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15 public questions tagged with this topic.

Electric Current, Drift Velocity and Mobility

A copper wire of cross-sectional area \( 8 \times 10^{-7} \, \text{m}^2 \) carries a current of \( 2 \, \text{A} \). If

**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 = (2/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 8 × 10⁻⁷) . Calculate: v_d = (2/1.088 × 10⁴) ≈ 1.84 × 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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Torque on Current Loop, Magnetic Moment and Galvanometer

An electron moves at \( 7 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.2 \, \text{T} \). What is the ma

**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. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 7 × 10⁶ × 0.2 = 2.24 × 10⁻¹³ N . Using F = q v B sinθ, F = I l B

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

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Torque on Current Loop, Magnetic Moment and Galvanometer

An electron moves at \( 6.5 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.2 \, \text{T} \). What is the

**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. r = (mv/qB) . r = (9.1 × 10⁻³¹ × 6.5 × 10⁶/1.6 × 10⁻¹⁹ × 0.2) = (5.915 × 10⁻²⁴/3.2 × 10⁻²⁰) = 1.848 × 10⁻⁴ m ≈ 0.0185 cm . Using F = q v

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

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Torque on Current Loop, Magnetic Moment and Galvanometer

An electron moves at \( 2.5 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.5 \, \text{T} \). What is the

**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. r = (mv/qB) . r = (9.1 × 10⁻³¹ × 2.5 × 10⁶/1.6 × 10⁻¹⁹ × 0.5) = (2.275 × 10⁻²⁴/8 × 10⁻²⁰) = 2.84375 × 10⁻⁵ m ≈ 2.84 × 10⁻³ cm . Using F = q v B sinθ, F = I l B sinθ, B

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

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Solenoid, Toroid and Ampere's Law

An electron moves at \( 5 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.2 \, \text{T} \). What

**Solenoid field** inside long solenoid is B = μ₀ n I, n = N/L turns per meter (m⁻¹), uniform and parallel to axis, outside negligible for long solenoid because fields from opposite sides cancel. For n = 1200 m⁻¹, I = 1 A, B = 4π×10⁻⁷×1200 = 1.51×10⁻³ T = 1.51 mT. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 5 × 10⁶ × 0.2 = 1.6 × 10⁻¹³ N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Solenoid, Toroid and Ampere's Law

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Magnetic Force on Moving Charge - Lorentz Force and Motion

An electron moves at \( 6 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.4 \, \text{T} \). What is the ma

**Effect of doubling velocity** on magnetic force F = q v B sinθ is linear increase, F doubles for same θ and B. Electron with charge 1.6×10⁻¹⁹ C, v = 4.5×10⁶ m/s, B = 0.35 T, θ = 90°, F = 1.6×10⁻¹⁹×4.5×10⁶×0.35 = 2.52×10⁻¹³ N, illustrating magnitude for typical lab values. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 6 × 10⁶ × 0.4 = 3.84 × 10⁻¹³ N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r),

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Magnetic Force on Moving Charge - Lorentz Force and Motion

An electron moves at \( 4.5 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.35 \, \text{T} \). What is the

**Lorentz force** on charge q moving with velocity v in magnetic field B is F = q v × B, magnitude F = q v B sinθ, θ angle between v and B (degrees), unit N. Direction perpendicular to both v and B via right-hand rule. When v ⊥ B, motion circular with radius r = m v/(q B), centripetal force provided by magnetic force. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 4.5 × 10⁶ × 0.35 = 2.52 × 10⁻¹³ N . Using F = q

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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CHEMISTRY

Calculate the de Broglie wavelength of an electron moving at 2.05 × __10POW₇__m s^{-1 . ( h = 6.626 × 10⁻³⁴J s, m_e = 9.

Given: Calculate the de Broglie wavelength of an electron moving at 2.05 × __10POW₇__m s^{-1 . ( h = 6.626 × 10⁻³⁴J s, m_e = 9.1 × 10⁻³¹kg ) These values define the system as per NCERT data. Formula: lambda = h/mv = frac6.626 × 10⁻³⁴⁹.1 × 10⁻³¹ × 2.05 × __10POW₇__= 3.55 × 10⁻¹¹m .. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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