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WBB • Class XII • Physics • Ch 3
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Magnetic Effect of Current and Magnetism

Magnetic Effect of Current and Magnetism bridges classical electrodynamics and macroscopic magnetic phenomena, establishing how steady electric currents act as fundamental sources of magnetic fields and how external magnetic fields exert decisive mechanical forces on moving charges and current-carrying conductors. Starting from Hans Christian Oersted's 1820 discovery and formulated quantitatively through the Biot-Savart Law and Ampere's Circuital Law, this chapter explores the magnetic field geometries of straight conductors, circular coils, solenoids, and toroids. It examines the helical trajectories of charged particles under the Lorentz force, the resonance mechanics of the Cyclotron, the fundamental definition of the SI unit Ampere, the operational physics of the Moving Coil Galvanometer, and culminates in terrestrial magnetism, atomic magnetic dipoles, and the classification of magnetic materials into diamagnetic, paramagnetic, and ferromagnetic substances.

Why This Chapter Matters

Mastery of electromagnetic and magnetic principles is essential for understanding both fundamental modern physics and everyday technology. The interaction between electric currents and magnetic fields forms the operating foundation of electric motors, particle accelerators, magnetic resonance imaging (MRI) systems, high-speed maglev transportation, and sensitive analog electrical measurement meters. In the WBCHSE Higher Secondary examination, Unit III consistently carries heavy weight, testing students on vector cross products, multi-step calculus derivations, circuit conversion of galvanometers, and conceptual distinctions between diamagnetic, paramagnetic, and ferromagnetic materials.

Chapter Roadmap & Progression

1 Biot-Savart Law and Magnetic Field...
2 Ampère's Circuital Law, Solenoid, a...
3 Motion of Charged Particles in Magn...
4 Magnetic Force on Conductors and Pa...
5 Torque on Current Loops, Magnetic D...
6 Magnetism, Terrestrial Magnetism, a...

Complete Concept Guide (100% Curriculum Coverage)

Biot-Savart Law and Magnetic Field Calculations

1.1 Historical Context and Vector Formulation

In 1820, Hans Christian Oersted discovered that an electric current deflects a nearby magnetic compass needle. Jean-Baptiste Biot and Félix Savart subsequently quantified the magnetic field produced by a steady line current element I d𝐥. At a displacement vector 𝐫 from the element, the infinitesimal magnetic field d𝐁 is given by the Biot-Savart Law:

d𝐁 = (μ₀ / 4π) · [I (d𝐥 × 𝐫̂) / r²] = (μ₀ / 4π) · [I (d𝐥 × 𝐫) / r³]

where μ₀ = 4π × 10⁻⁷ T·m·A⁻¹ (or H/m) is the magnetic permeability of free space. In scalar magnitude:

dB = (μ₀ / 4π) · [I dl sin θ / r²]

where θ is the angle between the current element direction d𝐥 and the position vector 𝐫. The direction of d𝐁 is strictly perpendicular to the plane containing both d𝐥 and 𝐫, determined by the Right-Hand Screw Rule.

1.2 Comparison: Biot-Savart Law vs Coulomb's Law
Characteristic Coulomb's Law (Electrostatics) Biot-Savart Law (Magnetism)
Source Scalar source: electric charge dq Vector source: current element I d𝐥
Field Direction Along displacement vector 𝐫 (radial) Normal to plane of d𝐥 and 𝐫 (cross product)
Angle Dependence No angular dependence (isotropic) Depends on sin θ; zero along line of current element
Distance Dependence Inverse-square law: 1/r² Inverse-square law: 1/r²
1.3 Magnetic Field on the Axis of a Circular Current Loop

Consider a circular loop of radius R carrying steady current I in the y-z plane. For a point P on the loop axis (x-axis) at distance x from the center:

  • The distance from any element d𝐥 on the perimeter to point P is r = √(R² + x²).
  • Since d𝐥 is tangential to the circumference and 𝐫 lies in the plane connecting the element to the axis, d𝐥 ⊥ 𝐫, meaning sin 90° = 1.
  • The magnetic field component perpendicular to the axis cancels out by diametrical symmetry: ∮ dB_⊥ = 0.
  • The axial component is dB_x = dB cos α = dB · (R / r) = (μ₀ I dl / 4πr²) · (R / r) = (μ₀ I R dl) / [4π (R² + x²)^(3/2)].

Integrating around the entire circumference ∮ dl = 2πR for an N-turn coil:

B_axial = (μ₀ N I R²) / [2 (R² + x²)^(3/2)]   [Tesla (T)]

Special Cases:

  1. At the Center of the Loop (x = 0):
    B_center = (μ₀ N I) / (2R)
  2. Far from the Loop (x ≫ R):
    B_far = (μ₀ N I R²) / (2x³) = (μ₀ / 4π) · [2 (N I · πR²) / x³] = (μ₀ / 4π) · (2M / x³)
    where M = N I A is the magnetic dipole moment. This proves that a planar current loop behaves identically to a magnetic dipole on its axial line!

Ampère's Circuital Law, Solenoid, and Toroid

2.1 Statement and Mathematical Formulation

Ampère's Circuital Law states that the line integral of the magnetic field 𝐁 around any closed arbitrary loop (called an Amperian loop) in free space equals μ₀ times the net steady current enclosed by the loop:

∮_C 𝐁 · d𝐥 = μ₀ I_enclosed = μ₀ ∑ I_net

Sign Convention: Curl the fingers of the right hand in the direction of traversal along the closed Amperian loop; currents passing through the enclosed surface in the direction of the extended thumb are considered positive, while opposing currents are negative.

2.2 Derivation: Long Straight Current-Carrying Conductor

Consider an infinitely long, thin straight wire carrying steady current I. By cylindrical symmetry, magnetic field lines form concentric circles centered on the wire. Choose a circular Amperian loop of radius r concentric with the wire:

  • At every point on the path, 𝐁 is tangential to the circle, so 𝐁 ∥ d𝐥 (cos 0° = 1).
  • By symmetry, the magnitude of B is constant at all points equidistant from the wire.
∮ 𝐁 · d𝐥 = B ∮ dl = B (2πr) = μ₀ I  ⟹  B = (μ₀ I) / (2πr)
2.3 Magnetic Field of an Ideal Long Solenoid

A solenoid consists of a long helical winding of insulated wire. Let n = N / L represent the number of turns per unit length carrying current I. In an ideal long solenoid (L ≫ R):

  1. The field outside the solenoid is practically zero: B_ext ≈ 0.
  2. The field inside is uniform, intense, and strictly parallel to the solenoid axis.

Choose a rectangular Amperian loop abcd of length h, where side ab lies inside along the axis, sides bc and da extend radially outwards, and side cd lies outside:

∮ 𝐁 · d𝐥 = ∫_a^b 𝐁·d𝐥 + ∫_b^c 𝐁·d𝐥 + ∫_c^d 𝐁·d𝐥 + ∫_d^a 𝐁·d𝐥 = B·h + 0 + 0 + 0 = B·h

The total current enclosed by the loop is I_enc = (n · h) · I. Applying Ampère's Law:

B·h = μ₀ (n·h·I)  ⟹  B_internal = μ₀ n I

Important Boundary Note: At the extreme open ends of a semi-infinite solenoid, half the magnetic field lines escape into space, yielding exactly half the central value: B_end = (1/2) μ₀ n I.

2.4 Magnetic Field of a Toroid

A toroid is an endless solenoid bent into the shape of a doughnut (doughnut core of inner radius r₁ and outer radius r₂). With mean radius r and total turns N:

  • Inside the Toroidal Core: Choosing a circular Amperian loop of radius r inside the core: ∮ 𝐁·d𝐥 = B (2πr) = μ₀ (N I) ⟹ B = (μ₀ N I) / (2πr) = μ₀ n I, where n = N / (2πr).
  • In the Open Space Inside the Toroid (r < r₁): I_enc = 0 ⟹ B = 0.
  • In the Space Outside the Toroid (r > r₂): Each turn enters and leaves the Amperian surface equally, so I_enc = N I - N I = 0 ⟹ B = 0.

Motion of Charged Particles in Magnetic Fields, Lorentz Force, and Cyclotron

3.1 Lorentz Force Equation

When a particle carrying electric charge q moves with velocity 𝐯 through a region containing both an electric field 𝐄 and a magnetic field 𝐁, the net vector force experienced is the Lorentz Force:

𝐅 = 𝐅_electric + 𝐅_magnetic = q 𝐄 + q (𝐯 × 𝐁)

Fundamental Properties of the Magnetic Force (𝐅_B = q(𝐯 × 𝐁)):

  • Magnitude: F_B = |q| v B sin θ. The force vanishes if the particle is at rest (v = 0), uncharged (q = 0), or moves parallel/antiparallel to 𝐁 (θ = 0° or 180°).
  • Direction: Determined by the right-hand cross product rule for positive charges, and inverted for negative charges (e.g. electrons).
  • Zero Work Done: Since 𝐅_B ⊥ 𝐯 at every instant, the rate of work done is P = 𝐅_B · 𝐯 = q (𝐯 × 𝐁) · 𝐯 = 0. Therefore, a static magnetic field cannot alter the speed or kinetic energy of a charged particle—it acts purely as a deflecting agency!
3.2 Trajectories of Charged Particles in Uniform Magnetic Fields
  1. Perpendicular Entry (θ = 90°): When 𝐯 ⊥ 𝐁, the constant normal force provides centripetal acceleration: q v B = m v² / r.
    Radius: r = (m v) / (q B) = p / (q B) = √(2 m K) / (q B)
    Time Period: T = (2πr) / v = (2πm) / (q B)
    Cyclotron Frequency: ν_c = 1 / T = (q B) / (2πm)
    Vital Exam Insight: Neither the orbital period T nor the cyclotron frequency ν_c depends on the velocity v or kinetic energy of the particle!
  2. Oblique Entry (0° < θ < 90°): The velocity decomposes into v_∥ = v cos θ (parallel to 𝐁) and v_⊥ = v sin θ (perpendicular to 𝐁). The parallel component remains unaffected while the normal component causes circular motion, tracing a helix of radius r = (m v sin θ) / (q B) and linear pitch:
    Pitch = v_∥ · T = (v cos θ) · [(2πm) / (q B)] = (2πm v cos θ) / (q B)
3.3 Velocity Selector (Wien Filter)

When perpendicular electric and magnetic fields are arranged such that 𝐅_E = -𝐅_B, particles with a specific velocity pass through undeflected:

q E = q v B  ⟹  v = E / B
3.4 The Cyclotron: Principle, Operation, and Limitations

Invented by Ernest Lawrence and M. Stanley Livingston (1934), the Cyclotron accelerates positive ions (protons, deuterons, α-particles) to millions of electron-volts (MeV):

  • Principle: Relies on the fact that the period of revolution of an ion inside a magnetic field is independent of its speed. An alternating radiofrequency (RF) electric field reverses polarity exactly when the ion completes a semicircle inside hollow D-shaped metal chambers ("Dees"), repeatedly accelerating the ion across the gap.
  • Resonance Condition: The RF oscillator frequency f_osc must equal the cyclotron frequency: f_osc = ν_c = (q B) / (2πm).
  • Maximum Kinetic Energy: If R is the radius of the Dees:
    v_max = (q B R) / m  ⟹  K_max = (1/2) m v_max² = (q² B² R²) / (2m)
  • Limitations:
    1. Relativistic Mass Increase: At high velocities approaching speed of light c, relativistic mass increases: m = m₀ / √(1 - v²/c²). This causes the revolution period T to increase, throwing ions out of resonance with the fixed-frequency RF field.
    2. Cannot Accelerate Electrons: Electrons have very small rest mass, reaching relativistic speeds after modest acceleration, rapidly losing resonance.
    3. Cannot Accelerate Neutrons: Neutrons carry zero charge (q = 0) and cannot be accelerated by electric or magnetic fields.

Magnetic Force on Conductors and Parallel Current Interactions

4.1 Force on a Current-Carrying Conductor

A current I in a conductor of cross-sectional area A consists of mobile charge carriers (density n, charge q) drifting with average velocity 𝐯_d. The magnetic force on a segment d𝐥 containing n A dl carriers is:

d𝐅 = (n A dl) · [q (𝐯_d × 𝐁)] = (n q A v_d) · (d𝐥 × 𝐁) = I (d𝐥 × 𝐁)

For a straight conductor of uniform length 𝐋 in a uniform magnetic field 𝐁:

𝐅 = I (𝐋 × 𝐁)  ⟹  F = B I L sin θ

The direction is determined by Fleming's Left-Hand Rule: stretch the thumb, forefinger, and middle finger mutually perpendicular. If the Forefinger points along Field 𝐁, and the Middle finger represents Current I, then the Thumb gives the direction of Force (Motion) 𝐅.

4.2 Mutual Force Between Two Long Parallel Current-Carrying Conductors

Consider two infinitely long, thin parallel wires separated by distance d in vacuum, carrying steady currents I₁ and I₂:

  • Wire 1 creates a magnetic field at the position of Wire 2: B₁ = (μ₀ I₁) / (2πd), directed into the page by the right-hand rule.
  • Wire 2 experiences a force on length L due to B₁: F₂₁ = I₂ L B₁ sin 90° = (μ₀ I₁ I₂ L) / (2πd).
  • By Newton's third law, Wire 1 experiences an equal and opposite force: 𝐅₁₂ = -𝐅₂₁.
Force per Unit Length: f = F / L = (μ₀ I₁ I₂) / (2πd)   [N/m]

Nature of the Force:

  • Parallel Currents (Same Direction): Attract each other.
  • Antiparallel Currents (Opposite Directions): Repel each other.
4.3 Classical Definition of the SI Unit of Current: The Ampere
The Ampere (Standard Definition):
One Ampere is that constant current which, if maintained in two straight parallel conductors of infinite length and negligible circular cross-section, placed 1 metre apart in vacuum, produces between these conductors a force equal to exactly 2 × 10⁻⁷ Newtons per metre of length.

Torque on Current Loops, Magnetic Dipole Moment, and Moving Coil Galvanometer

5.1 Torque on a Planar Current Loop in a Uniform Field

Consider a rectangular loop of dimensions a × b carrying current I in a uniform magnetic field 𝐁:

  • Forces on opposite sides of length a are collinear, equal, and opposite, cancelling each other with zero net force and zero net torque.
  • Forces on sides of length b have magnitude F = I b B. While they sum to zero net translational force (∑ 𝐅 = 0), their lines of action are separated by lever arm a sin θ (where θ is the angle between the normal vector 𝐧̂ of the loop and field 𝐁).
Torque: τ = F · (a sin θ) = (I b B) · (a sin θ) = I (a b) B sin θ = I A B sin θ

Defining the Magnetic Dipole Moment as 𝐌 = N I 𝐀 (where N is turn count, A is area, and 𝐀 points along right-hand normal):

𝛕 = 𝐌 × 𝐁  ⟹  τ = M B sin θ   [N·m]

Potential Energy of Dipole in Field:

U = -𝐌 · 𝐁 = -M B cos θ
  • Stable Equilibrium (θ = 0°): U_min = -MB, τ = 0.
  • Unstable Equilibrium (θ = 180°): U_max = +MB, τ = 0.
5.2 Magnetic Dipole Moment of an Atom: The Bohr Magneton

In Niels Bohr's atomic model, an electron (charge -e, mass m_e) revolves in a circular orbit of radius r with speed v:

  • Equivalent electric current: I = e / T = e / (2πr / v) = (e v) / (2πr).
  • Orbital magnetic moment: μ_l = I A = [(e v) / (2πr)] · (πr²) = (e v r) / 2.
  • Relating to orbital angular momentum L = m_e v r:
    𝛍_l = - (e / 2m_e) 𝐋  ⟹  Gyromagnetic Ratio = |μ_l| / L = e / (2m_e) = 8.79 × 10¹⁰ C/kg
  • Applying Bohr's quantization condition L = n ℏ = n (h / 2π) for the ground state (n = 1):
    Bohr Magneton: μ_B = (e ℏ) / (2m_e) = (e h) / (4π m_e) = 9.274 × 10⁻²⁴ A·m² (or J/T)
5.3 Moving Coil Galvanometer (MCG)

The Moving Coil Galvanometer is an electromagnetic instrument designed to detect and measure minuscule electric currents.

Essential Components:

  1. A rectangular coil of fine insulated copper wire wound on an aluminum frame.
  2. Cylindrical concave pole pieces of a permanent magnet that produce a radial magnetic field.
  3. A central soft iron cylinder core that intensifies the field and ensures field lines remain strictly parallel to the coil plane at all rotational angles (θ = 90° always, so sin θ = 1).
  4. A phosphor-bronze suspension strip providing restoring torque with small torsional constant k.

Working Principle: Deflecting torque τ_def = N I A B is counterbalanced by restoring torsional torque τ_res = k θ:

N I A B = k θ  ⟹  θ = (N A B / k) · I = (1 / G_c) · I  ⟹  θ ∝ I (Linear Scale!)
5.4 Galvanometer Sensitivity and Circuit Conversion
Parameter Ammeter Conversion Voltmeter Conversion
Purpose Measures line current I without loading circuit Measures potential difference V without draining current
Connected Element Low resistance (Shunt S) in Parallel with coil High resistance (Multiplier R_s) in Series with coil
Design Formula S = [I_g / (I - I_g)] · G R_s = (V / I_g) - G
Total Instrument Resistance R_A = (G · S) / (G + S) < S R_V = G + R_s > R_s
Ideal Device Resistance R_ideal = 0 Ω R_ideal = ∞ Ω

Magnetism, Terrestrial Magnetism, and Classification of Magnetic Materials

6.1 Gauss's Law for Magnetism

Gauss's Law for Magnetism states that the net magnetic flux Φ_B passing through any closed Gaussian surface is identically zero:

∮_S 𝐁 · d𝐀 = 0

Physical Significance: Magnetic field lines are continuous closed loops with no beginning or end. Unlike electrostatics where isolated positive and negative charges exist, isolated magnetic monopoles do not exist in nature. Every magnet is an inseparable dipole.

6.2 Earth's Magnetic Field (Terrestrial Magnetism)

The geomagnetic field approximates an imaginary magnetic dipole located at Earth's center, inclined at approximately 11.3° to the geographic rotational axis, with the magnetic South pole located near the geographic North pole. The field is fully specified at any location by three Magnetic Elements:

  1. Magnetic Declination (θ or D): The angle between the geographic meridian and the magnetic meridian at a given point on Earth's surface.
  2. Magnetic Inclination or Dip (δ or I): The angle between the direction of the total geomagnetic field 𝐁 and the horizontal plane in the magnetic meridian.
    • At the Magnetic Equator: δ = 0° (field is purely horizontal).
    • At the Magnetic Poles: δ = 90° (field is purely vertical).
  3. Horizontal Component of Earth's Field (B_H):
    B_H = B cos δ
    B_V = B sin δ
    tan δ = B_V / B_H  ⟹  B = √(B_H² + B_V²)
6.3 Magnetization and Magnetic Susceptibility

When a material is subjected to an external magnetizing field 𝐇 (in A/m):

  • Intensity of Magnetization (𝐌): Magnetic dipole moment induced per unit volume: 𝐌 = 𝐦_net / V (A/m).
  • Magnetic Susceptibility (χ_m): Dimensionless ratio measuring material response: χ_m = M / H.
  • Total Magnetic Field (Induction 𝐁): 𝐁 = μ₀ (𝐇 + 𝐌) = μ₀ 𝐇 (1 + χ_m) = μ 𝐇.
  • Relative Magnetic Permeability (μ_r): μ_r = μ / μ₀ = 1 + χ_m.
6.4 Classification: Diamagnetic, Paramagnetic, and Ferromagnetic Substances
Property Diamagnetic Paramagnetic Ferromagnetic
Atomic Origin Paired electrons; zero net atomic dipole moment Unpaired electrons; permanent individual atomic dipoles Quantum exchange interaction forming aligned domains
Response to External B Feebly repelled; moves from stronger to weaker field Feebly attracted; moves from weaker to stronger field Strongly attracted; magnetizes strongly along external field
Susceptibility (χ_m) Small & negative (-10⁻⁵ to -10⁻⁹) Small & positive (+10⁻³ to +10⁻⁵) Very large & positive (10² to 10⁵)
Relative Permeability (μ_r) 0 < μ_r < 1 μ_r slightly > 1 μ_r ≫ 1
Temperature Dependence Independent of temperature Curie's Law: χ = C / T Curie-Weiss Law (above T_c): χ = C / (T - T_c)
Examples Bi, Cu, Ag, Au, Water, Superconductors Al, Na, Pt, Liquid O₂, Cr Fe, Co, Ni, Alnico, Gadolinium
6.5 Magnetic Hysteresis and Material Selection

Plotting magnetic flux density B versus magnetizing field H through a full cycle of AC magnetization reveals a closed Hysteresis Loop:

  • Retentivity (Remanence): The residual magnetic induction B_r remaining in the specimen when magnetizing field H is reduced to zero.
  • Coercivity: The reverse magnetizing field H_c required to completely demagnetize the specimen (reduce B to zero).
  • Hysteresis Loss: The area enclosed by the B-H loop equals the thermal energy dissipated per unit volume per magnetization cycle: W = ∮ B dH.
Crucial WBCHSE Board Material Selection Comparison:
• Electromagnet & Transformer Cores (Soft Iron): Requires high permeability μ_r, high retentivity, very low coercivity, and a narrow hysteresis loop to minimize wasteful heat loss during rapid 50 Hz AC cycles.
• Permanent Magnets (Steel / Alnico): Requires high retentivity and exceptionally high coercivity with a broad loop so that external stray fields, mechanical shocks, and thermal agitation cannot easily demagnetize the magnet.

Key Formulas, Laws & Physical Constants

Biot-Savart Law & Circular Loop
Calculates magnetic field from current elements; specialized for axial and central points of circular coils.
Ampère's Circuital Law & Solenoid
Enclosed line integral relation for steady current configurations with high geometric symmetry.
Lorentz Force & Cyclotron Dynamics
Deflection mechanics of charged particles; radius and kinetic energy in uniform perpendicular magnetic fields.
Conductor Force & Parallel Wires
Laplace force on current conductors and mutual electromagnetic attraction/repulsion between parallel currents.
Torque, Dipole & Galvanometer Conversions
Current loop dipole torque, moving coil galvanometer deflection, and shunt/multiplier conversion formulas.
Geomagnetism & Magnetic Materials
Earth's magnetic elements, magnetic induction relations, and Curie's susceptibility laws.

Conceptual Solved Examples & Case Studies

Example 1
A straight wire carrying a current of 12 A is bent into a semi-circular arc of radius 2.0 cm. What is the magnetic field at the center of the arc?
Step-by-Step Solution:
For a complete circular loop of radius R carrying current I, the magnetic field at the center is B = (μ₀ I) / (2R). Since the wire is bent into a semi-circle, the arc subtends an angle of π radians at the center (exactly half of a complete loop): B_semi = (1/2) · [ (μ₀ I) / (2R) ] = (μ₀ I) / (4R) Given: I = 12 A R = 2.0 cm = 0.02 m μ₀ = 4π × 10⁻⁷ T·m·A⁻¹ Substituting the values: B_semi = [ (4π × 10⁻⁷) × 12 ] / [ 4 × 0.02 ] B_semi = [ 48π × 10⁻⁷ ] / 0.08 = 600π × 10⁻⁷ T = 6π × 10⁻⁵ T ≈ 1.885 × 10⁻⁴ T Direction: According to the Right-Hand Rule, if current flows clockwise, B is directed perpendicularly into the page; if counter-clockwise, perpendicularly outwards.
Example 2
State Ampère's Circuital Law. Use it to derive an expression for the magnetic field inside an infinitely long straight solenoid carrying current I with n turns per unit length.
Step-by-Step Solution:

Statement: Ampère's Circuital Law states that the line integral of magnetic field B around any closed loop in free space equals μ₀ times the net steady current enclosed by the loop: ∮ 𝐁 · d𝐥 = μ₀ I_enclosed

Derivation: Consider an ideal long solenoid with n turns per unit length carrying current I. The magnetic field outside is negligible (B ≈ 0), and the internal field B is uniform and parallel to the axis. Choose a rectangular Amperian loop abcd of length h where:

  • side ab of length h lies inside along the solenoid axis;
  • sides bc and da are perpendicular to the axis;
  • side cd lies completely outside.

Evaluating the line integral along the closed loop: ∮ 𝐁 · d𝐥 = ∫_a^b 𝐁·d𝐥 + ∫_b^c 𝐁·d𝐥 + ∫_c^d 𝐁·d𝐥 + ∫_d^a 𝐁·d𝐥

  1. Along ab: 𝐁 is parallel to d𝐥 (θ = 0°), so ∫_a^b 𝐁·d𝐥 = B · h.
  2. Along bc and da: 𝐁 is perpendicular to d𝐥 (θ = 90°), so ∫ 𝐁·d𝐥 = 0.
  3. Along cd: Outside the solenoid 𝐁 = 0, so ∫_c^d 𝐁·d𝐥 = 0. Hence, ∮ 𝐁 · d𝐥 = B · h.

The number of turns enclosed by loop length h is N = n · h. Therefore, total enclosed current is I_enc = (n · h) · I. Applying Ampère's Circuital Law: B · h = μ₀ · (n · h · I) Dividing both sides by h: B = μ₀ n I This confirms the uniform magnetic field inside an ideal solenoid.

Example 3
A proton, a deuteron, and an alpha particle are accelerated through the same potential difference V and then enter a uniform magnetic field B perpendicular to their velocity. Find the ratio of the radii of their circular paths.
Step-by-Step Solution:

When a particle of charge q and mass m is accelerated through potential difference V, its kinetic energy acquired is: K = q V

The radius of circular motion in a magnetic field B (where 𝐯 ⊥ 𝐁) is: r = (m v) / (q B) = √(2 m K) / (q B) = √(2 m q V) / (q B) = (1 / B) · √(2 V) · √(m / q)

Therefore, for a constant accelerating potential V and magnetic field B: r ∝ √(m / q)

Let mass and charge of proton (p) be m_p = m, q_p = e. For deuteron (d): m_d = 2m, q_d = e. For alpha particle (α): m_α = 4m, q_α = 2e.

Calculating the ratio √(m / q) for each particle:

  1. Proton: r_p ∝ √(m / e)
  2. Deuteron: r_d ∝ √(2m / e) = √2 · √(m / e)
  3. Alpha particle: r_α ∝ √(4m / 2e) = √(2m / e) = √2 · √(m / e)

Taking the ratio: r_p : r_d : r_α = 1 : √2 : √2.

Example 4
Two long parallel horizontal wires separated by 4.0 cm carry currents of 10 A and 25 A in opposite directions. (a) Calculate the magnitude and nature of the force per unit length between them. (b) Find the position of a third parallel wire carrying current I₃ where it experiences zero net magnetic force.
Step-by-Step Solution:
(a) Force per unit length between two parallel current-carrying wires: f = F / L = (μ₀ I₁ I₂) / (2π d) Given: I₁ = 10 A, I₂ = 25 A d = 4.0 cm = 0.04 m μ₀ / (2π) = 2 × 10⁻⁷ T·m·A⁻¹ f = [ (2 × 10⁻⁷) × 10 × 25 ] / 0.04 = [ 500 × 10⁻⁷ ] / 0.04 = 5 × 10⁻⁵ / 0.04 = 1.25 × 10⁻³ N/m Nature: Since currents flow in opposite directions (antiparallel), the force is strictly REPULSIVE. (b) Finding the zero net force point: For a third wire carrying current I₃ to experience zero net force, the magnetic fields produced by Wire 1 and Wire 2 must be equal in magnitude and opposite in direction: B₁ = B₂. Since I₁ (10 A) < I₂ (25 A) and the currents are antiparallel, the neutral point cannot lie between the wires (where both fields point in the same direction). It must lie outside the wires, closer to the weaker current I₁. Let the point be at distance x from Wire 1 (so distance from Wire 2 is d + x = 4 + x cm): B₁ = B₂ (μ₀ I₁) / (2π x) = (μ₀ I₂) / [ 2π (d + x) ] I₁ / x = I₂ / (d + x) 10 / x = 25 / (4 + x) Cross-multiplying: 10 (4 + x) = 25 x 40 + 10x = 25x 15x = 40 ⟹ x = 40 / 15 = 8 / 3 ≈ 2.67 cm Thus, the third wire must be placed 2.67 cm outside Wire 1 (on the side opposite to Wire 2).
Example 5
A galvanometer has a coil of resistance 50 Ω and shows full-scale deflection for a current of 4 mA. How can it be converted into: (a) an ammeter of range 0 to 6 A? (b) a voltmeter of range 0 to 12 V? Calculate the required resistance values and draw circuit diagrams for each.
Step-by-Step Solution:
Given: Galvanometer coil resistance G = 50 Ω Full-scale deflection current I_g = 4 mA = 4 × 10⁻³ A = 0.004 A (a) Conversion into Ammeter (Range I = 6 A): A low resistance shunt S must be connected in parallel with the galvanometer coil so that when total current I enters, only I_g passes through G and the remaining (I - I_g) bypasses through S. Potential difference across G equals potential difference across S: I_g · G = (I - I_g) · S S = (I_g · G) / (I - I_g) S = (0.004 × 50) / (6 - 0.004) = 0.20 / 5.996 ≈ 0.03336 Ω = 33.36 mΩ Result: Connect a shunt resistance of approximately 0.0334 Ω in parallel with the galvanometer. (b) Conversion into Voltmeter (Range V = 12 V): A high resistance R_s must be connected in series with the galvanometer coil so that when total voltage V is applied across the combination, the current does not exceed I_g: V = I_g · (G + R_s) G + R_s = V / I_g R_s = (V / I_g) - G R_s = (12 / 0.004) - 50 = 3000 - 50 = 2950 Ω Result: Connect a series multiplier resistance of 2950 Ω with the galvanometer.
Example 6
Explain the principle and construction of a Moving Coil Galvanometer. Why is a radial magnetic field used, and how is it produced? Define current sensitivity and voltage sensitivity. Explain why increasing current sensitivity may not necessarily increase voltage sensitivity.
Step-by-Step Solution:
  1. Principle: A current-carrying coil placed in an external magnetic field experiences a mechanical deflecting torque τ = N I A B sin θ. If the field is made radial (θ = 90°), τ_def = N I A B. This deflecting torque is balanced by the restoring torque developed in a suspension phosphor-bronze strip τ_res = k θ, where k is the torsional restoring torque per unit twist. At equilibrium: N I A B = k θ ⟹ θ = (N A B / k) · I Since (N A B / k) is constant, θ ∝ I. The angular deflection is directly proportional to current, ensuring a linear scale.

  2. Radial Magnetic Field (Production & Purpose):

  • Production: It is produced by using cylindrical concave magnetic pole pieces and placing a soft iron cylindrical core symmetrically between them.
  • Purpose: A radial magnetic field ensures that the plane of the coil remains parallel to the magnetic field lines in all positions of rotation. Consequently, the angle between the normal to the coil and the field lines is always 90° (sin 90° = 1), keeping the deflecting torque strictly independent of angle θ and making the scale perfectly uniform.
  1. Sensitivity Definitions: (a) Current Sensitivity (I_s): Angular deflection produced per unit electric current: I_s = θ / I = (N A B) / k [rad/A or div/A] To increase current sensitivity, one can increase N, A, B or decrease torsional constant k.

(b) Voltage Sensitivity (V_s): Angular deflection produced per unit potential difference: V_s = θ / V = θ / (I · R) = I_s / R = (N A B) / (k · R) [rad/V or div/V] where R is the resistance of the galvanometer coil.

  1. Why Increasing Current Sensitivity May Not Increase Voltage Sensitivity: Suppose the current sensitivity I_s is doubled by doubling the number of turns from N to 2N (I_s' = 2 I_s). However, doubling the number of turns doubles the total length of the wire, which also doubles the resistance of the coil: R' = 2R. The new voltage sensitivity is: V_s' = (2N A B) / [k · (2R)] = (N A B) / (k R) = V_s Hence, the voltage sensitivity remains completely unchanged! Therefore, increasing current sensitivity by adding turns does not necessarily increase voltage sensitivity because the coil resistance increases proportionately.

Common Misconceptions & Examiner Traps

Common Misconception

Thinking that a static magnetic field does work on a charged particle to increase its kinetic energy.

Scientific Reality & Correction

The magnetic Lorentz force 𝐅 = q(𝐯 × 𝐁) is always strictly perpendicular to the velocity vector 𝐯 (𝐅 · 𝐯 = 0). Thus, instantaneous power P = dW/dt = 0. A static magnetic field changes only the particle's direction of motion, never its speed or kinetic energy.

Common Misconception

Stating that like currents repel and opposite currents attract, by analogy with static charges.

Scientific Reality & Correction

Parallel currents in the same direction ATTRACT each other because their magnetic fields produce inward-directed Lorentz forces. Antiparallel currents in opposite directions REPEL each other.

Common Misconception

Applying B_center = μ₀I / (2R) to find the field of a solenoid instead of B = μ₀nI.

Scientific Reality & Correction

For a single circular coil at its center, B = μ₀I / (2R). For an extended solenoid of n turns per unit length, the superposition of all turns gives B = μ₀nI inside the core.

Common Misconception

Believing that an ammeter requires a high resistance and a voltmeter requires a low resistance.

Scientific Reality & Correction

An ammeter connects in series, so it must have very LOW resistance (shunt S in parallel with coil) to avoid altering branch current. A voltmeter connects in parallel, so it must have very HIGH resistance (multiplier R_s in series) to avoid drawing branch current.

Common Misconception

Assuming magnetic susceptibility of diamagnetic materials depends on temperature according to Curie's law.

Scientific Reality & Correction

Diamagnetism arises from induced orbital electron currents (Lenz's law) and is independent of temperature. Only paramagnetism (χ ∝ 1/T) and ferromagnetism follow temperature-dependent Curie/Curie-Weiss laws.

Master Concept and Formula Architecture Blueprint for Magnetic Effect of Current and Magnetism

MAGNETIC EFFECT OF CURRENT & MAGNETISM: ARCHITECTURE WBCHSE Class 12 Physics • Chapter 3 • Biot-Savart, Ampère, Lorentz, MCG & Earth Magnetism B⃗ Tesla (T) 1. Biot-Savart & Ampère's Circuital Law ▸ Biot-Savart Law: dB = (μ₀/4π) · [I(dl × r̂) / r²] ▸ Circular Loop Center: B = μ₀I / (2R) ▸ Circular Loop Axis: B = μ₀IR² / [2(R² + x²)^(3/2)] ▸ Ampère's Law: ∮ B · dl = μ₀ I_enclosed ★ Long Solenoid: B = μ₀nI | Toroid: B = μ₀nI 2. Lorentz Force, Cyclotron & Conductor Dynamics ▸ Lorentz Force: F = q(E + v × B) | Magnetic: F = q(v × B) ▸ Helical/Circular Radius: r = mv/(qB) | Frequency: ν = qB/(2πm) ▸ Cyclotron Energy: K_max = (q²B²R²) / (2m) ▸ Conductor Force: F = I (L × B) = BIL sin θ ★ Parallel Wire Force: F/L = (μ₀ I₁ I₂) / (2πd) (Ampere Def.) 3. Magnetic Dipole & Moving Coil Galvanometer ▸ Magnetic Dipole: M = NIA | Torque: τ = M × B = MB sin θ ▸ Revolving Electron: Bohr Magneton μ_B = eℏ / (2m_e) ▸ Galvanometer Deflection: θ = (NAB / k) · I (Radial Field) ▸ Ammeter Conversion (Shunt): S = [I_g / (I - I_g)] · G ★ Voltmeter Conversion: R_s = (V / I_g) - G Ideal Ammeter R = 0 | Ideal Voltmeter R = ∞ 4. Terrestrial Magnetism & Magnetic Materials ▸ Earth's Field: B_H = B cos δ, B_V = B sin δ, tan δ = B_V / B_H ▸ Gauss's Law for Magnetism: ∮ B · dA = 0 (No Monopoles) ▸ Susceptibility: χ_m = M / H | Permeability: μ_r = 1 + χ_m ▸ Curie's Law (Para): χ = C / T | Curie-Weiss: χ = C / (T - T_c) ★ Hysteresis (B-H): Retentivity, Coercivity & Loop Energy Loss Soft Iron: High μ, Low Hysteresis | Steel: High Coercivity WBCHSE Class 12 Physics • TargetExams Gold Standard Master Architecture

Chapter Summary & 10 Key Takeaways

Takeaway 1
The Biot-Savart Law expresses the differential magnetic field dB = (μ₀/4π) · [I dl sin θ / r²], where dB is perpendicular to both dl and r.
Takeaway 2
The magnetic field at the center of an N-turn circular loop of radius R is B = (μ₀ N I) / (2R), and along its axis is B = (μ₀ N I R²) / [2(R² + x²)^(3/2)].
Takeaway 3
Ampère's Circuital Law states ∮ B · dl = μ₀ I_enc, enabling rapid derivation of fields for long straight wires (B = μ₀I / 2πr) and solenoids (B = μ₀nI).
Takeaway 4
The magnetic Lorentz force F = q(v × B) does zero mechanical work on a charged particle because force and velocity are always mutually perpendicular.
Takeaway 5
In a uniform magnetic field, a charged particle entering perpendicularly executes uniform circular motion with frequency ν = (qB)/(2πm), independent of speed.
Takeaway 6
The Cyclotron exploits magnetic deflection and radiofrequency electric resonance to accelerate positive ions to high kinetic energies K_max = (q²B²R²)/(2m).
Takeaway 7
Parallel currents flowing in the same direction attract each other with force per unit length f = (μ₀ I₁ I₂)/(2πd), which defines the standard SI Ampere.
Takeaway 8
A planar current loop acts as a magnetic dipole with moment M = N I A, experiencing torque τ = M × B and potential energy U = -M · B in uniform fields.
Takeaway 9
A moving coil galvanometer achieves a linear scale (θ ∝ I) by utilizing a radial magnetic field created by concave pole pieces and a soft iron core.
Takeaway 10
Magnetic materials are classified by susceptibility χ: diamagnetic (χ < 0, temperature-independent), paramagnetic (χ > 0, χ ∝ 1/T), and ferromagnetic (χ ≫ 1, domain structure, B-H hysteresis).

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
An electron and a proton enter a uniform magnetic field with equal kinetic energies at right angles to the field lines. Which particle describes a circular path of larger radius, and why?
Reveal Answer & Explanation
Answer: The proton describes a path of much larger radius. Since r = √(2mK) / (qB), for identical K, q, and B, r ∝ √m. Because the mass of a proton is ~1836 times the mass of an electron, r_p / r_e = √(m_p / m_e) ≈ √1836 ≈ 42.8.
Recall that radius r = √(2mK) / (qB). Compare masses m_p and m_e since charges q and kinetic energies K are equal.
2
Why is the core of a transformer or electromagnet made of soft iron instead of steel?
Reveal Answer & Explanation
Answer: Soft iron has very high magnetic permeability (easily magnetized), high retentivity, very low coercivity, and a narrow hysteresis loop. This ensures that it demagnetizes almost instantaneously when current stops and minimizes thermal energy dissipation during continuous AC cycles.
Consider permeability, retentivity, coercivity, and the area of the B-H hysteresis loop.
3
What is the angle of dip at a location on Earth where the horizontal component of the geomagnetic field equals the vertical component?
Reveal Answer & Explanation
Answer: The angle of dip is 45°. Since tan δ = B_V / B_H and B_V = B_H, tan δ = 1, which gives δ = 45°.
Use tan δ = B_V / B_H.
4
Can a cyclotron be used to accelerate electrons or neutrons? State specific physical reasons for each.
Reveal Answer & Explanation
Answer: No. Electrons have a very small rest mass and reach relativistic speeds rapidly, causing relativistic mass increase m = m₀/√(1-v²/c²) to quickly disrupt resonance with the fixed-frequency oscillator. Neutrons have zero electric charge (q = 0) and cannot experience electric or magnetic forces.
Think about relativistic mass increase for light particles, and the nature of electric/magnetic forces on neutral particles.
5
A closely wound solenoid of 800 turns and area 0.05 m² carries a current of 3.0 A. What is its magnetic dipole moment and the torque experienced when placed at 30° to an external field of 0.25 T?
Reveal Answer & Explanation
Answer: M = 800 × 3.0 × 0.05 = 120 A·m² (or J/T). Torque τ = M B sin 30° = 120 × 0.25 × 0.5 = 15 N·m.
Use M = N I A and τ = M B sin θ.
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