#parallel wires

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

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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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 =

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 \( 0.01 \, \text{m} \) 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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Force on Current-Carrying Conductor and Between Parallel Wires

Two parallel wires \( 0.07 \, \text{m} \) apart carry \( 8 \, \text{A} \) and \( 6 \, \text{A} \) in the same direction.

**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 = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 8 × 6/2 π × 0.07) = (192 × 10⁻⁷/0.14) = 1.3714 × 10⁻⁵ ≈ 1.37 × 10⁻⁵ N/m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B =