#calculation

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

View Details

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

View Details

Kirchhoff's Laws and Combination of Resistors

Two cells in parallel have emf \( 12 \, \text{V} \) and \( 6 \, \text{V} \) with internal resistances \( 4 \, \Omega \)

**Series combination** R_eq = R₁+R₂+..., same current I through each, voltage divides proportionally V_i = I R_i. Parallel combination 1/R_p = 1/R₁+1/R₂+..., same voltage V across each, current divides inversely, equivalent R_p = (R₁ R₂)/(R₁+R₂) for two resistors. For parallel: εₑq = (ε₁ r₂ + ε₂ r₁/r₁ + r₂) . Substitute: εₑq = (12 × 2 + 6 × 4/4 + 2) = (24 + 24/6) = (48/6) = 8 V . 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 8.0 V,

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

View Details

Temperature Dependence of Resistance and Resistivity

A wire has a resistance of \( 25 \, \Omega \) at \( 20^\circ \text{C} \) and \( 27.5 \, \Omega \) at \( 90^\circ \text{C

**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: 27.5 = 25 [1 + α (90 - 20)] . Solve: 27.5 = 25 + 1750α ⇒ 1750α = 2.5 ⇒ α = (2.5/1750) ≈ 1.43 × 10⁻³ °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

View Details

Torque on Current Loop, Magnetic Moment and Galvanometer

Two parallel wires \( 0.12 \, \text{m} \) apart carry currents of \( 4 \, \text{A} \) and \( 7 \, \text{A} \) in the sam

**Torque on current loop** in magnetic field B is τ = N I A × B, magnitude τ = N I A B sinθ, N turns, I current (A), A area (m²) = l×b for rectangular, θ angle between normal to plane and B. Maximum when plane parallel to B (θ=90°), zero when perpendicular (θ=0°), magnetic moment m = N I A direction along normal via right-hand rule. Force per unit length f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 4 × 7/2 π × 0.12) = (112 × 10⁻⁷/0.24) = 4.67 × 10⁻⁶ N/m . Using F

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Torque on Current Loop, Magnetic Moment and Galvanometer

View Details

Torque on Current Loop, Magnetic Moment and Galvanometer

A solenoid with 1100 turns per meter carries \( 1.8 \, \text{A} \). What is the magnetic field inside? (\( \mu_0 = 4 \pi

**SI unit of magnetic field** is tesla (T), defined as force 1 N on 1 A·m wire perpendicular to field. Moving coil galvanometer uses torque τ = N I A B balanced by spring torque k φ, so deflection φ ∝ I, enabling current measurement, with radial field ensuring τ = N I A B always maximum. B = μ₀ n I . B = 4 π × 10⁻⁷ × 1100 × 1.8 = 7.92 π × 10⁻⁴ ≈ 2.49 × 10⁻³ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Torque on Current Loop, Magnetic Moment and Galvanometer

View Details

Torque on Current Loop, Magnetic Moment and Galvanometer

A solenoid has 1250 turns per meter and carries a current of \( 1.5 \, \text{A} \). What is the magnetic field inside it

**Magnetic moment of loop** m = N I A (A·m²), potential energy U = -m·B = -N I A B cosθ, torque tends to align m with B. For square side 0.18 m, A = 0.0324 m², N=30, I=2 A, B=0.4 T, θ=60°, τ =30×2×0.0324×0.4×sin60° =0.7776×0.866=0.673 N·m, illustrating large torque for modest parameters. Magnetic field B = μ₀ n I . B = 4 π × 10⁻⁷ × 1250 × 1.5 = 7.5 π × 10⁻⁴ ≈ 2.36 × 10⁻³ 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 > Torque on Current Loop, Magnetic Moment and Galvanometer

View Details

Torque on Current Loop, Magnetic Moment and Galvanometer

A solenoid has 600 turns per meter and carries a current of \( 3 \, \text{A} \). What is the magnetic field inside it? (

**Torque on current loop** in magnetic field B is τ = N I A × B, magnitude τ = N I A B sinθ, N turns, I current (A), A area (m²) = l×b for rectangular, θ angle between normal to plane and B. Maximum when plane parallel to B (θ=90°), zero when perpendicular (θ=0°), magnetic moment m = N I A direction along normal via right-hand rule. Magnetic field B = μ₀ n I . B = 4 π × 10⁻⁷ × 600 × 3 = 7.2 π × 10⁻⁴ ≈ 2.26 × 10⁻³ T . Using F = q v B sinθ, F

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Torque on Current Loop, Magnetic Moment and Galvanometer

View Details

Earth's Magnetism and Magnetic Declination

The magnetic potential energy of a dipole with \( m = 0.8 \, \text{A m}^2 \) in a field \( B = 0.25 \, \text{T} \) at \(

**Elements of Earth's field** include declination D, inclination I, horizontal component B_H, total field B = √(B_H² + B_V²). B_H provides compass direction, declination varies with location, important for navigation, inclination 0° at magnetic equator, 90° at poles. U_m = -m B cosθ . Given: m = 0.8 A m² , B = 0.25 T , θ = 180° , cos 180° = -1 . Substitute: U_m = -0.8 × 0.25 × (-1) = 0.2 J . Substituting values gives 0.2 J, 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

View Details

Earth's Magnetism and Magnetic Declination

A magnetic dipole of moment \( 0.2 \, \text{A m}^2 \) is in a uniform field of \( 0.9 \, \text{T} \) at \( 30^\circ \).

**Earth's magnetism** approximated as dipole inclined to rotation axis, magnetic declination is angle between geographic north and magnetic north, inclination or dip angle is angle between total field and horizontal. Horizontal component B_H = B cosδ, vertical B_V = B sinδ, δ dip angle, B_H ≈ 3-4×10⁻⁵ T in India. Torque is tau = m B sinθ . Given: m = 0.2 A m² , B = 0.9 T , θ = 30° , sin 30° = 0.5 . Substitute: tau = 0.2 × 0.9 × 0.5 = 0.09 N m . Substituting values gives 0.09 N m, which matches expected magnitude for this magnetic

Ref: NCERT > Physics Book > Magnetism and Matter > Earth's Magnetism and Magnetic Declination

View Details

Earth's Magnetism and Magnetic Declination

A paramagnetic material with \( \chi = 4 \times 10^{-4} \) in \( H = 2000 \, \text{A m}^{-1} \) has magnetization \( M \

**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. M = chi H . Given: chi = 4 × 10⁻⁴ , H = 2000 A m⁻¹ . M = 4 × 10⁻⁴ × 2000 = 0.8 A m⁻¹ . Substituting values gives 0.8 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 > Earth's Magnetism and Magnetic Declination

View Details

Earth's Magnetism and Magnetic Declination

A dipole with \( m = 0.5 \, \text{A m}^2 \) in a field \( B = 0.6 \, \text{T} \) at \( 180^\circ \) has potential energy

**Earth's magnetism** approximated as dipole inclined to rotation axis, magnetic declination is angle between geographic north and magnetic north, inclination or dip angle is angle between total field and horizontal. Horizontal component B_H = B cosδ, vertical B_V = B sinδ, δ dip angle, B_H ≈ 3-4×10⁻⁵ T in India. U_m = -m B cosθ . Given: m = 0.5 A m² , B = 0.6 T , θ = 180° , cos 180° = -1 . U_m = -0.5 × 0.6 × (-1) = 0.3 J . Substituting values gives 0.3 J, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence

Ref: NCERT > Physics Book > Magnetism and Matter > Earth's Magnetism and Magnetic Declination

View Details