The solenoid, the force on a wire and parallel wires
Week 5 · Chapter 2 (B): the solenoid (B = μnI = μNI/ℓ, poles by the screw rule), the force on a current-carrying wire in a magnetic field (F = BIℓ sin θ, Fleming’s left-hand rule, the tesla) and the force between two parallel wires · 3 lessons
1
The solenoid and its magnetic field
Physics Sec 3 Term 1, Chapter 2: pp. 36–41
GoalCalculate the flux density inside a long solenoid with B = μ0NI ÷ ℓ and find its poles.
New words
solenoid · a long coil of many turns of wire wound closely in a spiral, like a springturns per metre (n) · the number of turns divided by the length: n = N ÷ ℓcore · the material inside the coil; a core of soft iron makes the field much stronger
Explanation
A solenoid is a long, closely wound coil. Each turn makes the field of a loop, and inside the solenoid all the turns help each other, so the field is strong and nearly uniform, with straight field lines along the axis. Outside, the lines spread out and return from one end to the other, like those of a bar magnet. In the middle of a long solenoid in air, B = μ0NI ÷ ℓ = μ0nI: B grows with the current and with the turns per metre n = N ÷ ℓ, and it does not depend on the radius of the coil. The formula is for the inside of a long solenoid, not for every point near it. If the coil is wound on a soft-iron core, μ0 is replaced by the much larger permeability μ of the core. The poles follow the screw rule of Week 4: looking at one end, a clockwise current makes it a south pole and an anticlockwise current makes it a north pole.
Worked examples
N = 500 turns, ℓ = 0.25 m and I = 1.5 A, in air (μ0 = 1.256 × 10⁻⁶ T·m/A). B = 1.256 × 10⁻⁶ × 500 × 1.5 ÷ 0.25 = 1.256 × 10⁻⁶ × 3000 = 3.77 × 10⁻³ T.
A solenoid has 2000 turns per metre and carries 0.8 A, in air. B = μ0nI = 1.256 × 10⁻⁶ × 2000 × 0.8 = 2.01 × 10⁻³ T.
Core
Permeability (T·m/A)
Current for B = 0.2 T
air
1.256 × 10⁻⁶
about 32 A
iron (invented value)
5 × 10⁻³
8 mA
An iron core with an invented permeability μ = 5 × 10⁻³ T·m/A fills a solenoid with N = 1000 turns and ℓ = 0.2 m. To get B = 0.2 T, I = Bℓ ÷ (μN) = (0.2 × 0.2) ÷ (5 × 10⁻³ × 1000) = 0.04 ÷ 5 = 8 × 10⁻³ A = 8 mA. In air the same B would need I = 0.04 ÷ (1.256 × 10⁻⁶ × 1000) ≈ 32 A, about 4000 times more.
Class activity
Paper practice, no wires or cells. Draw a solenoid as five loops with arrows showing a current. Use the right-hand screw rule to decide which end is the north pole, mark it N and the other end S, and compare with the bar magnet drawn above. Then reverse the arrows and say what changes.
Practice
1
A solenoid of 800 turns and length 0.4 m carries 2 A in air (μ0 = 1.256 × 10⁻⁶ T·m/A). Find B inside.
B = 1.256 × 10⁻⁶ × 800 × 2 ÷ 0.4 = 5.02 × 10⁻³ T.
2
A solenoid has 5000 turns per metre and carries 0.2 A in air (μ0 = 1.256 × 10⁻⁶ T·m/A). Find B inside.
B = μ0nI = 1.256 × 10⁻⁶ × 5000 × 0.2 = 1.26 × 10⁻³ T.
3
A solenoid has an invented iron core with μ = 2 × 10⁻³ T·m/A, N = 500, ℓ = 0.2 m and I = 0.04 A. Find B.
B = μNI ÷ ℓ = 2 × 10⁻³ × 500 × 0.04 ÷ 0.2 = 0.2 T.
4
Looking at the left end of a solenoid, the current goes anticlockwise. What pole is the left end?
A north pole: an anticlockwise current makes the face you look at a north pole.
