Moving Charge and Magnetism

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

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Torque on Current Loop, Magnetic Moment and Galvanometer

A square loop of side \( 0.15 \, \text{m} \) with 50 turns carries \( 1 \, \text{A} \) in a magnetic field of \( 0.6 \,

**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. Torque tau = N I A B sin θ , where A = 0.15 × 0.15 = 0.0225 m² . tau = 50 × 1 × 0.0225 × 0.6 × sin 30° = 0.675 × 0.5 =

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Torque on Current Loop, Magnetic Moment and Galvanometer

Which property of magnetic field lines distinguishes them from electric field lines?

**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. Magnetic field lines form closed loops because there are no magnetic monopoles, unlike electric field lines, which originate from positive charges and terminate at negative charges or extend to infinity. Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r),

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Torque on Current Loop, Magnetic Moment and Galvanometer

A long wire carries \( 15 \, \text{A} \). At what distance is the magnetic field \( 3 \times 10^{-6} \, \text{T} \)? (\(

**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. B = (μ₀ I/2 π r) , so r = (μ₀ I/2 π B) . r = (4 π × 10⁻⁷ × 15/2 π × 3 × 10⁻⁶) = (60 × 10⁻⁷/6 × 10⁻⁶) = 1 m . Using F = q v B sinθ, F = I l B sinθ, B

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Torque on Current Loop, Magnetic Moment and Galvanometer

A circular loop of radius \( 0.08 \, \text{m} \) with 60 turns carries a current of \( 0.75 \, \text{A} \). 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. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 60 × 0.75/2 × 0.08) = (18 π × 10⁻⁶/0.16) = 1.125 π × 10⁻⁴ ≈ 3.53 × 10⁻⁴ T .

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Torque on Current Loop, Magnetic Moment and Galvanometer

A circular coil of 45 turns and radius \( 6 \, \text{cm} \) carries a current of \( 1.2 \, \text{A} \). What is the magn

**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. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 45 × 1.2/2 × 0.06) = (21.6 π × 10⁻⁶/0.12) = 1.8 π × 10⁻⁴ ≈ 5.65 × 10⁻⁴ T . Using F = q v B sinθ, F = I l

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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

A solenoid with 1400 turns per meter carries a current of \( 2.5 \, \text{A} \). 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 × 1400 × 2.5 = 14 π × 10⁻⁴ ≈ 4.40 × 10⁻³ T . Using F = q v B sinθ, F

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Torque on Current Loop, Magnetic Moment and Galvanometer

A long wire carries \( 14 \, \text{A} \). At what distance is the magnetic field \( 7 \times 10^{-6} \, \text{T} \)? (\(

**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 I/2 π r) , so r = (μ₀ I/2 π B) . r = (4 π × 10⁻⁷ × 14/2 π × 7 × 10⁻⁶) = (56 × 10⁻⁷/14 × 10⁻⁶) = 0.4 m . Using F = q v B sinθ, F = I

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Torque on Current Loop, Magnetic Moment and Galvanometer

A rectangular loop of area \( 0.06 \, \text{m}^2 \) with 15 turns carries \( 2.5 \, \text{A} \) in a 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. tau = N I A B sin θ , where θ = 60° to plane means sin 30° with normal. tau = 15 × 2.5 × 0.06 × 0.8 × sin 60° = 1.8 × 0.866 = 1.5588 ≈ 1.56 N m . Using F = q v B sinθ, F =

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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

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