#current density

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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Wheatstone Bridge and Meter Bridge

A copper wire carries \( 4 \, \text{A} \) with a drift speed of \( 1.0 \times 10^{-4} \, \text{m/s} \). If \( n = 8.5 \t

**Meter bridge** uses uniform wire of length 1 m, balance length l gives R_unknown = R_known·l/(100-l). Principle same as Wheatstone, with wire resistances proportional to lengths, allowing unknown resistance determination from length ratio. Drift speed: v_d = (I/n e A) . Rearrange: A = (I/n e v_d) . Substitute: A = (4/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.0 × 10⁻⁴) . Calculate: A = (4/1.36 × 10⁵) ≈ 2.94 × 10⁻⁵ m² . 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 > Wheatstone Bridge and Meter Bridge

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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

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

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 \( 5 \times 10^{-7} \, \text{m}^2 \) 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

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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Electric Current, Drift Velocity and Mobility

A copper wire carries \( 3.4 \, \text{A} \) with a drift speed of \( 1.25 \times 10^{-4} \, \text{m/s} \). If \( n = 8.5

**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. Drift speed: v_d = (I/n e A) . Rearrange: A = (I/n e v_d) . Substitute: A = (3.4/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.25 × 10⁻⁴) . Calculate: A = (3.4/1.7 × 10⁵) = 2.0 × 10⁻⁵ m² . 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

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

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EMF, Internal Resistance and Cells Combination

Why does a conductor’s current density increase when its length is halved while keeping the potential difference constan

**Internal resistance** causes voltage drop I r inside battery, so V = ε - I r decreases with I. For 16 V battery, r=2 Ω, I=2 A, V=16-4=12 V, external R = V/I =6 Ω. Measurement of V and I yields r = (ε - V)/I. Resistance R = rho l / A . Halving length ( l' = l/2 ) halves R ( R' = R/2 ). Current I = V / R , so I' = V / (R/2) = 2I . Current density j = I / A , so j' = 2I / A = 2j . Applying I = n

Ref: NCERT > Physics Book > Current Electricity > EMF, Internal Resistance and Cells Combination

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