#material properties

Wheatstone Bridge and Meter Bridge

A conductor of length \( 1 \, \text{m} \) and cross-sectional area \( 1 \times 10^{-6} \, \text{m}^2 \) has a resistance

**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. Resistance is given by R = (rho l/A) . Rearrange: rho = (R A/l) . Given: R = 2 Ω , A = 1 × 10⁻⁶ m² , l = 1 m . Substitute: rho = (2 × 1 × 10⁻⁶/1) = 2 × 10⁻⁶ Ω m . 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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Temperature Dependence of Resistance and Resistivity

A conductor has a resistivity of \( 7 \times 10^{-8} \, \Omega \text{m} \) and \( \alpha = 4 \times 10^{-3} \, ^\circ\te

**Temperature dependence** of resistance R_t = R₀[1+α(T-T₀)], α temperature coefficient (per °C), R₀ resistance at T₀ (Ω). For metals α positive ≈10⁻³ /°C, resistance increases with temperature because τ decreases due to increased phonon scattering, n nearly constant. Use: rho_t = rho₀ [1 + α (T - T₀)] . Substitute: rho_t = 7 × 10⁻⁸ [1 + 4 × 10⁻³ (85 - 25)] . Calculate: rho_t = 7 × 10⁻⁸ [1 + 0.24] = 7 × 10⁻⁸ × 1.24 = 8.68 × 10⁻⁸ Ω m . 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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Temperature Dependence of Resistance and Resistivity

A platinum wire has a resistance of \( 6 \, \Omega \) at \( 0^\circ \text{C} \) and \( 6.48 \, \Omega \) at \( 100^\circ

**Temperature dependence** of resistance R_t = R₀[1+α(T-T₀)], α temperature coefficient (per °C), R₀ resistance at T₀ (Ω). For metals α positive ≈10⁻³ /°C, resistance increases with temperature because τ decreases due to increased phonon scattering, n nearly constant. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: 6.48 = 6 [1 + α (100 - 0)] . Solve: 6.48 = 6 + 600α ⇒ 600α = 0.48 ⇒ α = (0.48/600) = 8 × 10⁻⁴ °C⁻¹ . 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 > Temperature Dependence of Resistance and Resistivity

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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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Electric Current, Drift Velocity and Mobility

A conductor has a resistivity of \( 6 \times 10^{-8} \, \Omega \text{m} \) and \( \alpha = 4 \times 10^{-3} \, ^\circ\te

**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 = 6 × 10⁻⁸ [1 + 4 × 10⁻³ (90 - 20)] . Calculate: rho_t = 6 × 10⁻⁸ [1 + 0.28] = 6 × 10⁻⁸ × 1.28 = 7.68 × 10⁻⁸ Ω m . Applying I = n e A v_d, R = ρ l/A, R_t

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

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Resistance, Resistivity and Ohm's Law

A wire of length \( 3 \, \text{m} \) and resistance \( 6 \, \Omega \) is stretched to \( 6 \, \text{m} \). What is the n

**Ohm's law deviation** at high fields occurs when τ or n vary with E, resistivity ρ = m/(n e² τ) changes, non-ohmic behaviour seen in semiconductors, electrolytes. At moderate fields, linear V-I holds, slope = R. 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 × 6 = 24 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 24 Ω, consistent

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

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Resistance, Resistivity and Ohm's Law

A conductor has a resistivity of \( 1.2 \times 10^{-7} \, \Omega \text{m} \) and \( \alpha = 4 \times 10^{-3} \, ^\circ\

**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. Use: rho_t = rho₀ [1 + α (T - T₀)] . Substitute: rho_t = 1.2 × 10⁻⁷ [1 + 4 × 10⁻³ (80 - 20)] . Calculate: rho_t = 1.2 × 10⁻⁷ [1 + 0.24] = 1.2 × 10⁻⁷ × 1.24 = 1.488 × 10⁻⁷ Ω m .

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

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Resistance, Resistivity and Ohm's Law

A wire of length \( 3 \, \text{m} \) and cross-sectional area \( 3 \times 10^{-6} \, \text{m}^2 \) has a resistance of \

**Resistivity** depends on material and temperature, not geometry. For metallic conductor, ρ ≈10⁻⁸ Ω·m for copper. Given ρ=4×10⁻⁸ Ω·m, l=2 m, A=π r², R calculation uses R=ρ l/A. Stretching wire conserves volume V = l A = l' A', so A' = A l/l', new R' = ρ l'²/V. Resistance: R = (rho l/A) . Rearrange: rho = (R A/l) . Given: R = 5 Ω , A = 3 × 10⁻⁶ m² , l = 3 m . Substitute: rho = (5 × 3 × 10⁻⁶/3) = 5 × 10⁻⁶ Ω m . Applying I = n e A v_d, R = ρ l/A,

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

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Resistance, Resistivity and Ohm's Law

A wire has a resistance of \( 30 \, \Omega \) at \( 25^\circ \text{C} \) and \( 31.5 \, \Omega \) at \( 75^\circ \text{C

**Resistivity** depends on material and temperature, not geometry. For metallic conductor, ρ ≈10⁻⁸ Ω·m for copper. Given ρ=4×10⁻⁸ Ω·m, l=2 m, A=π r², R calculation uses R=ρ l/A. Stretching wire conserves volume V = l A = l' A', so A' = A l/l', new R' = ρ l'²/V. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: 31.5 = 30 [1 + α (75 - 25)] . Solve: 31.5 = 30 + 1500α ⇒ 1500α = 1.5 ⇒ α = (1.5/1500) = 1.0 × 10⁻³ °C⁻¹ . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT],

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

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Electric Current, Drift Velocity and Mobility

What causes the resistivity of a conductor to deviate from Ohm’s law at very high electric fields?

**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. At high fields, resistivity ( rho = m / (n e² tau) ) may change as tau or n varies (e.g., due to electron saturation or heating), making the I -versus- V relationship non-linear and violating Ohm’s law. 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

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

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Resistance, Resistivity and Ohm's Law

A nichrome wire has a resistance of \( 60 \, \Omega \) at \( 30^\circ \text{C} \) and \( \alpha = 1.7 \times 10^{-4} \,

**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. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: R_t = 60 [1 + 1.7 × 10⁻⁴ (330 - 30)] . Calculate: R_t = 60 [1 + 1.7 × 10⁻⁴ × 300] = 60 [1 + 0.051] = 60 × 1.051 = 63.06 Ω .

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

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Earth's Magnetism and Magnetic Declination

A material with susceptibility \( \chi = 5 \times 10^{-3} \) has a relative permeability \( \mu_r \) of:

**Geomagnetic field** arises from outer core dynamo, field lines emerge near geographic south pole. Understanding D and I allows conversion between geographic and magnetic coordinates, with B_H = B cos(inclination) used in experiments with tangent galvanometer. μ_r = 1 + chi . Given: chi = 5 × 10⁻³ . Substitute: μ_r = 1 + 5 × 10⁻³ = 1.005 . Substituting values gives 1.005, 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 > Earth's Magnetism and Magnetic Declination

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