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Unit IX · Magnetism and magnetic effect of electric current (lesson 18)
10 min readConcept sheetSection A MCQ (1 mark) on the direction rules; Section B short answer (3 marks) or part of a long answer (5 marks): state the Biot-Savart law and derive the field at the centre of a circular coil

Magnetic field of a current: Oersted's experiment and the Biot-Savart law

A current produces a magnetic field (Oersted, 1820). The Biot-Savart law gives the field of each small piece of wire; at a coil's centre B = μ₀NI/2r.

Biot-Savart lawΔB = (μ₀/4π) I Δl sin θ/r², for a small piece Δl at a distance r
Permeability of free spaceμ₀ = 4π × 10⁻⁷ T m A⁻¹; air has almost the same value
Field at the centre of a circular coilB = μ₀NI/2r, for N turns of radius r (your NIOS lesson writes n for the number of turns)
SI unit of Btesla (T); 1 T = 1 N A⁻¹ m⁻¹

Activity: Oersted's effect with a matchbox

You need a 1.5 V cell, about 1 m of insulated wire, a small compass and an empty matchbox.

  1. 1
    Wind 10 to 15 turns of the wire round the matchbox.
  2. 2
    Slide the compass under the turns. Place the box so that the turns run along the needle, north to south.
  3. 3
    Touch the wire ends to the cell for a few seconds. Observe: the needle turns away from north-south.
  4. 4
    Reverse the cell connections. Observe: the needle turns the other way.
  5. 5
    Disconnect the cell. Observe: the needle returns to north-south.
  6. 6★
    Result: a current produces a magnetic field around it, and reversing the current reverses the field. Hans Christian Oersted found this in 1820.

★ key step

The Biot-Savart law

Think of a wire as a chain of tiny pieces, each of length Δl. Each piece carrying a current I makes a small field ΔB at a point P a distance r away. Experiments show that ΔB is proportional to the current I, to the length Δl and to sin θ. Here θ is the angle between the piece and the line joining it to P. ΔB is also inversely proportional to r². So ΔB = (μ₀/4π) I Δl sin θ/r². This is the Biot-Savart law. The constant μ₀ is the permeability of free space (vacuum), 4π × 10⁻⁷ T m A⁻¹. Permeability tells us how easily a material lets a magnetic field form in it; air is almost the same as vacuum. In another medium, μ₀ is replaced by that medium's permeability μ. The total field of the wire is the vector sum of the fields of all its pieces. A piece gives no field at points along its own line, where θ = 0.

Derivation: field at the centre of a circular coil

A circular coil of radius r carries a current I. Find the field at its centre O.

  1. 1
    Take a small piece Δl of the coil. Its distance from O is r.
  2. 2
    The piece is at right angles to the radius, so θ = 90° and sin θ = 1.
  3. 3
    Biot-Savart law: ΔB = (μ₀/4π) I Δl/r².
  4. 4
    Every piece gives a field at O in the same direction, at right angles to the plane of the coil. So the fields simply add.
  5. 5
    The pieces together make the circumference: ΣΔl = 2πr. So B = (μ₀/4π)(I/r²) × 2πr.
  6. 6★
    So, B = μ₀I/2r at the centre. For a coil of N turns, B = μ₀NI/2r.

★ key step

Finding the direction: right-hand grip rule and end rule

  • Right-hand grip rule: hold the wire in your right hand with the thumb along the current. Your curled fingers show the direction of the field lines.
  • Round a long straight wire, the field lines are concentric circles, that is, circles with the wire at their common centre. They lie in a plane at right angles to the wire.
  • On paper, a dot ⊙ means a current coming out of the page, and a cross ⊗ means one going into it.
  • End rule for a coil: look at one face. If the current there flows clockwise, that face acts as a south pole and B points into it.
  • If the current flows anticlockwise, the face is a north pole and B comes out of it. So a current-carrying coil behaves like a short bar magnet.

Worked example: field at the centre of a coil

Problem

A circular coil of 50 turns and radius 10 cm carries a current of 2.0 A. (a) Find the field at its centre. (b) The same wire is rewound into a coil of radius 5.0 cm. Find the new field at the centre for the same current. (μ₀ = 4π × 10⁻⁷ T m A⁻¹)

Working
  1. Given: N = 50, r = 10 cm = 0.10 m, I = 2.0 A.
  2. (a) Formula: B = μ₀NI/2r.
  3. B = (4π × 10⁻⁷ × 50 × 2.0)/(2 × 0.10) T = (4π × 10⁻⁵)/0.20 T = 2π × 10⁻⁴ T ≈ 6.3 × 10⁻⁴ T.
  4. (b) The length of wire is the same. Halving the radius halves the length of each turn, so N doubles to 100.
  5. B = (4π × 10⁻⁷ × 100 × 2.0)/(2 × 0.050) T = (8π × 10⁻⁵)/0.10 T = 8π × 10⁻⁴ T ≈ 2.5 × 10⁻³ T.
  6. The field is now 4 times as large, because N is doubled and r is halved.
Answer

(a) 2π × 10⁻⁴ T, about 6.3 × 10⁻⁴ T; (b) 8π × 10⁻⁴ T, about 2.5 × 10⁻³ T, four times as large

How to draw the field of a circular coil

  • Draw the coil seen edge-on: two small circles one above the other, a distance 2r apart, where the wire cuts the page.
  • Put ⊙ in the top circle and ⊗ in the bottom one to show the current.
  • Round each small circle, draw closed loops: anticlockwise round ⊙ and clockwise round ⊗.
  • Near the centre, draw lines that run almost straight through the coil, along its axis, with arrows.
  • Mark the centre O, the radius r, the current I and the field B along the axis at O.
  • Label the face that the lines come out of as N, and the other face as S.
In real life

Every current-carrying wire makes a magnetic field. A compass held close to a car's battery cable swings while the engine is being started, because the cable then carries a large current. Coils in fans, mixers and loudspeakers use this effect to make motion.

Exam angle

2 marks: “State the Biot-Savart law and write its formula.” The field ΔB of a small current piece is proportional to I, Δl and sin θ, and inversely proportional to r²: ΔB = (μ₀/4π) I Δl sin θ/r².

3 marks: “Use the Biot-Savart law to derive the magnetic field at the centre of a circular coil of N turns and radius r carrying a current I.” θ = 90° for every piece; ΔB = (μ₀/4π) IΔl/r², all in the same direction; ΣΔl = 2πr gives B = μ₀I/2r; for N turns B = μ₀NI/2r, at right angles to the plane (end rule).

2 marks: “A wire carrying a steady current is bent first into one circular turn, and then into two turns using the same length. Compare the fields at the centre.” Two turns: radius r/2 and N = 2. B₂ = μ₀(2)I/(2 × r/2) = 4 × μ₀I/2r. So the field becomes 4 times as large.

Remember it

Grip the wire, thumb along the current: the fingers show B. A clockwise face is South, an anticlockwise face is North.