#Ampere's law

Production of EM Waves and Hertz Experiment

What is the role of displacement current in Maxwell's generalization of Ampere's law?

**Hertz experiment** used induction coil connected to two rods with gap, spark produced oscillating charge, emitted EM wave, received by loop with gap sparking when E induced, measured wavelength by standing wave, demonstrated EM wave properties, validating Maxwell. Maxwell introduced displacement current to account for the magnetic field produced by a time-varying electric field, ensuring consistency in Ampere's circuital law when applied to capacitors. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields It produces a magnetic field due to a chang

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Electromagnetic Spectrum - Radio Waves and Microwaves

What role does the displacement current play in a region where there is no conduction current?

**Hertz experiment** produced radio waves using spark gap LC oscillator, detected with loop antenna, confirming Maxwell's prediction, frequency ≈10⁸ Hz, wavelength ≈3 m, demonstrating EM waves travel at c, transverse, can be reflected, refracted, polarized. The document explains that in regions with no conduction current but a time-varying electric field, displacement current acts as a source of magnetic field, enabling electromagnetic wave propagation. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields It acts as a source of magnetic field, illus

Ref: NCERT > Physics Book > Electromagnetic Waves > Electromagnetic Spectrum - Radio Waves and Microwaves

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EM Wave Characteristics - Transverse Nature and E/B Ratio

What does the introduction of displacement current by Maxwell explain about electromagnetic phenomena?

**EM wave in vacuum** transverse, E and B perpendicular to propagation and to each other, E×B along propagation, in phase, E/B = c =3×10⁸ m/s, c =1/√(μ₀ ε₀), μ₀=4π×10⁻⁷ H/m, ε₀=8.85×10⁻¹² F/m. For E₀=45 V/m, B₀=E₀/c=45/3×10⁸=1.5×10⁻⁷ T=150 nT, illustrating B much smaller than E. The document explains that displacement current explains the generation of magnetic fields by time-varying electric fields, crucial for the propagation of electromagnetic waves and consistency in Ampere's law. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields Generation o

Ref: NCERT > Physics Book > Electromagnetic Waves > EM Wave Characteristics - Transverse Nature and E/B Ratio

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Torque on Current Loop, Magnetic Moment and Galvanometer

Two parallel wires \( 0.08 \, \text{m} \) apart carry currents of \( 5 \, \text{A} \) and \( 3 \, \text{A} \) in the sam

**Magnetic moment of loop** m = N I A (A·m²), potential energy U = -m·B = -N I A B cosθ, torque tends to align m with B. For square side 0.18 m, A = 0.0324 m², N=30, I=2 A, B=0.4 T, θ=60°, τ =30×2×0.0324×0.4×sin60° =0.7776×0.866=0.673 N·m, illustrating large torque for modest parameters. Force per unit length f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 5 × 3/2 π × 0.08) = (60 × 10⁻⁷/0.16) = 3.75 × 10⁻⁶ N/m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Torque on Current Loop, Magnetic Moment and Galvanometer

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Torque on Current Loop, Magnetic Moment and Galvanometer

A long wire carries \( 15 \, \text{A} \). At what distance is the magnetic field \( 3 \times 10^{-6} \, \text{T} \)? (\(

**Magnetic moment of loop** m = N I A (A·m²), potential energy U = -m·B = -N I A B cosθ, torque tends to align m with B. For square side 0.18 m, A = 0.0324 m², N=30, I=2 A, B=0.4 T, θ=60°, τ =30×2×0.0324×0.4×sin60° =0.7776×0.866=0.673 N·m, illustrating large torque for modest parameters. B = (μ₀ I/2 π r) , so r = (μ₀ I/2 π B) . r = (4 π × 10⁻⁷ × 15/2 π × 3 × 10⁻⁶) = (60 × 10⁻⁷/6 × 10⁻⁶) = 1 m . Using F = q v B sinθ, F = I l B sinθ, B

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Torque on Current Loop, Magnetic Moment and Galvanometer

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Torque on Current Loop, Magnetic Moment and Galvanometer

A long wire carries \( 14 \, \text{A} \). At what distance is the magnetic field \( 7 \times 10^{-6} \, \text{T} \)? (\(

**SI unit of magnetic field** is tesla (T), defined as force 1 N on 1 A·m wire perpendicular to field. Moving coil galvanometer uses torque τ = N I A B balanced by spring torque k φ, so deflection φ ∝ I, enabling current measurement, with radial field ensuring τ = N I A B always maximum. B = (μ₀ I/2 π r) , so r = (μ₀ I/2 π B) . r = (4 π × 10⁻⁷ × 14/2 π × 7 × 10⁻⁶) = (56 × 10⁻⁷/14 × 10⁻⁶) = 0.4 m . Using F = q v B sinθ, F = I

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Torque on Current Loop, Magnetic Moment and Galvanometer

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Force on Current-Carrying Conductor and Between Parallel Wires

Two parallel wires \( 0.11 \, \text{m} \) apart carry currents of \( 6 \, \text{A} \) and \( 2 \, \text{A} \) in the sam

**Force between parallel wires** per unit length is f = μ₀ I₁ I₂/(2π d), μ₀/2π = 2×10⁻⁷ T·m/A, d separation (m), attractive if currents same direction, repulsive if opposite. For I₁=5 A, I₂=7 A, d=0.04 m, f = 2×10⁻⁷×35/0.04 = 1.75×10⁻⁴ N/m, sign indicates repulsion for opposite directions. Force per unit length f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 6 × 2/2 π × 0.11) = (48 × 10⁻⁷/0.22) = 2.18 × 10⁻⁶ N/m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N

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Force on Current-Carrying Conductor and Between Parallel Wires

Two parallel wires \( 0.07 \, \text{m} \) apart carry currents of \( 9 \, \text{A} \) and \( 2 \, \text{A} \) in the sam

**Force between parallel wires** per unit length is f = μ₀ I₁ I₂/(2π d), μ₀/2π = 2×10⁻⁷ T·m/A, d separation (m), attractive if currents same direction, repulsive if opposite. For I₁=5 A, I₂=7 A, d=0.04 m, f = 2×10⁻⁷×35/0.04 = 1.75×10⁻⁴ N/m, sign indicates repulsion for opposite directions. Force per unit length f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 9 × 2/2 π × 0.07) = (72 × 10⁻⁷/0.14) = 5.14 × 10⁻⁶ N/m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Force on Current-Carrying Conductor and Between Parallel Wires

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A long wire carries \( 18 \, \text{A} \). At what distance is the magnetic field \( 6 \times 10^{-6} \, \text{T} \)? (\(

**Motion of charged particle perpendicular to uniform B** is circular because force F ⊥ v, no work done, speed constant, radius r = m v/(q B), period T = 2π m/(q B) independent of v. If v has component parallel to B, helical path results, pitch = v_parallel·T. B = (μ₀ I/2 π r) , so r = (μ₀ I/2 π B) . r = (4 π × 10⁻⁷ × 18/2 π × 6 × 10⁻⁶) = (72 × 10⁻⁷/12 × 10⁻⁶) = 0.6 m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A long wire carries \( 20 \, \text{A} \). At what distance from the wire is the magnetic field \( 4 \times 10^{-6} \, \t

**Effect of doubling velocity** on magnetic force F = q v B sinθ is linear increase, F doubles for same θ and B. Electron with charge 1.6×10⁻¹⁹ C, v = 4.5×10⁶ m/s, B = 0.35 T, θ = 90°, F = 1.6×10⁻¹⁹×4.5×10⁶×0.35 = 2.52×10⁻¹³ N, illustrating magnitude for typical lab values. B = (μ₀ I/2 π r) , so r = (μ₀ I/2 π B) . r = (4 π × 10⁻⁷ × 20/2 π × 4 × 10⁻⁶) = (8 × 10⁻⁶/8 × 10⁻⁶) = 1 m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Magnetic Force on Moving Charge - Lorentz Force and Motion

Why do parallel currents attract each other?

**Effect of doubling velocity** on magnetic force F = q v B sinθ is linear increase, F doubles for same θ and B. Electron with charge 1.6×10⁻¹⁹ C, v = 4.5×10⁶ m/s, B = 0.35 T, θ = 90°, F = 1.6×10⁻¹⁹×4.5×10⁶×0.35 = 2.52×10⁻¹³ N, illustrating magnitude for typical lab values. Parallel currents produce magnetic fields that interact such that the force on one wire due to the field of the other is directed toward it, resulting in attraction. Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ =

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Force on Current-Carrying Conductor and Between Parallel Wires

Two parallel wires \( 0.05 \, \text{m} \) apart carry currents of \( 8 \, \text{A} \) and \( 3 \, \text{A} \) in the sam

**Force on current-carrying wire** in magnetic field is F = I l × B, magnitude F = I l B sinθ, I current (A), l length (m), B field (T), θ angle between current direction and B. Direction perpendicular to plane containing wire and B, given by Fleming's left-hand rule. Force per unit length f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 8 × 3/2 π × 0.05) = (96 × 10⁻⁷/0.1) = 9.6 × 10⁻⁶ N/m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B =

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Force on Current-Carrying Conductor and Between Parallel Wires

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