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Chapter 1 - Electric Current, Ohms Law and Kirchhoffs Laws

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Chapter 2 - Magnetic Effects of Electric Current and Measuring Instruments

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Chapter 4 - Alternating Current AC Circuits

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Chapter 7 - Lasers and Principles

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Chapter 8 - Modern Electronics and Semiconductors

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Text lesson

Lesson 1: Magnetic Field of a Straight Wire, Circular Coil and Solenoid

Physics

Lesson 1: Magnetic Field of a Straight Wire, Circular Coil and Solenoid

Overview

Overview

🌳 Concept tree

  • Lesson 1 Magnetic Field of a Straight Wire, Circular Coil and Solenoid
  • ├── Oersted's discovery
  • ├── The magnetic flux density
  • ├── Field of a long straight wire
  • ├── Field of a circular loop (coil)
  • └── Field of a solenoid
Lesson 1: Magnetic Field of a Straight Wire, Circular Coil and Solenoid · 1 / 7
Oersted’s discovery

Oersted’s discovery

  • In 1819 the Danish physicist Hans Christian Oersted placed a compass near a wire carrying current.
  • The needle deflected while the current flowed and returned to its original position when the current was switched off.
  • A compass needle is deflected only by a magnetic field, so a current produces a magnetic field around the conductor.
  • This discovery linked electricity and magnetism and started a chain of events that led to generators, motors, transformers and many instruments.
We study the field of three shapesa straight wire, a circular loop (coil) and a solenoid.
In each case we ask three questionswhat is the shape of the field lines, what does the strength depend on, and how do we find the direction.
Lesson 1: Magnetic Field of a Straight Wire, Circular Coil and Solenoid · 2 / 7
The magnetic flux density

The magnetic flux density

    • The magnetic flux density B (also called magnetic induction) measures the strength of the field at a point.
    • It is the magnetic flux per unit area perpendicular to the field:

    B = Φm / A (so Φm = B A for an area A held perpendicular to the field)

    Unit: weber/m2 = tesla (T) = N/(A·m).

  • The magnetic permeability μ of a medium tells how easily the field is set up in it. For air (and vacuum) μ₀ = 4π × 10−7 Wb/(A·m) (T·m/A). Materials such as iron have a permeability thousands of times larger than that of air.

    • The magnetic field is drawn as lines of flux.
    • The closer together the lines, the stronger the field.
    • Lines never cross, and each line is a closed curve.
Example 1.0.→A coil of area 0.05 m2 is held perpendicular to a uniform field of 0.30 T. The flux through it is Φ = B A = 0.30 × 0.05 = 0.015 Wb.
Lesson 1: Magnetic Field of a Straight Wire, Circular Coil and Solenoid · 3 / 7
Field of a long straight wire

Field of a long straight wire

  • Experiment pass a wire vertically through a card and sprinkle iron filings on the card.
  • After tapping gently, the filings arrange themselves in concentric circles around the wire.
  • Observations:
  • the circles (flux lines) are crowded close to the wire and become farther apart as the distance increases, so the field is weaker far from the wire;
  • when the current is increased and the card is tapped again, the circles become more crowded, so the field grows with the current;
  • reversing the current reverses the direction of the field (a small compass placed on the card turns round).
  • The flux density at a point at perpendicular distance d from a long straight wire carrying current I is

B = μ I / (2π d)

  • (the book calls this relation Ampere’s circuital law.)
  • B ∝ I and B ∝ 1/d. It does not depend on the wire’s thickness.
  • In air, μ₀/(2π) = 2 × 10−7, so a handy form is B = 2 × 10−7 I / d.
  • Practical remark because B falls with distance, it is advisable not to live very close to high-voltage cables, which carry large currents.
  • Direction — Ampere’s right-hand rule (grip rule):
  • grasp the wire in the right hand with the thumb along the current.
  • The curled fingers show the direction of the magnetic field lines.
  • Example 1. A long wire carries 25 A in air. Flux density 5 cm from the wire:
  • B = 4π×10−7 × 25 / (2π × 0.05) = 2×10−7 × 25 / 0.05 = 1×10−4 T.
  • Example 1a. A long straight cable carries 40 A. Find B at 8 cm from it. B = 2×10−7 × 40 / 0.08 = 1.0×10−4 T. At 16 cm (twice as far) the field halves to 5×10−5 T.
  • Example 1b — finding the current. The field 20 cm from a long wire in air is 2.5×10−5 T. I = 2π d B / μ = B d / (2×10−7) = 2.5×10−5 × 0.2 / (2×10−7) = 25 A.
  • Example 1c — two wires. Two long parallel wires 20 cm apart carry 10 A each. At the point midway between them (10 cm from each wire) each wire produces B = 2×10−7 × 10 / 0.1 = 2×10−5 T.
  • If the currents are in opposite directions the two fields point the same way at the midpoint and add: B = 4×10−5 T.
  • If the currents are in the same direction the two fields are opposite and cancel: B = 0.
Lesson 1: Magnetic Field of a Straight Wire, Circular Coil and Solenoid · 4 / 7
Field of a circular loop (coil)

Field of a circular loop (coil)

Comparison aspectMain ideaKeywords
•
  • A loop carrying current produces a field like a short bar magnet (a magnetic dipole). The face from
  • the lines emerge is a north pole
  • the face into
  • they enter is a south pole. In effect the loop behaves like a thin disc magnet with a pole on each face
loopcarryingcurrentproduces
Viewing the loop from one face
  • if the current is clockwise that face is a south pole
  • if anticlockwise it is a north pole
currentclockwisesouthpole
With iron filings (the loop passes through a board)
  • near the centre the lines are straight
  • parallel and perpendicular to the plane of the coil
  • which means the field there is nearly uniform
  • away from the centre the lines are curved and B varies from point to point
  • very close to the wire itself the lines are circles around the wire
nearcentrelinesstraight
  1. For a coil of N turns and radius r carrying current I
  2. the flux density at its centre is

B = μ N I / (2 r)

Comparison aspectMain ideaKeywords
B depends on three factors
  • the number of turns (B ∝ N)
  • the current (B ∝ I) and the radius (B ∝ 1/r). Direction at the centre: right-hand screw rule — if a right-handed screw is turned in the direction of the current
  • the direction in
  • it advances is the direction of the field at the centre
numberturnscurrentradius
No magnetic monopole exists
  • poles always appear as N–S pairs
  • so a loop is a dipole
polesalwaysappearpairs
Example 2.
  • A coil of 50 turns
  • radius 8 cm
  • carries 0.6 A in air. At its centre
coilturnsradiuscarries
•B = 4π×10−7 × 50 × 0.6 / (2 × 0.08) = 2.36×10−4 T
Example 2a.A circular coil of 30 turns and radius 5 cm carries 2 A. B = 4π×10−7 × 30 × 2 / (2 × 0.05) = 7.54×10−4 Tcircularcoilturnsradius
Comparison aspectMain ideaKeywords
Example 2b — scaling.
  • If the radius of the coil in Example 2a is halved and the number of turns is doubled (same current)
  • B becomes 2 × 2 = 4 times larger: B = 3.02×10−3 T
radiuscoilhalvednumber
Example 2c — finding the current.A single circular loop of radius 12 cm must produce B = 5×10−5 T at its centre. I = 2 r B / (μ N) = 2 × 0.12 × 5×10−5 / (4π×10−7 × 1) = 9.55 Asinglecircularloopradius
Lesson 1: Magnetic Field of a Straight Wire, Circular Coil and Solenoid · 5 / 7
Field of a solenoid

Field of a solenoid

  • A solenoid is a long cylindrical spiral coil of insulated wire.
  • Its field resembles that of a bar magnet inside, the lines are parallel to the axis and nearly uniform; the lines are closed curves that pass inside the coil from the south pole to the north pole and return outside through the air.
  • The end where the lines leave is the north pole.
  • At a point on the axis, inside a long solenoid, the flux density depends on the current and the number of turns per unit length n = N/L:

B = μ n I = μ N I / L

  • B ∝ I; B ∝ n; the core material changes μ. An iron core multiplies the field because μiron is much larger than μ₀.
  • Polarity is found by the right-hand screw rule (or the grip rule applied to the coil: curl the fingers of the right hand in the direction of the current, the thumb points to the north pole).
  • Why does a solenoid give a strong, uniform field? Each turn contributes a field in the same direction inside the coil, so the contributions add; outside the coil the contributions mostly cancel. Compressing the turns (larger n) strengthens the field.
  • The field is strongest inside the solenoid and weaker near the ends (about half the central value at the very end of a long solenoid).
  • Example 3. A solenoid 40 cm long has 1000 turns and carries 2.0 A in air:
  • B = 4π×10−7 × 1000 × 2.0 / 0.4 = 6.28×10−3 T.
  • Example 3a. A solenoid has 500 turns spread over 25 cm and carries 1.2 A. n = 500/0.25 = 2000 turns/m, so B = 4π×10−7 × 2000 × 1.2 = 3.02×10−3 T.
  • Example 4 (iron core). To get B = 0.5 T inside a 25 cm solenoid with 500 turns and an iron core of permeability 2.0×10−3 Wb/(A·m):
  • I = B L / (μ N) = 0.5 × 0.25 / (2.0×10−3 × 500) = 0.125 A.
  • Example 4a (iron core, second case). A solenoid of 600 turns and length 30 cm has an iron core of permeability 1.5×10−3 Wb/(A·m). To produce B = 0.6 T the current is I = 0.6 × 0.30 / (1.5×10−3 × 600) = 0.18/0.9 = 0.2 A. The same solenoid in air would need a current μiron/μ₀ ≈ 1190 times larger.
Lesson 1: Magnetic Field of a Straight Wire, Circular Coil and Solenoid · 6 / 7
Part 7

Part 7

  • 📖

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

Physics Thanaweya Amma (2026/2027)

  • Lesson 1 – Magnetic Field of a Straight Wire, Circular Coil and Solenoid
  • Hans Christian Oersted discovered that an electric current produces an associated magnetic field.
  • We derive the magnetic flux density (B) for a straight conductor (Ampere's Circular Law), a flat circular coil, and a cylindrical solenoid.

🌳 Hierarchy Classification & Concept Tree

Magnetic Flux Density (B)

├── 1. Straight Conducting Wire

│ ├── B = (μ · I) / (2π · d) [Ampere's Law]

Comparison aspectMain ideaKeywords
•
  • │ ├── Concentric circular field lines centered on wire
  • spacing decreases near wire
Concentriccircularfieldlines
│ ├── Direction determined byRight-Hand Grip RuleRightHandGripRule
│ └── Neutral Point
  • Point where opposing fields cancel (B1 = B2)
  • located closer to smaller current
opposingfieldscancellocated

├── 2. Flat Circular Coil (N turns, radius r)

│ ├── B = (μ · N · I) / (2r)

  1. │ └── Center field lines are straight
  2. parallel
  3. perpendicular to coil plane

└── 3. Cylindrical Solenoid (Length L)

  • ├── B = (μ · N · I) / L = μ · n · I (where n = N/L is turns per unit meter)
  • └── Field inside solenoid along axis is uniform and parallel

📋 Master Teacher's Blackboard & Exam Secrets

Comparison aspectMain ideaKeywords
RULE 1Magnetic flux Φm = B · A · sin θ (where θ is the angle between field B and plane of area A)Magneticfluxanglebetween
RULE 2
  • Neutral point condition between two parallel straight wires: (I1 / d1) = (I2 / d2). If currents are in same direction
  • neutral point lies between wires
  • if opposite
  • it lies outside closer to the weaker current
Neutralpointconditionbetween
RULE 3If a circular coil is stretched uniformly along its axis to become a solenoid: Bsolenoid / Bcoil = 2r / Lcircularcoilstretcheduniformly
Conductor Geometry Magnetic Flux Density Formula (B) Key Factors Affecting B Field Line Pattern
Long Straight Wire B = (μ · I) / (2π · d) Current I, distance d, permeability μ Concentric circular rings around wire
Flat Circular Coil B = (μ · N · I) / (2r) Current I, turns N, coil radius r Straight parallel lines at axis center
Helical Solenoid B = (μ · N · I) / L Current I, turns N, length L Uniform parallel axial lines resembling bar magnet
Lesson 1: Magnetic Field of a Straight Wire, Circular Coil and Solenoid · 7 / 7