#electric current

Displacement Current and Ampere-Maxwell Law

In a charging capacitor, if the conduction current in the wires is \( 3 \, \text{A} \), what is the displacement current

**Displacement current** I_d = ε₀ dΦ_E/dt, Φ_E = ∫ E·dA electric flux (V·m), ε₀=8.85×10⁻¹² F/m, ensures continuity of current in charging capacitor where conduction current stops between plates, I_d equals conduction current in wires, 3 A conduction ⇒ 3 A displacement, maintaining Ampere's law ∮ B·dl = μ₀(I_c+I_d). The document states that inside a charging capacitor, the displacement current equals the conduction current in the wires. Thus, i_d = 3 A . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 3 A, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Displacement Current and Ampere-Maxwell Law

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Potentiometer, Conductivity and Special Cases

Why does the current density in a conductor remain uniform across its cross-section under steady-state conditions?

**Potential difference drop** across resistor when current flows because charges lose potential energy qV = I² R t as heat, field E = V/l drives drift, maintaining current. At very high E, velocity saturation or heating changes τ, causing non-ohmic behaviour. Current density ( j = I / A ) is uniform if the current distributes evenly. In steady state, charge conservation (Kirchhoff’s junction rule) ensures a constant current through a uniform conductor, making j consistent across the area. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and

Ref: NCERT > Physics Book > Current Electricity > Potentiometer, Conductivity and Special Cases

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Kirchhoff's Laws and Combination of Resistors

A \( 30 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, ea

**Kirchhoff's loop rule** (energy conservation) states algebraic sum of potential differences around closed loop zero, Σ ε - Σ I R =0, ensures total voltage drop across series resistors equals source emf. Junction rule (charge conservation) Σ I_in = Σ I_out ensures current continuity. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 5 = (25/6) ≈ 4.17 Ω . Total current: I = (V/Rₑq) = (30/(25/6)) = 30 × (6/25) = 7.2 A . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P = I²R, evaluation

Ref: NCERT > Physics Book > Current Electricity > Kirchhoff's Laws and Combination of Resistors

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Wheatstone Bridge and Meter Bridge

In a Wheatstone bridge with \( R_1 = 30 \, \Omega \), \( R_2 = 60 \, \Omega \), \( R_3 = 15 \, \Omega \), and \( R_4 = 3

**Wheatstone bridge balance** condition R₁/R₂ = R₃/R₄, R₄ = R₂ R₃/R₁, when galvanometer current zero, potentials at midpoints equal. At balance, no current through galvanometer, enabling precise resistance measurement independent of source voltage. Apply Kirchhoff’s rules. Let currents be I₁ (AB), I₂ (AD), I_g (BD). Junction B: I₁ = I_g + I₄ , Junction D: I₂ = I_g + I₃ . Loop BADB: 30 I₁ + 10 I_g - 60 I₂ = 0 ⇒ 3 I₁ + I_g - 6 I₂ = 0 . Loop BCDB: 60 (I₁ - I_g) - 10 I_g - 15 (I₂ + I_g) = 0 ⇒ 4 I₁ - 2

Ref: NCERT > Physics Book > Current Electricity > Wheatstone Bridge and Meter Bridge

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Electric Current, Drift Velocity and Mobility

In a circuit with a battery, why does the current establish almost instantaneously when the circuit is closed, despite t

**Current and drift relation** I = n e A v_d shows current proportional to drift velocity and area. For A=6×10⁻⁷ m², I=1.8 A, n=8.5×10²⁸ m⁻³, v_d =1.8/(8.5×10²⁸×1.6×10⁻¹⁹×6×10⁻⁷)=2.2×10⁻⁴ m/s, illustrating small drift speed even for ampere currents. The electric field propagates through the conductor at near-light speed, causing all free electrons to start moving simultaneously. Drift velocity is slow, but the field’s rapid establishment initiates current instantly. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P = I²R, evaluation yields Electric field propagates quickly,

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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Electric Current, Drift Velocity and Mobility

What happens to the current density in a conductor if the conductor is stretched to twice its original length without ch

**Mobility** μ = v_d/E = e τ/m, τ relaxation time (s), measures ease of electron drift under field E (V/m). Conductivity σ = n e μ = 1/ρ, linking microscopic τ to macroscopic resistivity, explaining why metals conduct well due to large n and τ. Stretching doubles length ( l' = 2l ) and halves area ( A' = A/2 ) due to volume conservation. Resistance becomes R' = rho (2l) / (A/2) = 4R . Current I' = V / (4R) = I/4 . Current density j = I / A , so j' = I' / A' = (I/4) / (A/2) = I

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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Resistance, Resistivity and Ohm's Law

In a circuit with two resistors of different resistances in series, which resistor dissipates more power if the same cur

**Resistance** R = ρ l/A, ρ resistivity (Ω·m), l length (m), A area (m²), ρ = m/(n e² τ) from Drude model, τ average collision time. Ohm's law V = I R holds when ρ constant, independent of V. Volume constant stretching l→2l implies A→A/2, so R' = ρ·2l/(A/2)=4R, resistance quadruples when length doubles at constant volume. Power P = I² R . With the same current ( I ) in series, power is proportional to resistance ( R ). The resistor with higher resistance dissipates more power. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0,

Ref: NCERT > Physics Book > Current Electricity > Resistance, Resistivity and Ohm's Law

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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 Field Due to Current - Straight Wire and Circular Loop

A circular coil of radius \( 0.05 \, \text{m} \) with 30 turns carries \( 2.5 \, \text{A} \). What is the magnetic field

**Biot-Savart law** dB = μ₀/4π·I dl × r̂/r² underlies both straight wire and loop formulas. For square loop side a, area A = a², moment m = N I A, field pattern similar to dipole at large distances, with superposition for N turns. B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 30 × 2.5/2 × 0.05) = (30 π × 10⁻⁶/0.1) = 3 π × 10⁻⁴ ≈ 9.42 × 10⁻⁴ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Field Due to Current - Straight Wire and Circular Loop

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Solenoid, Toroid and Ampere's Law

A solenoid with 2000 turns per meter carries \( 0.8 \, \text{A} \). What is the magnetic field inside? (\( \mu_0 = 4 \pi

**Ampere's circuital law** ∮ B·dl = μ₀ I_enc explains solenoid field, choosing rectangular Amperian loop with one side inside, one outside where B≈0, giving B L = μ₀ n L I ⇒ B = μ₀ n I. Toroid B = μ₀ N I/(2π r) inside, zero outside, field confined. B = μ₀ n I . B = 4 π × 10⁻⁷ × 2000 × 0.8 = 6.4 π × 10⁻⁴ ≈ 2.01 × 10⁻³ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N I A

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Solenoid, Toroid and Ampere's Law

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A circular coil of radius \( 0.1 \, \text{m} \) with 20 turns carries \( 4 \, \text{A} \). What is the magnetic field at

**Magnetic field due to long straight wire** at distance r is B = μ₀ I/(2π r), μ₀ = 4π×10⁻⁷ T·m/A, direction circular around wire given by right-hand grip rule. For I = 18 A, r = 0.15 m, B = 2×10⁻⁷×18/0.15 = 2.4×10⁻⁵ T, illustrating 1/r dependence. B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 20 × 4/2 × 0.1) = (32 π × 10⁻⁶/0.2) = 16 π × 10⁻⁵ ≈ 5.03 × 10⁻⁴ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R)

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Field Due to Current - Straight Wire and Circular Loop

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 800 turns per meter carries a current of \( 2.5 \, \text{A} \). What is the magnetic intensity \( H \) i

**Magnetic field inside solenoid** with core B = μ₀ μ_r n I is uniform, direction along axis given by right-hand grip rule. For n = 2000 m⁻¹, μ_r = 400, B = 1.2 T, I = 1.2/(4π×10⁻⁷×400×2000) ≈ 1.19 A, showing modest current produces tesla-level field with high μ_r core. Magnetic intensity H = n I . Given: n = 800 m⁻¹ , I = 2.5 A . Substitute: H = 800 × 2.5 = 2000 A m⁻¹ . Substituting values gives 2000 A m⁻¹, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

Ref: NCERT > Physics Book > Magnetism and Matter > Solenoid with Magnetic Core and Magnetic Properties

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