#straight wire

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 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

What is the shape of the magnetic field lines produced by a straight current-carrying wire?

**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. The magnetic field lines around a straight current-carrying wire form concentric circles centered on the wire, as derived from the Biot-Savart law or Ampere’s law. Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N

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Torque on Current Loop, Magnetic Moment and Galvanometer

A long straight wire carries a current of \( 40 \, \text{A} \). What is the magnetic field at a distance of \( 0.5 \, \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. Magnetic field B = (μ₀ I/2 π r) . B = (4 π × 10⁻⁷ × 40/2 π × 0.5) = (160 × 10⁻⁷/1) = 1.6 × 10⁻⁵ T . 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 straight wire carries a current of \( 10 \, \text{A} \). What is the magnetic field at a point \( 0.1 \, \text{m}

**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. Magnetic field B = (μ₀ I/2 π r) . B = (4 π × 10⁻⁷ × 10/2 π × 0.1) = (4 × 10⁻⁶/0.2) = 2 × 10⁻⁵ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N

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Solenoid, Toroid and Ampere's Law

A long wire carries \( 12 \, \text{A} \). At what distance is the magnetic field \( 2.4 \times 10^{-6} \, \text{T} \)? (

**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. B = (μ₀ I/2 π r) , so r = (μ₀ I/2 π B) . r = (4 π × 10⁻⁷ × 12/2 π × 2.4 × 10⁻⁶) = (48 × 10⁻⁷/4.8 × 10⁻⁶) = 1 m . Using F = q v B sinθ, F = I l B sinθ, B =

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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 =

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A long straight wire carries a current of \( 25 \, \text{A} \). What is the magnetic field at a distance of \( 0.25 \, \

**Biot-Savart law** dB = μ₀/4π·I dl × r̂/r² underlies both straight wire and loop formulas. For square loop side a, area A = a², moment m = N I A, field pattern similar to dipole at large distances, with superposition for N turns. Magnetic field B = (μ₀ I/2 π r) . B = (4 π × 10⁻⁷ × 25/2 π × 0.25) = (100 × 10⁻⁷/0.5) = 2 × 10⁻⁵ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N I A B sinθ, result

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A long straight wire carries a current of \( 30 \, \text{A} \). What is the magnetic field at a distance of \( 0.3 \, \t

**Magnetic field due to long straight wire** at distance r is B = μ₀ I/(2π r), μ₀ = 4π×10⁻⁷ T·m/A, direction circular around wire given by right-hand grip rule. For I = 18 A, r = 0.15 m, B = 2×10⁻⁷×18/0.15 = 2.4×10⁻⁵ T, illustrating 1/r dependence. Magnetic field B = (μ₀ I/2 π r) . B = (4 π × 10⁻⁷ × 30/2 π × 0.3) = (120 × 10⁻⁷/0.6) = 2 × 10⁻⁵ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N I

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A long straight wire carries a current of \( 22 \, \text{A} \). What is the magnetic field at a distance of \( 0.2 \, \t

**Field at centre of circular loop** with N turns is B = μ₀ N I/(2R), R radius (m), direction along axis via right-hand rule, magnitude proportional to N I/R. For R = 0.09 m, N = 45, I = 1.2 A, B = 4π×10⁻⁷×45×1.2/(2×0.09) = 3.77×10⁻⁴ T, showing N enhancement. Magnetic field B = (μ₀ I/2 π r) . B = (4 π × 10⁻⁷ × 22/2 π × 0.2) = (88 × 10⁻⁷/0.4) = 2.2 × 10⁻⁵ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ

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Force on Current-Carrying Conductor and Between Parallel Wires

A straight wire of length \( 0.9 \, \text{m} \) carries a current of \( 7 \, \text{A} \) perpendicular to a uniform magn

**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. Force F = I l B sin θ , where θ = 90° , so sin θ = 1 . F = 7 × 0.9 × 0.2 = 1.26 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A long wire carries \( 16 \, \text{A} \). At what distance is the magnetic field \( 8 \times 10^{-6} \, \text{T} \)? (\(

**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. B = (μ₀ I/2 π r) , so r = (μ₀ I/2 π B) . r = (4 π × 10⁻⁷ × 16/2 π × 8 × 10⁻⁶) = (64 × 10⁻⁷/16 × 10⁻⁶) = 0.4 m . Using F

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