☰ خطة العام · الفيزياء (بالإنجليزية)

Torque on a coil and the moving-coil galvanometer

الأسبوع 6 · الفصل الثاني (٣): أمثلة على القوة المغناطيسية، وعزم الازدواج المؤثر على ملف مستطيل، والجلفانومتر ذو الملف المتحرك وحساسيته · 3 دروس
1

Force problems and a coil in a magnetic field

Physics Sec 3 Term 1, Chapter 2: pp. 41–45
GoalUse F = BIℓ sin θ in problems and explain why a coil whose plane is parallel to a field starts to turn.
New words
couple · two equal, opposite, parallel forces that do not act along the same line; they turn a body without pushing it alongplane of the coil · the flat surface that the turns of the coil surroundside along the field · a side of the coil that lies along the field lines; the force on it is zero because sin 0° = 0
Explanation

Week 5 gave F = BIℓ sin θ for a straight wire, where θ is the angle between the wire (the direction of the current) and the field. Now wind the wire into a rectangular coil with N turns and hold it with its plane parallel to a uniform field. The two long sides are at right angles to the field, so each feels the largest force, F = BINℓ, where ℓ is the length of a long side (the N turns act together, so their forces add). The current runs in opposite directions in the two long sides, so Fleming’s left-hand rule pushes one side up and the other down. The two short sides lie along the field, θ = 0, so they feel no force. The two long-side forces are equal, opposite and parallel but not on the same line: they form a couple. The total force on the coil is zero, yet the coil turns.

Side of the coilPosition relative to the fieldForce
long side 1at right anglesBINℓ, upward
long side 2 (current opposite)at right anglesBINℓ, downward
two short sidesalong the fieldzero
Worked examples
A wire of length ℓ = 0.40 m carries I = 5 A in a field of B = 0.15 T. At right angles to the field, F = BIℓ = 0.15 × 5 × 0.40 = 0.3 N. If the wire makes 30° with the field, F = 0.15 × 5 × 0.40 × sin 30° = 0.15 N.
A wire 0.25 m long carries 8 A at right angles to a uniform field and feels a force of 0.5 N. B = F ÷ (Iℓ) = 0.5 ÷ (8 × 0.25) = 0.5 ÷ 2 = 0.25 T.
A coil of N = 20 turns has long sides of 0.12 m and short sides of 0.05 m. It carries 3 A in a field of 0.4 T, with its plane parallel to the field. Each long side feels F = BINℓ = 0.4 × 3 × 20 × 0.12 = 2.88 N, the two forces are opposite, and the two short sides feel 0 N. The total force on the coil is 0 N, but the couple turns it.
Class activity

Paper practice, no magnets or wires. Sketch a rectangle for the coil and draw field arrows along its two short sides. Write “zero” beside each short side and “BINℓ” beside each long side, and draw the two forces as arrows in opposite directions. In one sentence, explain why the coil turns although the total force on it is zero.

Practice
1
A wire 0.6 m long carries 2.5 A at right angles to a field of 0.08 T. Find the force.
2
The same wire is turned until it makes 30° with the field. Find the force.
3
A wire 0.45 m long carries 4 A at right angles to a uniform field and feels a force of 0.9 N. Find B.
4
A coil of 50 turns has long sides of 0.08 m and carries 1.5 A in a field of 0.2 T, with its plane parallel to the field. Find the force on each long side and the total force on the coil.