Class 12 Physics Notes Chapter 4 (Moving Charges and Magnetism) – Examplar Problems (English) Book

Alright class, let's dive deep into Chapter 4: Moving Charges and Magnetism. This is a crucial chapter, not just for your board exams but also extensively tested in various government competitive exams. We'll focus on the core concepts, formulas, and applications as presented in the NCERT framework, keeping the competitive exam perspective in mind.

Chapter 4: Moving Charges and Magnetism - Detailed Notes

1. Introduction & Oersted's Discovery:

• Concept: Electric currents (moving charges) produce magnetic fields around them. This was accidentally discovered by Hans Christian Oersted.
• Key Idea: Electricity and Magnetism are interconnected phenomena (Electromagnetism).

2. Magnetic Force (Lorentz Force):

• Force on a Moving Charge: A charge 'q' moving with velocity 'v' in a magnetic field 'B' experiences a force Fm given by:
  Fm = q (v × B)
  • Magnitude: F = |q| v B sinθ, where θ is the angle between v and B.
  • Direction: Given by the Right-Hand Palm Rule or Fleming's Left-Hand Rule (for positive charge; opposite for negative charge). Force is perpendicular to both v and B.
• Properties:
  • Fm = 0 if q = 0 (neutral particle).
  • Fm = 0 if v = 0 (stationary charge).
  • Fm = 0 if θ = 0° or 180° (charge moves parallel or anti-parallel to B).
  • Fm is maximum (qvB) if θ = 90° (charge moves perpendicular to B).
• Work Done: Magnetic force does no work on the charged particle because Fm is always perpendicular to v (and hence displacement). W = ∫ Fm . dr = 0. Kinetic energy and speed of the particle remain constant, only the direction of velocity changes.
• Magnetic Field (B): Defined from the Lorentz force equation. If q=1, v=1, θ=90°, then B = F.
  • SI Unit: Tesla (T). 1 T = 1 N s / (C m) = 1 N / (A m).
  • CGS Unit: Gauss (G). 1 T = 10⁴ G.
  • Dimensions: [M T⁻² A⁻¹]
• Lorentz Force (Total): If both electric field E and magnetic field B are present, the total force on charge q is:
  F = qE + q(v × B)

3. Motion of a Charged Particle in a Uniform Magnetic Field:

• Case 1: v is perpendicular to B (θ = 90°):
  • Path: Circular.
  • Centripetal force is provided by the magnetic force: mv²/r = qvB
  • Radius of circular path: r = mv / qB = p / qB (p = momentum)
  • Angular frequency (cyclotron frequency): ω = v/r = qB / m
  • Time Period: T = 2π/ω = 2πm / qB (Independent of v and r)
  • Frequency: f = 1/T = qB / 2πm (Independent of v and r)
• Case 2: v is parallel or anti-parallel to B (θ = 0° or 180°):
  • Force F = 0.
  • Path: Particle continues to move along a straight line with constant velocity.
• Case 3: v enters at an angle θ (0° < θ < 90°):
  • Resolve velocity into components: v∥ = v cosθ (parallel to B) and v⊥ = v sinθ (perpendicular to B).
  • v∥ causes linear motion along the field lines (no force).
  • v⊥ causes circular motion perpendicular to the field lines (radius r = mv⊥ / qB = mv sinθ / qB).
  • Resultant Path: Helical (spiral).
  • Pitch of Helix (p): Linear distance covered in one revolution. p = v∥ × T = (v cosθ) × (2πm / qB).

4. Velocity Selector:

• A region with perpendicular uniform electric (E) and magnetic (B) fields.
• Force due to E: Fe = qE
• Force due to B: Fm = q(v × B)
• By adjusting E and B such that Fe and Fm are opposite and equal in magnitude: qE = qvB
• Particles with a specific velocity v = E/B pass undeflected. Used in mass spectrometers.

5. Cyclotron:

• Principle: A charged particle can be accelerated to very high energies by making it pass repeatedly through a moderate electric field, using a strong magnetic field to constrain its path.
• Construction: Two hollow D-shaped metallic chambers ('Dees'), high-frequency oscillator, strong electromagnet.
• Working: Magnetic field makes the ion move in a semi-circular path inside a Dee. Electric field accelerates the ion across the gap between the Dees. Radius increases as velocity increases (r = mv/qB). Time spent in each Dee is constant (T/2 = πm/qB).
• Resonance Condition: Frequency of oscillator (f_osc) must match the cyclotron frequency (f_c = qB/2πm).
• Maximum Kinetic Energy: K_max = (q²B²R²) / 2m, where R is the radius of the Dees.
• Limitations: Cannot accelerate electrons (relativistic mass increase), cannot accelerate neutral particles.

6. Magnetic Force on a Current-Carrying Conductor:

• Concept: A current is a flow of charges. Each charge experiences a Lorentz force, the sum of which is the net force on the conductor.
• Formula: For a straight conductor of length L carrying current I in a uniform field B:
  F = I (L × B)
  • L is a vector whose magnitude is the length L and direction is along the current flow.
  • Magnitude: F = I L B sinθ, where θ is the angle between L and B.
  • Direction: Fleming's Left-Hand Rule or Right-Hand Palm Rule.
• Force on an arbitrarily shaped wire: dF = I (dl × B). Integrate this over the length of the wire.
• Force between two parallel infinite straight conductors:
  • Let wires carry currents I₁ and I₂ and be separated by distance 'd'.
  • Field due to wire 1 at wire 2: B₁ = μ₀I₁ / (2πd)
  • Force per unit length on wire 2: F₂/L = I₂ B₁ sin(90°) = μ₀I₁I₂ / (2πd)
  • Direction: Attractive if currents are in the same direction, repulsive if currents are in opposite directions.
• Definition of Ampere (SI Unit of Current): One Ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to 2 × 10⁻⁷ newton per metre of length.

7. Torque on a Current Loop in a Uniform Magnetic Field:

• Rectangular Loop: Consider a loop PQRS with length 'l' and breadth 'b' carrying current 'I' in a uniform field B. Forces on arms QR and SP are equal and opposite along the axis of rotation, causing no torque. Forces on arms PQ and RS (F = I l B) are equal, opposite, and separated, forming a couple that produces torque.
• Torque (τ): τ = Force × perpendicular distance = (IlB) × (b sinθ) = I (lb) B sinθ = I A B sinθ
  • A = lb = Area of the loop.
  • θ is the angle between the normal to the plane of the loop and the magnetic field B.
• Magnetic Dipole Moment (M or μ): A current loop behaves like a magnetic dipole.
  • M = N I A
    • N = number of turns in the loop.
    • I = current.
    • A = Area vector (magnitude = Area, direction perpendicular to the plane of the loop, given by Right-Hand Thumb Rule - curl fingers along current, thumb gives direction of A and M).
  • SI Unit: Ampere-metre² (A m²)
• Torque in vector form: τ = M × B
• Potential Energy of a Magnetic Dipole (U):
  • U = - M . B = - MB cosθ
  • Stable Equilibrium: θ = 0° (M || B), U = -MB (minimum)
  • Unstable Equilibrium: θ = 180° (M anti-parallel to B), U = +MB (maximum)
  • Zero Energy: θ = 90° (M ⊥ B), U = 0

8. Moving Coil Galvanometer (MCG):

• Principle: Torque experienced by a current-carrying coil placed in a uniform magnetic field (τ = NIAB sinθ).
• Construction: Coil wound on a non-magnetic frame, suspended/pivoted in a radial magnetic field, spring, pointer, scale.
• Radial Magnetic Field: Achieved using concave pole pieces and a soft iron core. Ensures that the plane of the coil is always parallel to the magnetic field lines (θ = 90°, sinθ = 1). This makes the deflecting torque directly proportional to current (τ_deflecting = NIAB).
• Working: Deflecting torque (τ_deflecting = NIAB) twists the suspension/spring. Restoring torque develops in the spring (τ_restoring = kφ), where k is the torsional constant and φ is the deflection angle.
• In equilibrium: τ_deflecting = τ_restoring => NIAB = kφ => φ = (NAB/k) I
• Deflection is proportional to current: φ ∝ I. Hence, a linear scale is used.
• Sensitivity:
  • Current Sensitivity (IS): Deflection per unit current. IS = φ/I = NAB/k. (Unit: rad/A or div/A). Increase by increasing N, A, B or decreasing k (use phosphor-bronze).
  • Voltage Sensitivity (VS): Deflection per unit voltage. VS = φ/V = φ/(IR) = (NAB/k) * (1/R) = IS / R. (Unit: rad/V or div/V). R is the galvanometer resistance. Increasing N may increase R, so increasing IS might not necessarily increase VS.
• Conversion:
  • MCG to Ammeter: Connect a low resistance shunt (S) in parallel with the galvanometer. S = (Ig G) / (I - Ig), where G is galvanometer resistance, Ig is full-scale deflection current, I is the range of the ammeter. Effective resistance RA = GS / (G+S) (very low).
  • MCG to Voltmeter: Connect a high resistance (R) in series with the galvanometer. R = (V / Ig) - G, where V is the range of the voltmeter. Effective resistance RV = G + R (very high).

9. Biot-Savart Law:

• Statement: Gives the magnetic field (dB) produced at a point P due to a small current element Idl.
  dB = (μ₀ / 4π) * (I dl × r) / r³
  • μ₀ = Permeability of free space = 4π × 10⁻⁷ T m / A.
  • dl = length vector of the element in the direction of current.
  • r = position vector from the element dl to the point P.
  • r = |r|.
  • Magnitude: dB = (μ₀ / 4π) * (I dl sinθ) / r², where θ is the angle between dl and r.
  • Direction: Given by Right-Hand Thumb Rule (Thumb along Idl, curl fingers give field direction) or cross product rule. dB is perpendicular to the plane containing dl and r.
• Applications:
  • Magnetic field due to a straight finite wire: B = (μ₀I / 4πa) * (sinφ₁ + sinφ₂) (a = perpendicular distance from point to wire).
  • Magnetic field due to an infinite straight wire: B = μ₀I / 2πa (φ₁ = φ₂ = 90°).
  • Magnetic field at the center of a circular loop (radius R): B = μ₀I / 2R.
  • Magnetic field on the axis of a circular loop (radius R, distance x from center): B = (μ₀ N I R²) / [2 (R² + x²)^(3/2)]. (N = number of turns).

10. Ampere's Circuital Law:

• Statement: The line integral of the magnetic field B around any closed loop (Amperian loop) is equal to μ₀ times the total current (I_enclosed) passing through the area enclosed by the loop.
  ∮ B . dl = μ₀ I_enclosed
• Analogy: Similar to Gauss's Law in electrostatics. Useful for calculating B in situations with high symmetry.
• Applications:
  • Magnetic field due to an infinite straight wire: Choose a circular Amperian loop of radius 'a' centered on the wire. ∮ B dl cos(0°) = B (2πa) = μ₀I => B = μ₀I / 2πa.
  • Magnetic field inside a long Solenoid: A solenoid is a helical coil. Inside a long solenoid, the field is uniform and parallel to the axis; outside, it's nearly zero. Consider a rectangular Amperian loop. ∮ B . dl = BL = μ₀ (nL I), where n = number of turns per unit length, L = length of the side of the loop inside. => B = μ₀ n I.
  • Magnetic field inside a Toroid: A toroid is a solenoid bent into a circle. Field exists only inside the toroidal core. Consider a circular Amperian loop of radius r within the core. B (2πr) = μ₀ (N I), where N = total number of turns. => B = μ₀ N I / (2πr) = μ₀ n I (where n = N/2πr, turns per unit length). Field is not uniform across the cross-section.

Key Takeaways for Exams:

• Master the formulas for Lorentz force, motion in B-field (radius, time period), force on a wire, torque on a loop, magnetic moment.
• Understand the vector nature of quantities and direction rules (Right-Hand Rules, Fleming's Rule).
• Know the statements and applications of Biot-Savart Law and Ampere's Circuital Law, especially for standard geometries (wire, loop, solenoid, toroid).
• Understand the principle and conversion of MCG.
• Remember that magnetic forces do no work on free charges.
• Units and dimensions are important.

Multiple Choice Questions (MCQs)

1. A proton and an alpha particle enter a uniform magnetic field perpendicularly with the same speed. If the proton takes 25 μs to make 5 revolutions, then the periodic time for the alpha particle would be:
   (a) 50 μs
   (b) 25 μs
   (c) 10 μs
   (d) 5 μs

2. A charged particle is moving in a magnetic field of strength B perpendicular to the direction of the field. If q and m denote the charge and mass of the particle respectively, then the frequency of rotation of the particle is:
   (a) qB / (2πm)
   (b) qB / (πm)
   (c) 2πm / (qB)
   (d) qm / (2πB)

3. Two long parallel wires carry currents I₁ and I₂ (I₁ > I₂) in the same direction. The magnetic field midway between the wires is B. If the current I₂ is switched off, the magnetic field at that point will be:
   (a) B (I₁ + I₂) / (I₁ - I₂)
   (b) B (I₁ - I₂) / (I₁ + I₂)
   (c) B I₁ / (I₁ - I₂)
   (d) B I₂ / (I₁ - I₂)

4. A circular coil of radius R carrying current I lies in the x-y plane with its center at the origin. The total magnetic flux through the x-y plane (excluding the area of the coil) is:
   (a) Directly proportional to I
   (b) Directly proportional to R
   (c) Inversely proportional to I
   (d) Zero

5. A solenoid has a core of material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2 A. If the number of turns is 1000 per metre, the magnetic field B inside the solenoid is: (μ₀ = 4π × 10⁻⁷ Tm/A)
   (a) 1.0 T
   (b) 2.0 T
   (c) 0.5 T
   (d) π T

6. To convert a galvanometer into an ammeter, one needs to connect a:
   (a) Low resistance in parallel
   (b) High resistance in parallel
   (c) Low resistance in series
   (d) High resistance in series

7. A particle of charge q and mass m moves in a circular orbit of radius r with angular speed ω. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on:
   (a) ω and m
   (b) ω, q and m
   (c) q and m
   (d) ω and q

8. A current loop placed in a non-uniform magnetic field experiences:
   (a) a force but no torque
   (b) a torque but no force
   (c) a force and a torque
   (d) neither a force nor a torque

9. Ampere's circuital law is applicable for:
   (a) Any closed path
   (b) Only circular paths
   (c) Only paths with high symmetry
   (d) Only paths in vacuum

10. A long straight wire carries a current I. A proton travels with a speed v, parallel to the wire, distance d from it. The force on the proton is:
    (a) μ₀ I v / (2πd) , towards the wire
    (b) μ₀ q v I / (2πd) , towards the wire
    (c) μ₀ q v I / (2πd) , away from the wire
    (d) Zero

Answer Key for MCQs:

1. (c) 10 μs [T = 2πm/qB. T_p = 25μs/5 = 5μs. T_α / T_p = (m_α/q_α) / (m_p/q_p) = (4m_p/2q_p) / (m_p/q_p) = 2. T_α = 2 * T_p = 10μs]
2. (a) qB / (2πm) [Direct formula for cyclotron frequency]
3. (c) B I₁ / (I₁ - I₂) [B = B₁ - B₂ = (μ₀/2π(d/2)) (I₁ - I₂). When I₂ is off, B' = B₁ = (μ₀/2π(d/2)) I₁. Ratio B'/B = I₁ / (I₁ - I₂)]
4. (d) Zero [Magnetic field lines form closed loops. Flux entering the plane outside the coil must equal the flux exiting the plane through the coil area to form closed loops. Total flux through any infinite plane is zero.]
5. (a) 1.0 T [B = μ n I = μ₀ μᵣ n I = (4π × 10⁻⁷) × 400 × 1000 × 2 ≈ 1.0 T]
6. (a) Low resistance in parallel
7. (c) q and m [Magnetic moment M = IA = (q/T)πr² = (qω/2π)πr². Angular momentum L = mvr = m(ωr)r = mωr². Ratio M/L = (qωr²/2) / (mωr²) = q / 2m]
8. (c) a force and a torque [In a non-uniform field, forces on opposite sides may not cancel, leading to a net force. Torque is generally present unless M || B or M anti-parallel to B.]
9. (a) Any closed path [Although it's useful for calculation mainly for symmetric paths, the law itself holds true for any closed path.]
10. (b) μ₀ q v I / (2πd) , towards the wire [B field due to wire at distance d is B = μ₀I / (2πd) (into the plane, assuming wire current is up and proton is to the right). Force F = q(v × B). v is up, B is into the plane. v × B is towards the wire. F = qvB = qv(μ₀I / (2πd))]

Study these notes thoroughly, focusing on understanding the concepts behind the formulas. Practice applying these in various problems, especially from the Exemplar book. Good luck!