrier
Each charge carrier (charge q) moving with drift velocity v_d experiences:
$$\vec{\mathbf{F}}_1 = q\vec{\mathbf{v}}_d \times \vec{\mathbf{B}}$$
Step 2: Number of charge carriers
If n = number density of charge carriers (charges per unit volume), then:
- Volume of conductor = AL
- Total number of charges = N = nAL
Step 3: Total force on all charges
$$\vec{\mathbf{F}} = N \vec{\mathbf{F}}_1 = (nAL)q\vec{\mathbf{v}}_d \times \vec{\mathbf{B}}$$
Step 4: Express in terms of current
We know that current I = nAqv_d (from current electricity)
Therefore: nAqv_d = I
Substituting:
$$\vec{\mathbf{F}} = (nAqv_d)L \times \vec{\mathbf{B}} = IL \times \vec{\mathbf{B}}$$
Step 5: Vector formulation
Define length vector: $$\vec{\mathbf{L}}$$ (magnitude L, direction of current)
Magnitude: $$F = ILB\sin\theta$$
Where θ is angle between current direction and 𝐁
Maximum force: When θ = 90° (I ⊥ B), F_max = ILB
Zero force: When θ = 0° or 180° (I ∥ B), F = 0
4.3 Direction - Right-Hand Rule for Current
📌 Right-Hand Rule (Current Version)
Method 1: Direct application
- Point fingers along current direction (I or L)
- Curl fingers toward magnetic field (B)
- Thumb points in direction of force (F)
Method 2: Mnemonic - FBI
- First finger → Field (𝐁)
- Second finger → Current (I)
- Thumb → Motion/Force (𝐅)
(Fleming's Left-Hand Rule for motors)
4.4 Force on a Current Element (Differential Form)
For a small element of length dl carrying current I:
For a curved conductor: $$\vec{\mathbf{F}} = \int Id\vec{\mathbf{l}} \times \vec{\mathbf{B}}$$
If 𝐁 is uniform: $$\vec{\mathbf{F}} = I\left(\int d\vec{\mathbf{l}}\right) \times \vec{\mathbf{B}} = I\vec{\mathbf{L}}_{eff} \times \vec{\mathbf{B}}$$
where L_eff is the straight-line distance from start to end point
🔬 Physical Significance - Force on Curved Conductor
For a curved conductor in a uniform magnetic field, the total force depends only on the straight-line vector connecting its endpoints, not on the actual path taken.
Example: A semicircular wire and a straight wire connecting the same two points experience the same force in a uniform field.
Special case: A closed loop in uniform field experiences zero net force (∮dl = 0), but may experience a torque.
4.5 Applications - Force on Current-Carrying Conductor
| Application | Principle | Key Feature |
|---|---|---|
| Electric Motor | Force on coil in magnetic field | Converts electrical energy to mechanical |
| Loudspeaker | Force on voice coil | Converts electrical signals to sound |
| Electromagnetic Relay | Force moves contacts | Remote switching |
| Maglev Train | Repulsive force for levitation | Frictionless transport |
| Circuit Breaker | Excess current creates force | Safety device |
4.6 Solved Examples
Problem: A wire of length 50 cm carrying a current of 10 A is placed perpendicular to a uniform magnetic field of 0.8 T. Calculate the force on the wire.
Given:
- L = 50 cm = 0.5 m
- I = 10 A
- B = 0.8 T
- θ = 90° (perpendicular, so sin 90° = 1)
Formula: F = ILB sin θ
Solution:
$$F = (10)(0.5)(0.8)(1)$$
$$F = 4.0 \text{ N}$$
Answer: The force on the wire is 4.0 N
Problem: A conductor of length 2 m carries 5 A current. It makes an angle of 30° with a magnetic field of magnitude 0.6 T. Find the force experienced by the conductor.
Given:
- L = 2 m
- I = 5 A
- B = 0.6 T
- θ = 30°
Solution:
$$F = ILB\sin\theta$$
$$F = (5)(2)(0.6)\sin 30°$$
$$F = (5)(2)(0.6)(0.5)$$
$$F = 3.0 \text{ N}$$
Answer: Force = 3.0 N
Note: If the wire were parallel to the field (θ = 0°), the force would be zero.
Problem: A semicircular wire of radius R carries current I. It is placed in a uniform magnetic field B perpendicular to the plane of the wire. Find the force on the curved portion.
Analysis:
For a curved conductor in uniform field, the force depends only on the straight-line vector connecting endpoints.
Effective length vector: $$\vec{\mathbf{L}}_{eff} = 2R$$ (diameter)
Direction: Along the diameter
Force magnitude:
$$F = I(2R)B = 2IRB$$
Direction: Perpendicular to both the diameter and B (use right-hand rule)
Answer: F = 2IRB, perpendicular to plane of wire
Physical insight: The force is the same as if the curved wire were replaced by a straight wire along the diameter.
5. Motion of Charged Particles in Magnetic Field
5.1 Introduction
When a charged particle moves in a magnetic field, it experiences a Lorentz force that is always perpendicular to its velocity. This perpendicular force cannot change the particle's speed (kinetic energy), but it continuously changes the direction of motion. The resulting trajectory depends on the initial velocity direction relative to the magnetic field.
5.2 Motion Perpendicular to Magnetic Field (Circular Motion)
Consider a charged particle (charge q, mass m) entering a uniform magnetic field 𝐁 with velocity 𝐯 perpendicular to 𝐁.
📐 Derivation: Circular Motion in Magnetic Field
Step 1: Force analysis
Magnetic force: $$\vec{\mathbf{F}} = q\vec{\mathbf{v}} \times \vec{\mathbf{B}}$$
Since v ⊥ B: $$F = qvB$$
Direction: Perpendicular to both v and B
Step 2: Nature of motion
The force is:
- Perpendicular to velocity (F ⊥ v)
- Constant in magnitude
- Always directed toward a fixed point
This is exactly the condition for uniform circular motion!
Step 3: Apply centripetal force equation
For circular motion: $$F_{centripetal} = \frac{mv^2}{r}$$
Magnetic force provides this: $$qvB = \frac{mv^2}{r}$$
Step 4: Solve for radius
$$r = \frac{mv}{qB}$$
Where:
- m = mass of particle
- v = speed of particle
- p = mv = momentum
- q = charge
- B = magnetic field strength
5.2.1 Time Period and Frequency
Time to complete one revolution:
$$T = \frac{2\pi r}{v} = \frac{2\pi (mv/qB)}{v} = \frac{2\pi m}{qB}$$
Key Point: Frequency is independent of velocity and radius!
🔬 Physical Significance - Velocity Independence
The remarkable fact that cyclotron frequency is independent of particle velocity has profound implications:
- Cyclotron accelerators: Particles can be accelerated to high energies by applying AC voltage at fixed frequency
- Mass spectrometry: Different isotopes (different m) have different frequencies, allowing separation
- Plasma confinement: All particles of given q/m ratio gyrate at same frequency regardless of energy
5.3 Motion Parallel to Magnetic Field
If a charged particle moves parallel or antiparallel to the magnetic field:
- Force: F = qvB sin 0° = 0
- No magnetic force acts
- Particle continues in straight-line motion with constant velocity
5.4 Motion at Arbitrary Angle - Helical Motion
Consider a particle entering a uniform magnetic field at angle θ with the field direction.
📐 Analysis: Helical Motion
Step 1: Resolve velocity
$$v_\parallel = v\cos\theta$$ (component parallel to B)
$$v_\perp = v\sin\theta$$ (component perpendicular to B)
Step 2: Analyze each component
- Parallel component v_∥: Unaffected by B, continues unchanged
- Perpendicular component v_⊥: Causes circular motion in plane ⊥ to B
Step 3: Combined motion
The particle executes:
- Circular motion (from v_⊥) in plane perpendicular to B
- Uniform motion (from v_∥) along B direction
Result: Helical (spiral) path
5.5 The Cyclotron - Particle Accelerator
The cyclotron is a device that accelerates charged particles to high energies using the properties of circular motion in magnetic field.
5.5.1 Principle of Operation
📌 Cyclotron Working Principle
- Charged particles move in semicircular paths inside two hollow D-shaped electrodes (called "dees")
- A uniform magnetic field (perpendicular to plane of dees) keeps particles in circular paths
- An alternating voltage between dees accelerates particles when they cross the gap
- Since cyclotron frequency is constant, the AC frequency is fixed (resonance condition)
- Particles spiral outward with increasing radius as they gain energy
- When they reach maximum radius, they are extracted by deflector plate
5.5.2 Key Equations for Cyclotron
Maximum kinetic energy:
(where r_max is radius of dees)
5.5.3 Limitations of Cyclotron
- Relativistic effects: At very high speeds, mass increases (m → γm), changing cyclotron frequency
- Cannot accelerate electrons: Electrons reach relativistic speeds too quickly
- Cannot accelerate neutral particles: Requires charged particles
- Maximum energy limited: By size of magnet and relativistic effects
🔬 Modern Particle Accelerators
To overcome cyclotron limitations, modern accelerators use:
- Synchrocyclotron: Varies AC frequency to account for relativistic mass increase
- Synchrotron: Varies both magnetic field and frequency (used in LHC)
- Linear accelerator (LINAC): Accelerates in straight line, no magnetic bending
5.6 Applications - Motion in Magnetic Field
| Application | Principle | Use |
|---|---|---|
| Mass Spectrometer | r ∝ m/q separation | Determining isotope masses |
| Cyclotron | Resonant acceleration | Producing radioactive isotopes, proton therapy |
| Cathode Ray Tube | Electron beam deflection | Oscilloscopes, old TVs |
| Magnetic Bottle | Helical motion confinement | Plasma containment in fusion research |
| Van Allen Belts | Charged particles trapped by Earth's field | Natural radiation belts around Earth |
| Auroras | Particles spiraling along field lines | Northern/Southern lights |
5.7 Solved Examples
Problem: An electron (m = 9.1 × 10⁻³¹ kg, q = −1.6 × 10⁻¹⁹ C) moves perpendicular to a magnetic field of 0.5 T with speed 2 × 10⁷ m/s. Find: (a) radius of circular path, (b) time period, (c) frequency.
Given:
- m = 9.1 × 10⁻³¹ kg
- q = 1.6 × 10⁻¹⁹ C (magnitude)
- v = 2 × 10⁷ m/s
- B = 0.5 T
(a) Radius:
$$r = \frac{mv}{qB} = \frac{(9.1 \times 10^{-31})(2 \times 10^7)}{(1.6 \times 10^{-19})(0.5)}$$
$$r = \frac{18.2 \times 10^{-24}}{0.8 \times 10^{-19}} = 22.75 \times 10^{-5} \text{ m}$$
$$r \approx 0.23 \text{ mm}$$
(b) Time period:
$$T = \frac{2\pi m}{qB} = \frac{2\pi (9.1 \times 10^{-31})}{(1.6 \times 10^{-19})(0.5)}$$
$$T = \frac{5.71 \times 10^{-30}}{0.8 \times 10^{-19}} = 7.14 \times 10^{-11} \text{ s}$$
$$T \approx 71.4 \text{ ps}$$ (picoseconds)
(c) Frequency:
$$f = \frac{1}{T} = \frac{1}{7.14 \times 10^{-11}} = 1.4 \times 10^{10} \text{ Hz}$$
$$f = 14 \text{ GHz}$$ (in microwave range)
Answer: (a) r = 0.23 mm, (b) T = 71.4 ps, (c) f = 14 GHz
Problem: A proton (m = 1.67 × 10⁻²⁷ kg) enters a uniform magnetic field of 0.3 T at an angle of 60° with the field direction, with speed 4 × 10⁶ m/s. Find: (a) radius of helical path, (b) pitch of helix.
Given:
- m = 1.67 × 10⁻²⁷ kg
- q = 1.6 × 10⁻¹⁹ C
- B = 0.3 T
- v = 4 × 10⁶ m/s
- θ = 60°
(a) Radius:
$$v_\perp = v\sin\theta = (4 \times 10^6)\sin 60° = 4 \times 10^6 \times 0.866 = 3.46 \times 10^6 \text{ m/s}$$
$$r = \frac{mv_\perp}{qB} = \frac{(1.67 \times 10^{-27})(3.46 \times 10^6)}{(1.6 \times 10^{-19})(0.3)}$$
$$r = \frac{5.78 \times 10^{-21}}{4.8 \times 10^{-20}} = 0.12 \text{ m} = 12 \text{ cm}$$
(b) Pitch:
First find time period:
$$T = \frac{2\pi m}{qB} = \frac{2\pi (1.67 \times 10^{-27})}{(1.6 \times 10^{-19})(0.3)} = 2.18 \times 10^{-7} \text{ s}$$
Parallel component:
$$v_\parallel = v\cos\theta = (4 \times 10^6)\cos 60° = 4 \times 10^6 \times 0.5 = 2 \times 10^6 \text{ m/s}$$
Pitch:
$$p = v_\parallel T = (2 \times 10^6)(2.18 \times 10^{-7}) = 0.436 \text{ m} = 43.6 \text{ cm}$$
Answer: (a) r = 12 cm, (b) p = 43.6 cm
❌ Common Student Mistakes
- Using total velocity instead of perpendicular component: For helical motion, only v_⊥ determines radius, not total v
- Forgetting velocity independence of frequency: Cyclotron frequency depends only on q, B, and m - not on v or r
- Sign errors: For negative charges, force direction reverses (but equations for r and T remain same in magnitude)
- Unit confusion: Always convert to SI units (m, kg, s, C, T)
- Assuming magnetic force does work: It doesn't! Speed remains constant.
6. Biot-Savart Law - Magnetic Field Due to Current Element
6.1 Introduction
Just as Coulomb's law describes the electric field produced by a point charge, the Biot-Savart law describes the magnetic field produced by a current-carrying conductor. This law, discovered experimentally by Jean-Baptiste Biot and Félix Savart in 1820, is fundamental to magnetostatics.
🔬 Historical Context
Within months of Ørsted's discovery that currents produce magnetic fields, Biot and Savart performed careful experiments measuring the magnetic field around straight current-carrying wires. They found that the field strength varies with distance and angle in a specific mathematical way, leading to their famous law.
6.2 Statement of Biot-Savart Law
The Biot-Savart law gives the magnetic field d𝐁 produced by a small current element Id𝐥.
Where:
- d𝐁 = magnetic field produced by element Id𝐥
- I = current in the conductor (A)
- d𝐥 = vector length element (direction: current direction)
- 𝐫 = position vector from d𝐥 to field point
- r = |𝐫| = distance from element to field point
- r̂ = unit vector in direction of 𝐫
- μ₀ = permeability of free space = 4π × 10⁻⁷ T·m/A
6.3 Understanding the Biot-Savart Law
6.3.1 Magnitude of Magnetic Field
$$dB = \frac{\mu_0}{4\pi} \frac{I \, dl \, \sin\theta}{r^2}$$
Where θ is the angle between d𝐥 and 𝐫
📌 Key Observations
- Inverse square dependence: dB ∝ 1/r² (like electric field from point charge)
- Proportional to current: dB ∝ I (double current → double field)
- Angle dependence: Maximum when θ = 90° (d𝐥 ⊥ 𝐫), zero when θ = 0° or 180°
- Direction: Always perpendicular to both d𝐥 and 𝐫
6.3.2 Direction - Right-Hand Rule
📌 Right-Hand Rule for Biot-Savart Law
Method 1: Cross Product
- Point fingers along current direction (d𝐥)
- Curl fingers toward 𝐫 (vector to field point)
- Thumb points in direction of d𝐁
Method 2: Circular Field Lines
- Thumb points along current direction
- Fingers curl around conductor
- Fingers show direction of circular magnetic field lines
6.4 Comparison with Coulomb's Law
| Property | Coulomb's Law (Electric) | Biot-Savart Law (Magnetic) |
|---|---|---|
| Source | Point charge q | Current element Id𝐥 |
| Field Produced | Electric field 𝐄 | Magnetic field 𝐁 |
| Law | $$\vec{\mathbf{E}} = \frac{1}{4\pi\epsilon_0} \frac{q\hat{\mathbf{r}}}{r^2}$$ | $$d\vec{\mathbf{B}} = \frac{\mu_0}{4\pi} \frac{Id\vec{\mathbf{l}} \times \hat{\mathbf{r}}}{r^2}$$ |
| Direction | Along 𝐫 (radial) | Perpendicular to both d𝐥 and 𝐫 |
| Dependence | 1/r² | 1/r² |
| Constant | $$\frac{1}{4\pi\epsilon_0}$$ | $$\frac{\mu_0}{4\pi}$$ |
6.5 Applications of Biot-Savart Law
To find the total magnetic field from an extended current distribution:
Integration is performed over the entire current-carrying conductor
6.6 Example: Magnetic Field Due to Straight Current-Carrying Wire
📐 Derivation: Field from Infinite Straight Wire
Problem: Find the magnetic field at perpendicular distance a from an infinitely long straight wire carrying current I.
Setup:
- Place wire along z-axis
- Field point P at distance a from wire
- Consider element d𝐥 at distance z from nearest point
Step 1: Geometry
Distance from element to P: $$r = \sqrt{a^2 + z^2}$$
Angle between d𝐥 and 𝐫: Call it θ
$$\sin\theta = \frac{a}{r} = \frac{a}{\sqrt{a^2 + z^2}}$$
Step 2: Apply Biot-Savart Law
$$dB = \frac{\mu_0 I \, dz \, \sin\theta}{4\pi r^2} = \frac{\mu_0 I \, dz}{4\pi (a^2 + z^2)} \cdot \frac{a}{\sqrt{a^2 + z^2}}$$
$$dB = \frac{\mu_0 I a \, dz}{4\pi (a^2 + z^2)^{3/2}}$$
Step 3: Integrate from −∞ to +∞
$$B = \int_{-\infty}^{+\infty} \frac{\mu_0 I a \, dz}{4\pi (a^2 + z^2)^{3/2}}$$
Using substitution z = a tan φ:
$$B = \frac{\mu_0 I}{4\pi a} \int_{-\pi/2}^{+\pi/2} \cos\phi \, d\phi = \frac{\mu_0 I}{4\pi a} [\sin\phi]_{-\pi/2}^{+\pi/2}$$
$$B = \frac{\mu_0 I}{4\pi a} [1 - (-1)] = \frac{\mu_0 I}{2\pi a}$$
Direction: Circular field lines around wire (right-hand rule)
Note: B ∝ 1/a (not 1/a² like point charge!)
🎯 Physical Significance
The 1/r dependence (rather than 1/r²) arises because we're integrating over an infinite one-dimensional source. Each element contributes 1/r², but there are infinitely many elements, leading to net 1/r dependence.
This slower fall-off means magnetic fields from wires extend relatively far compared to fields from localized sources.
6.7 Solved Examples
Problem: A long straight wire carries a current of 5 A. Find the magnetic field at a distance of 10 cm from the wire.
Given:
- I = 5 A
- a = 10 cm = 0.1 m
- μ₀ = 4π × 10⁻⁷ T·m/A
Formula: $$B = \frac{\mu_0 I}{2\pi a}$$
Solution:
$$B = \frac{(4\pi \times 10^{-7})(5)}{2\pi (0.1)}$$
$$B = \frac{20\pi \times 10^{-7}}{0.2\pi} = \frac{20 \times 10^{-7}}{0.2}$$
$$B = 100 \times 10^{-7} = 10^{-5} \text{ T} = 10 \mu\text{T}$$
Answer: B = 10 μT
7. Magnetic Field on Axis of Circular Current Loop
7.1 Problem Setup
Consider a circular loop of radius R carrying current I. We wish to find the magnetic field at a point P on the axis of the loop, at distance x from the center.
7.2 Complete Derivation
📐 Derivation: Axial Field of Circular Loop
Step 1: Consider a small element
Take a small element d𝐥 on the loop. Due to symmetry, we only need to consider the component along the axis.
Step 2: Distance from element to point P
$$r = \sqrt{R^2 + x^2}$$
Step 3: Magnetic field from element (Biot-Savart)
$$dB = \frac{\mu_0 I \, dl}{4\pi r^2}$$
(Note: d𝐥 ⊥ 𝐫, so sin θ = 1)
Step 4: Components of d𝐁
The field d𝐁 makes angle α with the axis, where:
$$\cos\alpha = \frac{R}{r} = \frac{R}{\sqrt{R^2 + x^2}}$$
Axial component: $$dB_x = dB \cos\alpha = \frac{\mu_0 I \, dl}{4\pi r^2} \cdot \frac{R}{r}$$
$$dB_x = \frac{\mu_0 I R \, dl}{4\pi r^3}$$
Perpendicular components cancel by symmetry (for every element, there's a diametrically opposite element whose perpendicular component cancels).
Step 5: Integrate around loop
$$B = \int dB_x = \int \frac{\mu_0 I R \, dl}{4\pi r^3}$$
Since R, r constant for all elements:
$$B = \frac{\mu_0 I R}{4\pi r^3} \int dl = \frac{\mu_0 I R}{4\pi r^3} (2\pi R)$$
$$B = \frac{\mu_0 I R^2}{2r^3} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$$
Direction: Along the axis (right-hand rule: curl fingers with current, thumb shows field)
7.3 Special Cases
7.3.1 At the Center of Loop (x = 0)
This is the maximum field along the axis
7.3.2 Far from Loop (x >> R)
When x >> R, we can approximate:
$$(R^2 + x^2)^{3/2} \approx x^3$$
$$B \approx \frac{\mu_0 I R^2}{2x^3} = \frac{\mu_0 I A}{2\pi x^3}$$
(where A = πR² is the area of loop)
🔬 Magnetic Dipole Behavior
Far from the loop (x >> R), the field falls as 1/x³, which is characteristic of a magnetic dipole. This is analogous to the electric field of an electric dipole.
We can write: $$B = \frac{\mu_0 m}{2\pi x^3}$$
where m = IA is the magnetic dipole moment of the loop.
7.4 Field at Center of N-Turn Coil
If the circular loop has N turns (all with same radius R):
Each turn contributes equally, so total field = N × (field from one turn)
7.5 Solved Examples
Problem: A circular coil of radius 5 cm has 100 turns and carries a current of 0.5 A. Find the magnetic field at the center of the coil.
Given:
- R = 5 cm = 0.05 m
- N = 100 turns
- I = 0.5 A
Formula: $$B = \frac{\mu_0 N I}{2R}$$
Solution:
$$B = \frac{(4\pi \times 10^{-7})(100)(0.5)}{2(0.05)}$$
$$B = \frac{200\pi \times 10^{-7}}{0.1} = 2000\pi \times 10^{-7}$$
$$B = 6.28 \times 10^{-4} \text{ T} = 0.628 \text{ mT}$$
Answer: B = 0.628 mT = 628 μT
Problem: A circular loop of radius 10 cm carries 2 A current. Find the magnetic field at a point on the axis at distance 10 cm from the center.
Given:
- R = 10 cm = 0.1 m
- x = 10 cm = 0.1 m
- I = 2 A
Formula: $$B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$$
Solution:
$$R^2 + x^2 = (0.1)^2 + (0.1)^2 = 0.02$$
$$(R^2 + x^2)^{3/2} = (0.02)^{3/2} = 0.00283$$
$$B = \frac{(4\pi \times 10^{-7})(2)(0.1)^2}{2(0.00283)}$$
$$B = \frac{8\pi \times 10^{-9}}{0.00566} = 4.44 \times 10^{-6} \text{ T}$$
$$B \approx 4.44 \mu\text{T}$$
Answer: B ≈ 4.44 μT
Note: At center (x=0), B would be μ₀I/(2R) = 12.57 μT, which is larger.
Problem: At what distance from the center along the axis is the magnetic field of a circular loop (radius R) equal to half its value at the center?
At center: $$B_0 = \frac{\mu_0 I}{2R}$$
At distance x: $$B_x = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$$
Condition: B_x = B_0/2
$$\frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{2} \cdot \frac{\mu_0 I}{2R}$$
$$\frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{2R}$$
$$\frac{R^3}{(R^2 + x^2)^{3/2}} = \frac{1}{2}$$
$$2R^3 = (R^2 + x^2)^{3/2}$$
Raise both sides to power 2/3:
$$(2R^3)^{2/3} = R^2 + x^2$$
$$2^{2/3}R^2 = R^2 + x^2$$
$$x^2 = R^2(2^{2/3} - 1) = R^2(1.587 - 1) = 0.587R^2$$
$$x = 0.766R$$
Answer: x ≈ 0.77R or about 3R/4
❌ Common Student Mistakes
- Forgetting R² in numerator: The axial field formula has R² (area dependence), not just R
- Using center formula everywhere: B = μ₀I/(2R) is ONLY valid at center (x=0)
- Wrong distance in denominator: Use r = √(R²+x²), not just x
- Sign/direction confusion: Use right-hand rule consistently
- Forgetting number of turns: For N-turn coil, multiply by N
8. Ampere's Circuital Law
8.1 Introduction
While the Biot-Savart law can calculate the magnetic field for any current distribution, it often involves complex integrals. Ampère's circuital law provides an elegant alternative method for finding magnetic fields when there is high symmetry, analogous to how Gauss's law simplifies electric field calculations.
8.2 Statement of Ampère's Law
In words:
The line integral of magnetic field around any closed loop equals μ₀ times the total current passing through the loop.
Where:
- ∮ represents integration around a closed path (Amperian loop)
- 𝐁 = magnetic field
- d𝐥 = infinitesimal length element along the path
- I_enclosed = net current through the surface bounded by the loop
- μ₀ = 4π × 10⁻⁷ T·m/A
🔬 Physical Significance
Ampère's law reveals a fundamental relationship: magnetic fields circulate around currents. This is profoundly different from electric fields, which begin and end on charges.
The law states that currents are the sources of magnetic field circulation, just as charges are sources of electric field divergence.
8.3 Sign Convention for Current
📌 Right-Hand Rule for Ampère's Law
- Choose a direction to traverse the Amperian loop
- Curl fingers of right hand along this direction
- Thumb points in positive current direction
- Currents in thumb direction: +I_enclosed
- Currents opposite to thumb: −I_enclosed
Net enclosed current = ∑I_positive − ∑I_negative
8.4 Comparison with Gauss's Law
| Property | Gauss's Law (Electric) | Ampère's Law (Magnetic) |
|---|---|---|
| Law | $$\oint \vec{\mathbf{E}} \cdot d\vec{\mathbf{A}} = \frac{Q_{enc}}{\epsilon_0}$$ | $$\oint \vec{\mathbf{B}} \cdot d\vec{\mathbf{l}} = \mu_0 I_{enc}$$ |
| Integration | Surface integral (closed surface) | Line integral (closed loop) |
| Source | Enclosed charge Q | Enclosed current I |
| Applicability | Always valid | Always valid (steady currents) |
| Useful when | High symmetry (spherical, cylindrical, planar) | High symmetry (straight wire, solenoid, toroid) |
8.5 Application: Magnetic Field of Long Straight Wire
📐 Using Ampère's Law: Infinite Straight Wire
Problem: Find magnetic field at distance r from infinite straight wire carrying current I.
Step 1: Identify symmetry
By symmetry, magnetic field:
- Has same magnitude at all points equidistant from wire
- Forms circular field lines around wire
- Is tangent to circles centered on wire
Step 2: Choose Amperian loop
Circle of radius r centered on wire, in plane ⊥ to wire
Step 3: Apply Ampère's law
On this loop, 𝐁 is parallel to d𝐥 everywhere and has constant magnitude B.
$$\oint \vec{\mathbf{B}} \cdot d\vec{\mathbf{l}} = \oint B \, dl = B \oint dl = B(2\pi r)$$
Step 4: Enclosed current
$$I_{enclosed} = I$$
Step 5: Equate
$$B(2\pi r) = \mu_0 I$$
$$B = \frac{\mu_0 I}{2\pi r}$$
📌 Comparison: Biot-Savart vs Ampère
We derived the same result (B = μ₀I/(2πr)) using two different methods:
- Biot-Savart: Required complex integration over entire wire
- Ampère: Simple algebraic manipulation using symmetry
This demonstrates the power of Ampère's law for symmetric configurations!
8.6 Application: Magnetic Field Inside a Conductor
📐 Field Inside Current-Carrying Cylinder
Problem: A long cylindrical wire of radius R carries current I uniformly distributed. Find B at distance r from center (r < R).
Step 1: Current density
$$J = \frac{I}{\pi R^2}$$ (uniform distribution)
Step 2: Amperian loop
Circle of radius r (r < R) centered on axis
Step 3: Enclosed current
Current through area πr²:
$$I_{enc} = J \cdot \pi r^2 = \frac{I}{\pi R^2} \cdot \pi r^2 = I\frac{r^2}{R^2}$$
Step 4: Apply Ampère's law
$$B(2\pi r) = \mu_0 I\frac{r^2}{R^2}$$
$$B = \frac{\mu_0 I r}{2\pi R^2}$$
Note: Inside: B ∝ r (linear), Outside: B ∝ 1/r
At surface (r = R): Both formulas give same value
9. The Solenoid
9.1 Definition and Structure
A solenoid is a long coil of wire wound in the form of a helix. When current flows through it, the solenoid produces a remarkably uniform magnetic field in its interior, making it useful for many applications.
📌 Ideal Solenoid Properties
An ideal (infinitely long) solenoid has:
- Uniform, strong field inside the solenoid
- Zero field outside the solenoid
- Field lines parallel to axis inside
Real finite solenoids approximate these properties when length >> diameter
9.2 Magnetic Field Inside a Solenoid - Derivation
📐 Complete Derivation Using Ampère's Law
Setup: Consider a solenoid with:
- n = number of turns per unit length
- I = current through each turn
- Radius R
Step 1: Field pattern (from symmetry)
Inside long solenoid:
- Field is uniform and parallel to axis
- Magnitude same everywhere inside
Outside ideal solenoid: B = 0
Step 2: Choose Amperian loop
Rectangle with:
- One side (length L) inside solenoid, parallel to axis
- One side outside solenoid, parallel to axis
- Two sides perpendicular to axis
Step 3: Evaluate line integral ∮𝐁·d𝐥
Break into 4 segments:
- Inside parallel side: ∫B dl = BL (B parallel to dl)
- Outside parallel side: 0 (B = 0 outside)
- Two perpendicular sides: 0 (B ⊥ dl)
Total: ∮𝐁·d𝐥 = BL
Step 4: Calculate enclosed current
Number of turns in length L: N = nL
Current through each turn: I
Total enclosed current: I_enc = NI = nLI
Step 5: Apply Ampère's law
$$BL = \mu_0 (nLI)$$
$$B = \mu_0 nI$$
Where:
- n = number of turns per unit length (turns/m)
- I = current in solenoid (A)
If total N turns in length L:
Direction: Along axis (right-hand rule: curl fingers with current, thumb shows field)
🎯 Remarkable Properties of Solenoid Field
- Independent of position: Same field at all points inside (uniform field)
- Independent of diameter: Depends only on n and I, not on R
- Proportional to turn density: More tightly wound → stronger field
- Negligible outside: For long solenoid, external field ≈ 0
This makes solenoids ideal for:
- Creating uniform magnetic fields in laboratories
- Electromagnets
- Magnetic resonance imaging (MRI)
- Particle accelerators
9.3 Toroid - Solenoid Bent into Circle
A toroid is a solenoid bent into a circular (doughnut) shape.
📐 Magnetic Field in Toroid
Apply Ampère's law to circular loop of radius r (measured from toroid center):
Inside toroid (between inner and outer radius):
Amperian loop encloses N turns, each carrying current I
$$B(2\pi r) = \mu_0 NI$$
$$B = \frac{\mu_0 NI}{2\pi r}$$
Outside toroid: B = 0 (loop encloses zero net current)
Inside inner hollow: B = 0 (loop encloses zero net current)
Where:
- N = total number of turns
- r = distance from toroid center (inside toroid only)
Note: Field varies with r (unlike straight solenoid)
9.4 Applications of Solenoids
| Application | Principle | Example |
|---|---|---|
| Electromagnet | Strong uniform field with iron core | Cranes, magnetic locks |
| Electric Bell/Buzzer | Solenoid pulls iron armature | Doorbells, alarms |
| Relay/Actuator | Solenoid operates mechanical switch | Automotive starters, valves |
| MRI Machines | Superconducting solenoid (uniform field) | Medical imaging |
| Inductors | Store energy in magnetic field | Electronic circuits, power supplies |
| Particle Accelerators | Guide charged particle beams | LHC, synchrotrons |
9.5 Solved Examples
Problem: A solenoid of length 50 cm has 1000 turns and carries a current of 5 A. Find the magnetic field inside the solenoid.
Given:
- L = 50 cm = 0.5 m
- N = 1000 turns
- I = 5 A
Step 1: Find n (turns per unit length)
$$n = \frac{N}{L} = \frac{1000}{0.5} = 2000 \text{ turns/m}$$
Step 2: Apply formula
$$B = \mu_0 nI = (4\pi \times 10^{-7})(2000)(5)$$
$$B = 40000\pi \times 10^{-7} = 4\pi \times 10^{-3}$$
$$B = 0.0126 \text{ T} = 12.6 \text{ mT}$$
Answer: B = 12.6 mT (quite strong - comparable to small bar magnet)
Problem: What current is needed in a 2000 turns/m solenoid to produce a magnetic field of 2.0 mT inside it?
Given:
- n = 2000 turns/m
- B = 2.0 mT = 2.0 × 10⁻³ T
Formula: $$B = \mu_0 nI$$
Solve for I:
$$I = \frac{B}{\mu_0 n} = \frac{2.0 \times 10^{-3}}{(4\pi \times 10^{-7})(2000)}$$
$$I = \frac{2.0 \times 10^{-3}}{8\pi \times 10^{-4}} = \frac{2.0}{8\pi} \times 10^{1}$$
$$I = \frac{20}{8\pi} = 0.796 \text{ A}$$
Answer: I ≈ 0.8 A
Problem: A toroid has 3000 turns and mean radius 20 cm. If it carries 2 A current, find the magnetic field inside the toroid.
Given:
- N = 3000 turns
- r = 20 cm = 0.2 m (mean radius)
- I = 2 A
Formula: $$B = \frac{\mu_0 NI}{2\pi r}$$
Solution:
$$B = \frac{(4\pi \times 10^{-7})(3000)(2)}{2\pi (0.2)}$$
$$B = \frac{24000\pi \times 10^{-7}}{0.4\pi} = \frac{24000 \times 10^{-7}}{0.4}$$
$$B = 60000 \times 10^{-7} = 6 \times 10^{-3} \text{ T}$$
$$B = 6 \text{ mT}$$
Answer: B = 6 mT
10. Force Between Two Parallel Current-Carrying Conductors
10.1 The Interaction
When two parallel wires carry electric currents, they exert magnetic forces on each other. This fundamental interaction led to the modern definition of the ampere.
10.2 Derivation of Force
📐 Force Between Two Parallel Wires
Setup: Two long parallel wires separated by distance d, carrying currents I₁ and I₂.
Step 1: Magnetic field from wire 1
At distance d, wire 1 creates field:
$$B_1 = \frac{\mu_0 I_1}{2\pi d}$$
Step 2: Force on wire 2
Wire 2 (current I₂, length L) experiences force in field B₁:
$$F = I_2 L B_1 = I_2 L \cdot \frac{\mu_0 I_1}{2\pi d}$$
Step 3: Force per unit length
$$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$$
Direction:
- Parallel currents (same direction): Attractive force
- Antiparallel currents (opposite directions): Repulsive force
🔬 Definition of the Ampere
The ampere is defined using this force law:
One ampere is that constant current which, when maintained in two straight parallel conductors of infinite length and negligible cross-section, placed 1 meter apart in vacuum, produces between them a force of 2 × 10⁻⁷ newton per meter of length.
Setting I₁ = I₂ = 1 A, d = 1 m:
$$\frac{F}{L} = \frac{(4\pi \times 10^{-7})(1)(1)}{2\pi(1)} = 2 \times 10^{-7} \text{ N/m}$$ ✓
10.3 Physical Explanation
Why parallel currents attract:
- Wire 1 creates circular magnetic field around itself
- At location of wire 2, this field is perpendicular to wire 2
- Current in wire 2 experiences force (F = IL × B)
- By right-hand rule, force is toward wire 1 (attraction)
- By Newton's third law, wire 1 also attracted to wire 2
Problem: Two long parallel wires 10 cm apart carry currents of 5 A and 8 A in the same direction. Find force per unit length on each wire.
Given: d = 0.1 m, I₁ = 5 A, I₂ = 8 A
$$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{(4\pi \times 10^{-7})(5)(8)}{2\pi(0.1)}$$
$$= \frac{160\pi \times 10^{-7}}{0.2\pi} = 8 \times 10^{-5} \text{ N/m}$$
Answer: 8 × 10⁻⁵ N/m = 80 μN/m (attractive)
11. Torque on a Current Loop in Magnetic Field
11.1 Rectangular Current Loop
📐 Complete Derivation: Torque on Rectangular Loop
Setup: Rectangular loop (sides a and b), current I, in uniform field B, with loop's normal at angle θ to B.
Step 1: Forces on the four sides
- Sides parallel to B: F = 0 (θ = 0°)
- Sides perpendicular to B: F = IaB (both sides)
Step 2: Net force = 0
The two perpendicular forces are equal, opposite, and collinear → no net force
Step 3: Calculate torque
But these forces form a couple! Perpendicular distance between them = b sin θ
Torque = Force × perpendicular distance
$$\tau = (IaB) \times (b\sin\theta) = IAB\sin\theta$$
where A = ab = area of loop
Vector form: $$\vec{\tau} = I\vec{A} \times \vec{B}$$
For N-turn coil: $$\tau = NIAB\sin\theta$$
11.2 Torque Characteristics
| Orientation | θ | sin θ | Torque | Stability |
|---|---|---|---|---|
| Loop ⊥ to B | 90° | 1 | Maximum (IAB) | Unstable equilibrium |
| Loop ∥ to B | 0° | 0 | Zero | Stable equilibrium |
12. Magnetic Dipole and Dipole Moment
12.1 Magnetic Dipole Moment
Magnitude: m = NIA
Direction: Perpendicular to loop plane (right-hand rule)
Unit: A·m² or J/T
12.2 Torque in Terms of Dipole Moment
Magnitude: $$\tau = mB\sin\theta$$
12.3 Potential Energy
Minimum energy: θ = 0° (m ∥ B), U = −mB
Maximum energy: θ = 180° (m antiparallel to B), U = +mB
Problem: A circular coil of 100 turns, radius 5 cm, carries 2 A current. Find its magnetic dipole moment.
Given: N = 100, R = 0.05 m, I = 2 A
Area: $$A = \pi R^2 = \pi(0.05)^2 = 7.85 \times 10^{-3} \text{ m}^2$$
$$m = NIA = (100)(2)(7.85 \times 10^{-3}) = 1.57 \text{ A·m}^2$$
Answer: m = 1.57 A·m²
13. The Moving Coil Galvanometer
13.1 Introduction
A galvanometer is an instrument used to detect and measure small electric currents. It works on the principle of torque experienced by a current-carrying coil in a magnetic field.
13.2 Construction
📌 Main Components
- Coil: Rectangular coil of fine insulated copper wire wound on non-magnetic frame
- Magnet: Strong permanent horseshoe magnet with concave pole pieces
- Soft iron core: Cylindrical core inside coil (concentrates magnetic field)
- Spring: Phosphor bronze spring provides restoring torque
- Pointer: Attached to coil, moves over graduated scale
13.3 Working Principle
When current flows through the coil:
- Coil experiences torque: τ = NIAB
- Coil rotates
- Spring provides restoring torque: τ_spring = kθ
- At equilibrium: NIAB = kθ
- Deflection θ ∝ I
Current sensitivity: $$S_I = \frac{\theta}{I} = \frac{NAB}{k}$$
Voltage sensitivity: $$S_V = \frac{\theta}{V} = \frac{NAB}{kR_g}$$
where R_g = galvanometer resistance
13.4 Conversion to Ammeter
To convert galvanometer to ammeter, connect low resistance shunt (S) in parallel.
where I_g = full-scale galvanometer current, I = desired ammeter range
13.5 Conversion to Voltmeter
To convert galvanometer to voltmeter, connect high resistance R in series.
where V = desired voltmeter range
Problem: A galvanometer has resistance 50 Ω and gives full-scale deflection for 5 mA. Convert it to ammeter reading 5 A.
$$S = \frac{I_g R_g}{I - I_g} = \frac{(0.005)(50)}{5 - 0.005} = \frac{0.25}{4.995} = 0.05 \Omega$$
Answer: Shunt = 0.05 Ω
14. Multiple Choice Questions (50+)
Total 50+ MCQs with explanations in complete chapter
15. Very Short Answer Questions (1-2 marks)
Q1. State the SI unit of magnetic field.
Answer: Tesla (T) or Weber/m²
Q2. Why don't magnetic field lines intersect?
Answer: If they intersect, there would be two directions of B at that point, which is impossible.
Q3. Write the Lorentz force equation.
Answer: F = q(E + v × B)
Q4. State Biot-Savart law.
Answer: dB = (μ₀/4π) (Idl × r̂)/r²
Q5. Define magnetic dipole moment.
Answer: m = NIA (product of current, area, and number of turns)
50+ Very Short Answer questions in complete chapter
16. Short Answer Questions (3-4 marks)
Q1. Derive expression for cyclotron frequency.
Answer: For circular motion in magnetic field: qvB = mv²/r → r = mv/(qB)
Time period: T = 2πr/v = 2πm/(qB)
Frequency: f = 1/T = qB/(2πm) ✓
Q2. Why parallel currents attract?
Answer: Wire 1 creates circular magnetic field. At wire 2's location, this field is perpendicular to wire 2. Force F = IL×B on wire 2 points toward wire 1 (by right-hand rule), hence attraction.
Q3. State Ampère's circuital law.
Answer: ∮B·dl = μ₀I_enclosed. The line integral of magnetic field around closed loop equals μ₀ times net current through loop.
50+ Short Answer questions in complete chapter
17. Long Answer Questions (5 marks)
Q1. Derive expression for magnetic field on axis of circular current loop.
Answer: [Complete derivation as shown in Section 7]
Result: B = μ₀IR²/[2(R²+x²)^(3/2)]
At center (x=0): B = μ₀I/(2R)
Q2. Derive expression for magnetic field inside solenoid using Ampère's law.
Answer: [Complete derivation as shown in Section 9]
Result: B = μ₀nI (uniform field inside)
10+ Long Answer questions with full derivations in complete chapter
18. Complete Formula Sheet
18.1 Lorentz Force & Motion
| Formula | Description |
|---|---|
| F = q(E + v × B) | Lorentz force |
| F_B = qvB sin θ | Magnetic force magnitude |
| r = mv/(qB) | Radius of circular path |
| f = qB/(2πm) | Cyclotron frequency |
| p = 2πmv cos θ/(qB) | Pitch of helix |
18.2 Force on Conductor
| Formula | Description |
|---|---|
| F = IL × B | Force on straight conductor |
| F = ILB sin θ | Magnitude |
| F/L = μ₀I₁I₂/(2πd) | Force between parallel wires |
18.3 Magnetic Fields
| Formula | Source |
|---|---|
| dB = (μ₀/4π) Idl×r̂/r² | Biot-Savart law |
| B = μ₀I/(2πr) | Infinite straight wire |
| B = μ₀I/(2R) | Center of circular loop |
| B = μ₀IR²/[2(R²+x²)^(3/2)] | Axis of circular loop |
| ∮B·dl = μ₀I_enc | Ampère's law |
| B = μ₀nI | Inside solenoid |
| B = μ₀NI/(2πr) | Inside toroid |
18.4 Torque & Dipole
| Formula | Description |
|---|---|
| τ = NIAB sin θ | Torque on loop |
| m = NIA | Magnetic dipole moment |
| τ = m × B | Torque vector form |
| U = -m·B | Potential energy |
18.5 Galvanometer
| Formula | Use |
|---|---|
| S = I_gR_g/(I-I_g) | Ammeter shunt |
| R = V/I_g - R_g | Voltmeter series resistance |
18.6 Constants
| Constant | Value |
|---|---|
| μ₀ | 4π × 10⁻⁷ T·m/A |
| 1 Tesla | 1 Wb/m² = 10⁴ Gauss |
| 1 Ampere definition | Force = 2×10⁻⁷ N/m between parallel wires 1m apart |
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