Ampere’s circuital law and the solenoid
Ampere’s circuital law
Draw any closed loop in space. Walk round it in small steps dl. At each step, multiply the step by the part of B along it, B·dl, and add them all up round the loop.
Ampere’s circuital law says this sum, ∮B·dl, equals μ₀ times the total current I passing through the surface bounded by the loop. ∮B·dl equals μ₀ I.
Currents outside the loop do not count. And signs matter: curl the fingers of your right hand in the direction you walk round the loop; currents along your thumb count as positive.
This law is to Biot–Savart what Gauss’s law is to Coulomb’s law. It is always true, but it is most useful when symmetry lets us choose a loop on which B is constant and along the loop.
Then the sum becomes simply B times L, the length of the loop. So B L equals μ₀ I e, where I e is the current enclosed.
Take a long straight wire. Its field lines are circles around it, so choose a circle of radius r centred on the wire. B is the same all round it and along it.
B times 2πr equals μ₀ I. So B equals μ₀ I over 2πr. The field of a long straight wire falls as 1 over r.
For example, 5 amperes at 10 centimetres: B equals 2 × 10⁻⁷ times 5 over 0.1, which is 1 × 10⁻⁵ tesla. About as strong as the Earth’s field.
Direction: grasp the wire in your right hand with the thumb along the current. Your fingers curl the way the field lines go.
Inside a thick wire of radius a, a circle of radius r encloses only the fraction r² over a² of the current. Then B equals μ₀ I r over 2πa²: zero at the centre, rising to a maximum at the surface.
The solenoid
A solenoid is a long wire wound as a tight helix, turn after turn, like a spring. Each turn is very nearly a circular loop.
Cut it lengthwise and look at the field lines. Between neighbouring turns the fields cancel. Inside, the fields of all the turns add up along the axis.
For a long solenoid the field inside is uniform: straight, parallel, evenly spaced lines. Outside, the lines spread out so much that the field is nearly zero.
Now apply Ampere’s law to a rectangle. One long side, of length h, runs inside the solenoid, along the axis. The opposite side is outside.
Along the inside side, B·dl gives B h. Outside, B is about zero. On the two short sides, B is perpendicular to the path or zero, so they add nothing. So ∮B·dl equals B h.
If there are n turns per unit length, the rectangle encloses n h turns, each carrying I. The current through it is n h I.
So B h equals μ₀ n h I, and B equals μ₀ n I. The field depends only on the turns per metre and the current, not on the radius or on where you are inside.
Example: a solenoid 0.5 metre long, of radius 1 centimetre, has 500 turns and carries 5 amperes. n is 500 over 0.5, which is 1000 turns per metre.
B equals 4π × 10⁻⁷ times 1000 times 5, which is 6.28 × 10⁻³ tesla. Put a soft iron core inside, and the field becomes much stronger: an electromagnet.
Try it: Ampere’s law at work
- Wire: grow the white Amperian circle. B falls as 1/r, but 2πr grows, so B × 2πr always equals μ₀I.
- Tick “Reverse the current”: the field lines go round the other way.
- Switch to Solenoid and press “Side view”: the lines inside are straight and evenly spaced.
- Check B = μ₀nI with n = 1000 per metre and I = 5 A: 6.28 mT.
Field near a long wire
A long straight wire carries a current of 10 A. What is the magnetic field 20 cm from it? (μ₀ = 4π × 10⁻⁷ T m A⁻¹)
- 1.0 × 10⁻⁵ T
- 5.0 × 10⁻⁶ T
- 3.1 × 10⁻⁵ T
- 6.3 × 10⁻⁵ T
Answer
A. From Ampere’s law, B × 2πr = μ₀I, so B = μ₀I/2πr = (2 × 10⁻⁷ × 10)/0.2 = 1.0 × 10⁻⁵ T. Using 2r for r gives 5.0 × 10⁻⁶ T; using the loop formula μ₀I/2r gives 3.1 × 10⁻⁵ T; dropping the 2π gives 6.3 × 10⁻⁵ T.
Inside a solenoid
A solenoid of length 0.5 m and radius 1 cm has 500 turns and carries a current of 5 A. What is the magnetic field inside it?
- 1.57 × 10⁻³ T
- 3.14 × 10⁻³ T
- 6.28 × 10⁻³ T
- 0.157 T
Answer
C. n = 500/0.5 = 1000 turns per metre, so B = μ₀nI = 4π × 10⁻⁷ × 1000 × 5 = 6.28 × 10⁻³ T. Using 500 for n gives 3.14 × 10⁻³ T; multiplying by the length instead of dividing gives 1.57 × 10⁻³ T; using the coil formula μ₀NI/2R gives 0.157 T. The radius does not matter for a long solenoid.
Summary
- Ampere’s circuital law:
∮B·dl = μ₀ I - Long straight wire:
B = μ₀ I / 2πr - Long solenoid:
B = μ₀ n I - Inside a thick wire (r < a):
B = μ₀ I r / 2πa²
- ∮B·dl round any closed loop equals μ₀ times the current enclosed; currents outside do not count.
- It gives B easily when symmetry makes B constant along a chosen loop, so that ∮B·dl = BL.
- The field lines of a straight wire are circles; the right-hand thumb rule gives their direction.
- Inside a long solenoid the field is uniform and along the axis; outside it is nearly zero.