☰ Year plan · Physics

The magnetic field of a wire and of a loop

Week 4 · Chapter 2 Magnetic Effects of Electric Current and Measuring Instruments (A): Oersted’s observation, the field of a straight wire (flux density B = μI/2πd, Ampere’s right-hand rule) and of a circular loop (B = μNI/2r, the right-hand screw rule, the loop as a magnetic dipole) · 3 lessons
3

The field of a circular loop and the right-hand screw rule

Physics Sec 3 Term 1, Chapter 2: pp. 31–35
GoalCalculate the flux density at the centre of a circular coil with B = μ0NI ÷ 2r and find its poles with the right-hand screw rule.
New words
circular coil · N turns of wire wound into a circle of radius rright-hand screw rule · turn a right-handed screw the way the current turns; the screw moves along the fieldmagnetic dipole · a north and a south pole close together, like a short bar magnet
Explanation

Bend a straight wire into a circle and the field lines pass through the loop in the same direction, so the fields of all the turns add up. At the centre of a coil of N turns, radius r and current I, B = μ0NI ÷ (2r). More turns, more current or a smaller radius give a stronger field. With π = 3.14 this becomes B = 6.28 × 10⁻⁷ × NI ÷ r. The radius is measured in metres from the centre to the wire, so a diameter must be halved. The direction comes from the right-hand screw rule: if the current turns the way a screw is turned to drive it in, the field points the way the screw moves. The face the field lines leave is a north pole and the other face a south pole. Looking at a face, an anticlockwise current makes it a north pole and a clockwise current makes it a south pole. So a loop acts like a short bar magnet, a magnetic dipole, and poles always come in pairs.

NS
Worked examples
N = 50 turns, I = 2 A and r = 0.10 m. B = 6.28 × 10⁻⁷ × 50 × 2 ÷ 0.10 = 6.28 × 10⁻⁴ T.
ChangeEffect on B at the centre
N doubledB doubles
I doubledB doubles
r doubledB halves
r halvedB doubles
A single turn N = 1 carries I = 5 A with r = 0.05 m. B = 6.28 × 10⁻⁷ × 5 ÷ 0.05 = 6.28 × 10⁻⁵ T. Doubling the radius would halve B, and doubling the current would double it.
Looking at the front of a coil, the current goes anticlockwise. The front face is a north pole and the back face is a south pole. If the current is reversed, the poles swap.
Class activity

Curl the fingers of your right hand along an anticlockwise current in a loop drawn on paper: your thumb shows the field coming towards you, so you are looking at a north pole. Do the same for a clockwise loop. Then draw three loops with arrows and let a friend say which face towards them is north and which is south.

Practice
1
A coil has N = 40 turns, I = 2.5 A and r = 0.08 m. Find B at its centre (π = 3.14).
2
Looking at the front of a loop, the current goes clockwise. What pole is the front face?
3
How many turns of radius 0.05 m carrying 1 A give B = 3.14 × 10⁻⁴ T at the centre? Use μ0 = 1.256 × 10⁻⁶ T·m/A.
4
The radius of a coil is halved and its current is doubled, with the same N. How does B at the centre change?