Ampere's circuital law: straight wire, solenoid, toroid, electromagnets and displacement current

Ampere's circuital law, and the field of a long straight wire

Imagine any closed path round a current; we call it a **closed loop**. Break it into small pieces Δl, multiply each Δl by the part of B along it (B·Δl), and add these all round the loop. The sum, ΣB·Δl or ∮ **B**·d**l**, is the **line integral** of B. **Ampere's circuital law:** the line integral of B round any closed loop equals **μ₀ times the total current passing through the loop**. Only the current enclosed counts, not the size or shape of the loop. The law is easy to use when **symmetry** makes B the same size all along the loop. Like the Biot-Savart law, it gives the field produced by currents.

1. 1
   Draw a circle of radius r round the wire, centred on it, in a plane at right angles to it.
2. 2
   By symmetry, B has the same size at every point of this circle and points along it (along the tangent).
3. 3
   So ΣB·Δl = B × ΣΔl = B × 2πr.
4. 4
   The current through the circle is I.
5. 5
   Ampere's law: B × 2πr = μ₀I.
6. 6★
   So, B = μ₀I/2πr. The field halves when the distance doubles.

★ key step

Derivation: field inside a long solenoid and inside a toroid

A **solenoid** is a long coil of many closely wound turns along a straight line. A **toroid** is a solenoid bent round into a ring, like a doughnut.

1. 1
   Solenoid: inside, the field is uniform and along the axis. Outside, it is so weak that we take it as zero.
2. 2
   Take a rectangle abcd. Side ab, of length l, lies inside along the axis; side cd lies outside.
3. 3
   Along ab, B·Δl adds up to Bl. Along cd, B = 0. Along bc and da, B is zero outside and at right angles to the path inside. So ΣB·Δl = Bl.
4. 4
   If n is the number of turns per metre, the rectangle encloses nl turns. The current through it is nlI.
5. 5★
   Ampere's law: Bl = μ₀nlI. So B = μ₀nI. Near each end of a long solenoid, B is about half of this, μ₀nI/2.
6. 6
   Toroid: take a circle of radius r inside the ring, with the same centre. B is the same size all along it and along its tangent. So ΣB·Δl = B × 2πr.
7. 7★
   The circle is threaded by all N turns, so the current through it is NI. Then B × 2πr = μ₀NI, so B = μ₀NI/2πr. Outside the toroid, the field is zero.

★ key step

How to draw the solenoid, the rectangle abcd and the toroid

• Draw two parallel rows of small circles, top and bottom, for the turns cut along their length.
• Put ⊙ in every circle of the top row and ⊗ in every circle of the bottom row, to show the current.
• Inside, draw straight, parallel, equally spaced lines along the axis, with arrows.
• Outside, draw a few lines that leave one end, spread out and curve round to enter the other end. Label the end where the lines come out N and the other end S.
• For the derivation, draw the rectangle abcd: ab inside along the axis (length l), cd outside, and bc and da crossing the windings.
• For a toroid, draw a ring of turns and a dotted circle of radius r inside the ring, with B drawn along this circle.

Electromagnets and what makes them stronger

• A solenoid carrying a current behaves like a bar magnet. Its ends act as N and S poles, found by the end rule.
• It is a magnet only while the current flows, so it is called an electromagnet.
• Its strength, B = μ₀nI, grows with the number of turns per metre n and with the current I.
• A soft iron core inside the coil makes the field many times stronger.
• Soft iron loses its magnetism as soon as the current stops, so the magnet can be switched on and off.
• Uses: electric bells, cranes that lift scrap iron, relays, motors, fans, toys and transformers.

Displacement current (idea only)

Maxwell noticed a problem with Ampere's law. Think of a capacitor being charged. A current I flows in the wires, but no charge crosses the gap between the plates. Take one closed loop round the wire, with two different surfaces ending on it. A flat surface cut by the wire gives ΣB·Δl = μ₀I. A surface that bulges out through the gap carries no current and gives zero. Both cannot be right. Maxwell saw that the electric field between the plates is changing. The electric flux there is φ_E = q/ε₀, so ε₀ Δφ_E/Δt = Δq/Δt = I. He called ε₀ Δφ_E/Δt the displacement current. It equals the current in the wires, but it is not a flow of charges. The corrected law is ΣB·Δl = μ₀(I + ε₀ Δφ_E/Δt). So a changing electric field also produces a magnetic field.

Worked example: wire, solenoid and toroid