Magnetic field and Oersted's experiment
In 1820 Hans Christian Oersted saw a compass needle turn when a nearby wire carried current. So moving charges make a magnetic field. Stop the current and the needle goes back.
The magnetic field B tells how strongly and in which direction a magnet would be pushed at a point. Its SI unit is the tesla (T). Earth's field is only about 5 × 10⁻⁵ T; a fridge magnet is about 10⁻² T.
Direction: right-hand thumb rule
Hold the wire in your right hand with the thumb along the current. Your curled fingers show the direction of the field circles.
Field lines round a straight wire are closed circles centred on the wire. Reverse the current and the circles reverse.
Biot-Savart law
Cut a wire into tiny pieces of length dl. Each piece with current I makes a small field dB at a point P at distance r:
dB = (μ₀/4π) · I dl sinθ / r² (vector form: dB = (μ₀/4π) · I dl × r̂ / r²)
- θ is the angle between the piece dl and the line joining it to P.
- μ₀ = 4π × 10⁻⁷ T m A⁻¹ is the permeability of free space.
- dB is perpendicular to both dl and r (right-hand rule).
- It falls as 1/r², like Coulomb's law. But it depends on the angle: along the wire's own line (θ = 0) the field is zero.
Field of a circular loop (Biot-Savart)
At the centre
Every piece of the loop is at the same distance R from the centre and at 90° to the radius. All the small fields point the same way (along the axis). Add them round the whole circle (total length 2πR):
B = (μ₀/4π) · I · 2πR / R² = μ₀I / 2R. For N turns: B = μ₀NI / 2R.
On the axis, distance x from the centre
Each piece is now at distance √(R² + x²). The sideways parts of the small fields cancel in pairs; the parts along the axis add. The result is
B = μ₀NIR² / 2(R² + x²)^{3/2}
Check: at x = 0 this gives μ₀NI/2R. Far away (x ≫ R) it falls as 1/x³, just like a small bar magnet. Direction: curl right-hand fingers along the current; the thumb gives B.
Ampere's circuital law and the long straight wire
Walk round any closed path. Multiply the part of B along each small step by the step length and add up. The total equals μ₀ times the current passing through the path:
∮ B · dl = μ₀ Ienclosed
This law is most useful when the shape is very symmetric.
Long straight wire
Choose a circle of radius r centred on the wire. By symmetry B has the same size everywhere on it and points along the circle. So ∮B·dl = B × 2πr = μ₀I, giving
B = μ₀I / 2πr (= 2 × 10⁻⁷ I / r)
Double the distance and B halves. Currents outside the path do not count in Ienclosed (their effects cancel round the loop).
Solenoid (qualitative)
A solenoid is a long wire wound as a tight spiral of many turns. Inside, the fields of neighbouring turns point the same way and add up; between turns on the outside they mostly cancel.
- Inside a long solenoid the field is strong, uniform and along the axis.
- Outside it is almost zero.
- The size is B = μ₀nI, where n = turns per metre. It does not depend on the radius.
- At the ends the field is about half of the middle value.
- One end acts as a north pole, the other as a south pole, just like a bar magnet. Putting a soft iron core inside makes an electromagnet.
Exam corner
Common board questions: state Biot-Savart law (2 marks); derive B on the axis of a circular loop (3–5 marks); state Ampere's law and use it for a straight wire (3 marks); numericals on B = μ₀I/2πr and μ₀NI/2R. Always write the unit tesla and the direction.
Key formulas and definitions
- dB = (μ₀/4π) · I dl sinθ / r²
- Loop centre: B = μ₀NI / 2R
- Loop axis: B = μ₀NIR² / 2(R² + x²)^{3/2}
- Ampere's law: ∮B·dl = μ₀I_enclosed
- Long straight wire: B = μ₀I / 2πr
- Long solenoid (inside): B = μ₀nI, n = N/L
- μ₀ = 4π × 10⁻⁷ T m A⁻¹; μ₀/4π = 10⁻⁷; μ₀/2π = 2 × 10⁻⁷
Worked examples
1. A long straight wire carries 5 A. Find B at 10 cm from it.
B = μ₀I / 2πr = 2 × 10⁻⁷ × 5 / 0.10 = 1 × 10⁻⁵ T.
2. At 5 cm from a long wire the field is 4 × 10⁻⁶ T. Find the current.
I = B·2πr / μ₀ = B·r / (2 × 10⁻⁷) = 4 × 10⁻⁶ × 0.05 / (2 × 10⁻⁷) = 1 A.
3. A single circular loop of radius 10 cm carries 2 A. Find B at its centre.
B = μ₀I / 2R = 4π × 10⁻⁷ × 2 / 0.20 = 4π × 10⁻⁶ ≈ 1.26 × 10⁻⁵ T, along the axis.
4. A coil of 50 turns and radius 5 cm carries 1 A. Find B at the centre.
B = μ₀NI / 2R = 4π × 10⁻⁷ × 50 × 1 / 0.10 = 2π × 10⁻⁴ ≈ 6.3 × 10⁻⁴ T.
5. A loop has radius 3 cm. At a point on its axis 4 cm from the centre, what fraction of the centre field is present?
B_axis / B_centre = R³ / (R² + x²)^{3/2} = 27 / (9 + 16)^{3/2} = 27 / 125 = 0.216. So about 22 % of the centre field.
6. A long solenoid has 1000 turns per metre and carries 2 A. Find B inside.
B = μ₀nI = 4π × 10⁻⁷ × 1000 × 2 ≈ 2.5 × 10⁻³ T.
7. Two long parallel wires 10 cm apart carry 5 A each in opposite directions. Find B midway between them.
Each wire gives B = 2 × 10⁻⁷ × 5 / 0.05 = 2 × 10⁻⁵ T. With opposite currents both fields point the same way at the midpoint, so B = 4 × 10⁻⁵ T. (With currents in the same direction they would cancel: B = 0.)
Common mistakes
- Using μ₀I / 2R for a straight wire, or μ₀I / 2πr for a loop. Wire has 2π, loop centre has only 2.
- Forgetting to convert cm to m before putting r or R in the formula.
- Counting currents outside the Amperian loop in I_enclosed. Only the current passing through the loop counts.
- Thinking the solenoid field depends on its radius. B = μ₀nI depends only on turns per metre and current.