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Text lesson

Lesson 2: Magnetic Force on Currents, Torque and Magnetic Dipole Moment

Physics

Lesson 2: Magnetic Force on Currents, Torque and Magnetic Dipole Moment

Overview

Overview

🌳 Concept tree

Lesson 2→Magnetic Force on Currents, Torque and Magnetic Dipole Moment

├── Observation

├── Magnitude of the force

  1. ├── Force between two parallel current-carrying wires
  2. └── Torque on a rectangular coil in a magnetic field
Lesson 2: Magnetic Force on Currents, Torque and Magnetic Dipole Moment · 1 / 6
Observation

Observation

  • A straight wire carrying current between the poles of a magnet experiences a force.
  • The force is perpendicular to both the wire and the magnetic field.
  • The force reverses if either the current or the field is reversed.
  • If the wire is free to move it moves in the direction of the force.
  • (In diagrams a cross × means "into the page" and a dot · means "out of the page".)
  • Direction — Fleming’s left-hand rule stretch the thumb, forefinger and middle finger of the left hand mutually at right angles. The forefinger points along the magnetic field (N → S), the middle finger along the current; the thumb then points along the force (motion).
Lesson 2: Magnetic Force on Currents, Torque and Magnetic Dipole Moment · 2 / 6
Magnitude of the force

Magnitude of the force

  1. For a wire of length L carrying current I in a uniform field B
  2. the experiments show that
  • F ∝ L (a longer wire in the field feels a larger force),

F ∝ I (a larger current gives a larger force),

F ∝ B (a stronger field gives a larger force),

  • when the wire is perpendicular to the field.
  • Hence F = (constant) × B I L, and the unit of B is defined so that the constant equals 1:

F = B I L

Comparison aspectMain ideaKeywords
•
  • If the wire makes an angle θ with the field
  • resolve B into two components: B cos θ parallel to the wire (which exerts no force) and B sin θ perpendicular to the wire (which exerts the force). Therefore F = B I L sin θ
wiremakesanglefield
•
  • Maximum force when θ = 90° (F = BIL)
  • zero when the wire is parallel to the field (θ = 0°)
Maximumforcezerowire
Definition of the tesla
  • the flux density of a field
  • exerts a force of 1 newton on a 1 m long wire carrying 1 A perpendicular to the field. 1 T = 1 N/(A·m) = 1 Wb/m2
fluxdensityfieldexerts
Example 5.A 0.25 m wire carries 6 A at 30° to a field of 0.4 T. F = 0.4 × 6 × 0.25 × sin 30° = 0.3 Nwirecarriesfield
Example 5a.
  • A 0.40 m wire carries 7.5 A perpendicular to a 0.20 T field: F = 0.20 × 7.5 × 0.40 = 0.60 N. If the wire is turned so that it makes 37° with the field
  • F = 0.60 × sin 37° = 0.36 N
  • if it is turned parallel to the field
  • F = 0
wirecarriesperpendicularfield
Example 5b — finding B.→A force of 1.2 N acts on an 80 cm wire carrying 5 A perpendicular to the field. B = F / (I L) = 1.2 / (5 × 0.8) = 0.30 T.
Lesson 2: Magnetic Force on Currents, Torque and Magnetic Dipole Moment · 3 / 6
Force between two parallel current-carrying wires

Force between two parallel current-carrying wires

  • Wire 1 (current I₁) produces at wire 2 (distance d) a field B₁ = μ I₁ / (2π d). Wire 2 of length L in this field feels
  • F = B₁ I₂ L = μ I₁ I₂ L / (2π d)
  • Currents in the same direction → the wires attract.
  • Currents in opposite directions → the wires repel.
    (By Newton’s third law the forces on the two wires are equal and opposite.)
  • The force per unit length is F/L = μ I₁ I₂ / (2π d). This relation is the basis of the SI definition of the ampere: if two long parallel wires 1 m apart in vacuum carry equal currents and the force between them is 2×10−7 N per metre of length, each carries 1 A.
  • Example 5c. Two parallel wires 10 cm apart carry 15 A and 20 A in the same direction. For a 2 m length: F = 2×10−7 × 15 × 20 × 2 / 0.1 = 1.2×10−3 N, attractive (6×10−4 N per metre).
Lesson 2: Magnetic Force on Currents, Torque and Magnetic Dipole Moment · 4 / 6
Torque on a rectangular coil in a magnetic field

Torque on a rectangular coil in a magnetic field

  • Take a rectangular coil abcd (sides ab = cd = L, the other sides of width w) of N turns, carrying current I, free to rotate in a uniform field B.

  • The sides parallel to the field (ad and bc) feel no force, because θ = 0 for them.

  • The sides ab and cd, perpendicular to the field, feel equal and opposite parallel forces of magnitude F = BIL (per turn), forming a couple: the forces are equal, opposite and not along the same line, so they turn the coil about its axis without moving it as a whole.

  • When the plane of the coil is parallel to the field, the torque = force × perpendicular distance = BIL × w = BI A per turn (A = L w). For N turns:

    τ = B I A N

  • The coil has a magnetic dipole moment m = N I A, a vector perpendicular to the plane of the coil in the direction a right-hand screw advances when turned with the current (the direction from the south face to the north face of the coil). Then τ = B m.

  • If the normal to the coil (the direction of m) makes an angle θ with the field:

    τ = B I A N sin θ

    The torque is maximum (θ = 90°, plane parallel to the field) and zero when the plane is perpendicular to the field (θ = 0°, the normal along B). In that position the coil is in equilibrium, with the dipole moment lined up with the field.

  • This torque is the principle of the moving-coil galvanometer and the electric motor.
  • Example 6. A coil of 200 turns, 5 cm × 8 cm, carries 0.3 A and is placed with its plane parallel to a 0.5 T field. A = 4×10−3 m2. τ = 0.5 × 0.3 × 4×10−3 × 200 = 0.12 N·m.
  • Example 6a. A coil of 100 turns measures 6 cm × 4 cm (A = 2.4×10−3 m2) and carries 0.5 A in a field of 0.3 T with its plane parallel to the field: τ = 0.3 × 0.5 × 2.4×10−3 × 100 = 0.036 N·m. When the normal to the coil makes 30° with the field, τ = 0.036 × sin 30° = 0.018 N·m.
  • Example 6b — finding the current. A coil of 250 turns and area 3×10−3 m2 in a field of 0.2 T (plane parallel to the field) must experience a torque of 0.09 N·m. I = τ / (B A N) = 0.09 / (0.2 × 3×10−3 × 250) = 0.09 / 0.15 = 0.6 A.
Lesson 2: Magnetic Force on Currents, Torque and Magnetic Dipole Moment · 5 / 6
Part 6

Part 6

  • 📖

🔍 In-Depth Explanations, Deductions & Multi-Book Insights:

Physics Thanaweya Amma (2026/2027)

  • Lesson 2 – Magnetic Force on Currents, Torque and Magnetic Dipole Moment
  • When a current-carrying conductor is placed in an external magnetic field, it experiences a magnetic Lorentz force (F = B·I·L·sin θ).
  • A current loop experiences a deflecting magnetic torque (τ = B·I·A·N·sin θ) characterized by the magnetic dipole moment.

🌳 Hierarchy Classification & Concept Tree

Magnetic Forces and Torques

├── 1. Magnetic Force on Straight Conductor

│ ├── F = B · I · L · sin θ

Comparison aspectMain ideaKeywords
•
  • │ ├── Max force when perpendicular (θ = 90°)
  • Zero force when parallel (θ = 0°)
forceperpendicularZeroparallel
│ ├── Direction determined byFleming's Left-Hand RuleFlemingLeftHandRule
│ └── Mutual Force between two parallel wiresF = (μ · I1 · I2 · L) / (2π · d)

│ ├── Same current direction → Attractive force

  • │ └── Opposite current directions → Repulsive force

└── 2. Magnetic Deflecting Torque on Coil

  • ├── τ = B · I · A · N · sin θ (where θ is angle between normal to coil plane and field)
  • ├── Maximum torque when coil plane is parallel to field lines
  • ├── Zero torque when coil plane is perpendicular to field lines
  • └── Magnetic Dipole Moment (md = I · A · N) Vector perpendicular to coil plane, independent of B

📋 Master Teacher's Blackboard & Exam Secrets

Comparison aspectMain ideaKeywords
RULE 1
  • In the torque formula τ = B·I·A·N·sin θ
  • angle θ is measured between the NORMAL to the coil plane (or dipole vector md) and the magnetic field B (or cos of angle with coil plane)
torqueformulaanglemeasured
RULE 2
  • The mutual magnetic force between two parallel wires is ALWAYS equal in magnitude for both wires (Newton's 3rd law)
  • even if one wire carries 10 times more current than the other!
mutualmagneticforcebetween
RULE 3
  • Magnetic dipole moment md = I · A · N is an intrinsic property of the current loop
  • it depends ONLY on I
  • and N
  • and is completely INDEPENDENT of the external magnetic field B or coil tilt
Magneticdipolemomentintrinsic
📊 Comparative AnalysisForce vs. Torque Extremes ComparisonForceTorqueExtremesComparison
Orientation of Conductor / Coil to Field B Magnetic Force on Wire (F) Magnetic Torque on Coil (τ)
Parallel to Magnetic Field (0°) F = 0 (Zero force) τ = Maximum (τ_max = BIAN)
Perpendicular to Magnetic Field (90°) F = Maximum (Fmax = BIL) τ = 0 (Zero torque)
Inclined at 30° to Field F = 0.5 Fmax τ = 0.866 τ_max (using normal angle 60°)
Lesson 2: Magnetic Force on Currents, Torque and Magnetic Dipole Moment · 6 / 6