#magnetic force

Mutual Induction and Mutual Inductance

A conducting rod is stationary in a varying magnetic field. The induced emf in the rod arises due to which component of

**Mutual inductance calculation** M = e₂/(dI₁/dt), for 200 turns length 0.5 m nearby coil e=0.5 V dI=2 A dt=0.2 s dI/dt=10 A/s, M=0.5/10=0.05 H, depends on geometry, orientation, number of turns, area, separation, coupling coefficient k = M/√(L₁ L₂) ≤1. For a stationary rod ( v = 0 ), the emf is induced by the electric field generated by the time-varying magnetic field, not the magnetic force ( q v × B ), which requires motion. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M

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Induced EMF Due to Change in Magnetic Field

A metal ring is placed near a solenoid. When an AC current flows through the solenoid, the ring experiences a force. Thi

**Uniform field change** in coil produces emf proportional to area and turns, for circular coil radius 0.16 m area πr²=0.0804 m², B 0.12 T deformed to wire in 0.6 s, ΔΦ=0.12×0.0804=0.00965 Wb, e=0.00965/0.6=0.0161 V, illustrating area change also induces emf. The AC current creates a varying magnetic field, inducing currents in the ring that generate an opposing field, leading to a repulsive or attractive force based on Lenz’s law. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt) and U = ½ L

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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

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Torque on Current Loop, Magnetic Moment and Galvanometer

Two parallel wires \( 0.08 \, \text{m} \) apart carry currents of \( 5 \, \text{A} \) and \( 3 \, \text{A} \) in the sam

**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. Force per unit length f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 5 × 3/2 π × 0.08) = (60 × 10⁻⁷/0.16) = 3.75 × 10⁻⁶ N/m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π

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

A wire of length \( 0.8 \, \text{m} \) carrying \( 5 \, \text{A} \) is at \( 30^\circ \) to a magnetic field of \( 0.8 \

**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. Force F = I l B sin θ . F = 5 × 0.8 × 0.8 × sin 30° = 4 × 0.8 × 0.5 = 1.6 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R)

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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

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Solenoid, Toroid and Ampere's Law

A wire of length \( 1.5 \, \text{m} \) carrying \( 8 \, \text{A} \) is at \( 60^\circ \) to a magnetic field of \( 0.5 \

**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. F = I l B sin θ . F = 8 × 1.5 × 0.5 × sin 60° = 12 × 0.5 × 0.866 = 5.196 ≈ 5.2 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 > Solenoid, Toroid and Ampere's Law

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Solenoid, Toroid and Ampere's Law

A straight wire of length \( 1.2 \, \text{m} \) carries a current of \( 6 \, \text{A} \) perpendicular to a uniform magn

**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. Force F = I l B sin θ , where θ = 90° , so sin θ = 1 . F = 6 × 1.2 × 0.25 = 1.8 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 > Solenoid, Toroid and Ampere's Law

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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

Two parallel wires \( 0.1 \, \text{m} \) apart carry \( 8 \, \text{A} \) and \( 6 \, \text{A} \) in the same direction.

**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. f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 8 × 6/2 π × 0.1) = (192 × 10⁻⁷/0.2) = 9.6 × 10⁻⁵ N/m . Using F = q v B sinθ, F =

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Magnetic Force on Moving Charge - Lorentz Force and Motion

Two parallel wires \( 0.01 \, \text{m} \) apart carry \( 5 \, \text{A} \) and \( 6 \, \text{A} \) in opposite directions

**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. f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 5 × 6/2 π × 0.01) = (120 × 10⁻⁷/0.02) = 6 × 10⁻⁵ N/m . 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 > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Magnetic Force on Moving Charge - Lorentz Force and Motion

Two parallel wires \( 0.06 \, \text{m} \) apart carry \( 9 \, \text{A} \) and \( 5 \, \text{A} \) in the same direction.

**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. f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 9 × 5/2 π × 0.06) = (180 × 10⁻⁷/0.12) = 1.5 × 10⁻⁵ N/m . 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 > Magnetic Force on Moving Charge - Lorentz Force and Motion

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