#copper conductivity

Wheatstone Bridge and Meter Bridge

A copper wire of length \( 2 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) carries a curr

**Unbalanced bridge** has potential difference between galvanometer nodes, current direction determined by which node higher potential, i.e., if R₁/R₂ > R₃/R₄, left node higher, current flows one way, else opposite. Galvanometer deflection indicates imbalance magnitude. Drift speed is given by v_d = (I/n e A) . Given: I = 2 A , n = 8.5 × 10²⁸ m⁻³ , e = 1.6 × 10⁻¹⁹ C , A = 2 × 10⁻⁶ m² . Substitute: v_d = (2/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 2 × 10⁻⁶) . Calculate: v_d = (2/2.72 × 10⁴) = 7.35 × 10⁻⁵ m/s . Applying I = n e

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Temperature Dependence of Resistance and Resistivity

A copper wire of cross-sectional area \( 3 \times 10^{-7} \, \text{m}^2 \) carries a current of \( 0.9 \, \text{A} \). I

**Conductivity** σ=1/ρ decreases with temperature for metals, σ = n e² τ/m, τ ∝1/T due to lattice vibrations. For semiconductors, n increases exponentially with T, so σ increases, opposite to metals, explaining why metallic resistance rises with temperature. Drift speed: v_d = (I/n e A) . Substitute: v_d = (0.9/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 3 × 10⁻⁷) . Calculate: v_d = (0.9/4.08 × 10³) ≈ 2.21 × 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 - I r and P = I²R, evaluation yields 2.21

Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity

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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

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

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Electric Current, Drift Velocity and Mobility

A copper wire of cross-sectional area \( 4 \times 10^{-7} \, \text{m}^2 \) carries a current of \( 1.36 \, \text{A} \).

**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 = (1.36/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 4 × 10⁻⁷) . Calculate: v_d = (1.36/5.44 × 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

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

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Electric Current, Drift Velocity and Mobility

A copper wire of cross-sectional area \( 4 \times 10^{-7} \, \text{m}^2 \) carries a current of \( 1.2 \, \text{A} \). I

**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. Drift speed: v_d = (I/n e A) . Substitute: v_d = (1.2/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 4 × 10⁻⁷) . Calculate: v_d = (1.2/5.44 × 10³) ≈ 2.21 × 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 - I r

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

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