Electric Current, Drift Velocity and Mobility

Electric Current, Drift Velocity and Mobility

A \( 15 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance delivers a current of \( 2 \, \text{A} \) to a

**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. Terminal voltage: V = ε - I r = 15 - 2 × 3 = 9 V . Resistance: R = (V/I) = (9/2) = 4.5 Ω . 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 4.5

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Electric Current, Drift Velocity and Mobility

A conductor has a resistivity of \( 1.0 \times 10^{-7} \, \Omega \text{m} \) and \( \alpha = 4 \times 10^{-3} \, ^\circ\

**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. Use: rho_t = rho₀ [1 + α (T - T₀)] . Substitute: rho_t = 1.0 × 10⁻⁷ [1 + 4 × 10⁻³ (85 - 25)] . Calculate: rho_t = 1.0 × 10⁻⁷ [1 + 0.24] = 1.0 × 10⁻⁷ × 1.24 = 1.24 × 10⁻⁷ Ω m . Applying I = n e A v_d, R = ρ l/A, R_t

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Electric Current, Drift Velocity and Mobility

Why does a semiconductor’s resistance decrease with increasing temperature, unlike a metal?

**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. In semiconductors, as temperature rises, more electrons are thermally excited from the valence band to the conduction band, increasing the number of charge carriers ( n ). Since rho = m / (n e² tau) , a significant increase in n outweighs the decrease in tau , reducing rho and thus resistance. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0,

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Electric Current, Drift Velocity and Mobility

In a conductor, if the number of free electrons per unit volume doubles while the electric field and relaxation time rem

**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. Current density j = n e v_d , where v_d = e E tau / m . If n doubles and E , tau , and m remain constant, v_d is unchanged, so j' = 2n e v_d = 2j . Thus, current density doubles. 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

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Electric Current, Drift Velocity and Mobility

A copper wire of cross-sectional area \( 5 \times 10^{-7} \, \text{m}^2 \) carries a current of \( 1.7 \, \text{A} \). I

**Drift velocity** v_d = I/(n e A), I current (A), n number density of conduction electrons (m⁻³) ≈8.5×10²⁸ m⁻³ for copper, e =1.6×10⁻¹⁹ C, A cross-sectional area (m²). Typical v_d ≈10⁻⁴ m/s for 1 A in mm² wire, slow despite fast signal propagation due to electric field establishment. Drift speed: v_d = (I/n e A) . Substitute: v_d = (1.7/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 5 × 10⁻⁷) . Calculate: v_d = (1.7/6.8 × 10³) ≈ 2.5 × 10⁻⁴ m/s . 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

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Electric Current, Drift Velocity and Mobility

Under what condition does Ohm’s law fail to hold true for a conductor?

**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. Ohm’s law ( V = I R ) assumes a linear relationship between voltage and current. It fails if the resistance ( R ) changes with current or voltage (e.g., in non-ohmic devices like diodes), making the V -versus- I graph non-linear. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq

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Electric Current, Drift Velocity and Mobility

A wire of length \( 7 \, \text{m} \) and resistance \( 14 \, \Omega \) is stretched to \( 14 \, \text{m} \). What is the

**Drift velocity** v_d = I/(n e A), I current (A), n number density of conduction electrons (m⁻³) ≈8.5×10²⁸ m⁻³ for copper, e =1.6×10⁻¹⁹ C, A cross-sectional area (m²). Typical v_d ≈10⁻⁴ m/s for 1 A in mm² wire, slow despite fast signal propagation due to electric field establishment. Volume constant: l A = l' A' ⇒ A' = (A/2) . New resistance: R' = (rho l'/A') = (rho (2l)/(A/2)) = 4 (rho l/A) = 4R = 4 × 14 = 56 Ω . 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

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Electric Current, Drift Velocity and Mobility

What is the significance of the relaxation time in the context of electron drift in a conductor?

**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. Relaxation time ( tau ) is the average time between electron collisions with lattice ions. It determines drift velocity ( v_d = e E tau / m ), affecting how quickly electrons respond to the field and thus the conductor’s conductivity. 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,

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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,

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Electric Current, Drift Velocity and Mobility

What is the primary source of energy dissipation in a resistor carrying current?

**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. Energy dissipation in a resistor occurs as electrons collide with lattice ions, transferring kinetic energy gained from the electric field into thermal energy (heat) via lattice vibrations. 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 Collisions with

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Electric Current, Drift Velocity and Mobility

Why does the emf of a cell depend on its chemical composition rather than its size?

**Drift velocity** v_d = I/(n e A), I current (A), n number density of conduction electrons (m⁻³) ≈8.5×10²⁸ m⁻³ for copper, e =1.6×10⁻¹⁹ C, A cross-sectional area (m²). Typical v_d ≈10⁻⁴ m/s for 1 A in mm² wire, slow despite fast signal propagation due to electric field establishment. Emf is the potential difference generated by chemical reactions between electrodes and electrolyte, determined by the materials’ electrochemical properties, not physical dimensions like size. 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 Chemical reactions

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Electric Current, Drift Velocity and Mobility

A \( 6 \, \Omega \) resistor carries a current of \( 4 \, \text{A} \) for \( 5 \, \text{s} \). What is the energy dissip

**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. Energy: W = I² R t . Substitute: W = 4² × 6 × 5 = 16 × 30 = 480 J . 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 480 J,

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