#Lorentz force

Magnetic Flux and Faraday's Laws of Induction

A conducting rod is rotated about one end in a uniform magnetic field. The emf induced between the ends is due to what f

**Magnetic flux** Φ = B·A = B A cosθ, B magnetic field (T), A area (m²), θ angle between B and normal to area, unit Wb = T·m², Faraday's law induced emf e = -N dΦ/dt, N turns, negative sign Lenz's law indicating opposition, magnitude |e| = N |ΔΦ/Δt|, for 100 turns ΔΦ=0.03 Wb Δt=0.06 s e=100×0.03/0.06=50 V. The rotation causes charges to move through the field, experiencing a magnetic force (Lorentz force) that separates them, inducing an emf along the rod. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω

Ref: NCERT > Physics Book > Electromagnetic Induction > Magnetic Flux and Faraday's Laws of Induction

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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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Solenoid, Toroid and Ampere's Law

A proton moves with a speed of \( 2 \times 10^6 \, \text{m/s} \) perpendicular to a uniform magnetic field of \( 0.5 \,

**Field outside long solenoid** considered negligible because magnetic lines are concentrated inside and return path spreads over large area outside, making B_out ≈ 0. This justifies using solenoid for uniform field experiments, with n = 950 m⁻¹, I = 1.4 A giving B = 4π×10⁻⁷×950×1.4 = 1.67×10⁻³ T. Radius r = (mv/qB) . Substitute: r = (1.67 × 10⁻²⁷ × 2 × 10⁶/1.6 × 10⁻¹⁹ × 0.5) = (3.34 × 10⁻²¹/8 × 10⁻²⁰) = 4.175 × 10⁻² m = 4.18 cm . 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 > Solenoid, Toroid and Ampere's Law

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A proton moves at \( 2.5 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.15 \, \text{T} \). What is the ma

**Motion of charged particle perpendicular to uniform B** is circular because force F ⊥ v, no work done, speed constant, radius r = m v/(q B), period T = 2π m/(q B) independent of v. If v has component parallel to B, helical path results, pitch = v_parallel·T. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 2.5 × 10⁷ × 0.15 = 6 × 10⁻¹³ N . 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 > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Force on Current-Carrying Conductor and Between Parallel Wires

A wire of length \( 2.3 \, \text{m} \) carrying \( 4 \, \text{A} \) is at \( 60^\circ \) to a magnetic field of \( 0.6 \

**Force on current-carrying wire** in magnetic field is F = I l × B, magnitude F = I l B sinθ, I current (A), l length (m), B field (T), θ angle between current direction and B. Direction perpendicular to plane containing wire and B, given by Fleming's left-hand rule. F = I l B sin θ . F = 4 × 2.3 × 0.6 × sin 60° = 5.52 × 0.866 = 4.7803 ≈ 4.78 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Force on Current-Carrying Conductor and Between Parallel Wires

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Force on Current-Carrying Conductor and Between Parallel Wires

A wire of length \( 1 \, \text{m} \) with \( 10 \, \text{A} \) is at \( 45^\circ \) to a field of \( 0.2 \, \text{T} \).

**Force on current-carrying wire** in magnetic field is F = I l × B, magnitude F = I l B sinθ, I current (A), l length (m), B field (T), θ angle between current direction and B. Direction perpendicular to plane containing wire and B, given by Fleming's left-hand rule. F = I l B sin θ . F = 10 × 1 × 0.2 × sin 45° = 2 × (1/√(2)) = 1.414 ≈ 1.41 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Force on Current-Carrying Conductor and Between Parallel Wires

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Force on Current-Carrying Conductor and Between Parallel Wires

A straight wire of length \( 1.1 \, \text{m} \) carries a current of \( 4 \, \text{A} \) perpendicular to a uniform magn

**Parallel current interaction** arises because each wire's field B = μ₀ I/(2π d) exerts force F = I l B on other. Force per length f = B I, leading to f = μ₀ I₁ I₂/(2π d). This defines ampere: two wires 1 m apart carrying 1 A exert 2×10⁻⁷ N/m. Force F = I l B sin θ , where θ = 90° , so sin θ = 1 . F = 4 × 1.1 × 0.3 = 1.32 N . 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 > Force on Current-Carrying Conductor and Between Parallel Wires

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Magnetic Force on Moving Charge - Lorentz Force and Motion

An electron moves with a speed of \( 3.5 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.25 \, \t

**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. Radius r = (mv/qB) . r = (9.1 × 10⁻³¹ × 3.5 × 10⁶/1.6 × 10⁻¹⁹ × 0.25) = (3.185 × 10⁻²⁴/4 × 10⁻²⁰) = 7.9625 × 10⁻⁵ m = 7.96 × 10⁻³ cm . Using F = q v

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 \( 6 \times 10^6 \, \text{m/s} \) 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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Force on Current-Carrying Conductor and Between Parallel Wires

A wire of length \( 1.3 \, \text{m} \) carrying \( 6 \, \text{A} \) is at \( 45^\circ \) to a magnetic field of \( 0.8 \

**Parallel current interaction** arises because each wire's field B = μ₀ I/(2π d) exerts force F = I l B on other. Force per length f = B I, leading to f = μ₀ I₁ I₂/(2π d). This defines ampere: two wires 1 m apart carrying 1 A exert 2×10⁻⁷ N/m. F = I l B sin θ . F = 6 × 1.3 × 0.8 × sin 45° = 6.24 × 0.707 = 4.4117 ≈ 4.41 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ =

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Force on Current-Carrying Conductor and Between Parallel Wires

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Force on Current-Carrying Conductor and Between Parallel Wires

A wire of length \( 0.6 \, \text{m} \) carrying \( 9 \, \text{A} \) is at \( 60^\circ \) to a magnetic field of \( 0.4 \

**Parallel current interaction** arises because each wire's field B = μ₀ I/(2π d) exerts force F = I l B on other. Force per length f = B I, leading to f = μ₀ I₁ I₂/(2π d). This defines ampere: two wires 1 m apart carrying 1 A exert 2×10⁻⁷ N/m. F = I l B sin θ . F = 9 × 0.6 × 0.4 × sin 60° = 2.16 × 0.866 = 1.8706 ≈ 1.87 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ =

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Force on Current-Carrying Conductor and Between Parallel Wires

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