Power transmission, transformer efficiency and the DC motor
الأسبوع 11 · الفصل الثالث (٤): المحولات في نقل الطاقة الكهربية وفقد الطاقة وكفاءة المحول وأمثلة محلولة، والمحرك الكهربي (الموتور) وفكرة عمله ودورة كاملة له · 3 دروس
2
The DC motor: construction and principle
Physics Sec 3 Term 1, Chapter 3: pp. 80–85
GoalName the parts of a simple DC motor, say what each one does, and find the force on the coil sides and the torque when the coil’s plane is parallel to the field.
New words
DC motor · a machine that changes electric energy into mechanical (turning) energycommutator · a split ring of two halves that reverses the current in the coil twice in each turnbrush · a fixed contact that touches the turning commutator and carries current in or out
Explanation
A DC motor changes electric energy into mechanical energy. Its main part is a rectangular coil of many turns of insulated wire, wound on a laminated soft-iron core and free to turn between the poles of a strong magnet. The two ends of the coil are joined to the two halves of a split ring, the commutator, which turns with the coil. Two fixed brushes press on the halves and bring in the current from a DC source, often through a rheostat. The idea is the one you met in the galvanometer. The current in the two long sides of the coil flows in opposite directions, so by Fleming’s left-hand rule the forces on them are opposite, and they form a couple. Each side of length ℓ inside the field feels F = N·B·I·ℓ, and when the plane lies along the field the torque is F × (width of the coil) = BIAN, with A the area of the coil. The galvanometer is held back by a spring, but a motor must keep turning in one sense. That is why the current must be reversed twice in every turn: the commutator does it. We only draw and calculate with motors here; we do not build or open any.
Part
What it does
coil of many turns on a laminated soft-iron core
carries the current; the iron strengthens the field and the sheets cut eddy currents
strong magnet
gives the magnetic field B across the coil
split ring (commutator)
two halves joined to the two ends of the coil; turns with it
two brushes
fixed contacts that press on the halves and bring in the DC
rheostat
changes the current, and so the torque
Worked examples
Quantity
Value
turns N
25
ℓ (side in the field)
0.08 m
width
0.06 m
B
0.3 T
I
4 A
F on one side
2.4 N
torque
0.144 N·m
A coil has N = 25 turns. Each long side is ℓ = 0.08 m long inside the field, the width of the coil is 0.06 m, B = 0.3 T and I = 4 A. The force on one side is F = N·B·I·ℓ = 25 × 0.3 × 4 × 0.08 = 2.4 N. With the plane parallel to the field the torque is F × width = 2.4 × 0.06 = 0.144 N·m. Check with BIAN: A = 0.08 × 0.06 = 4.8 × 10⁻³ m², so 0.3 × 4 × 4.8 × 10⁻³ × 25 = 0.144 N·m.
Side
Field
Current
Force
ab
east
north
down
cd
east
south
up
The field points east (from the north pole on the west to the south pole on the east). In side ab the current flows north, and in side cd it flows south. Fleming’s left-hand rule for ab: forefinger east, middle finger north, thumb down. So the force on ab is downward. In cd the current is reversed, so the force is upward. The two sides are pushed in opposite ways, which turns the coil.
The torque is proportional to B, I, A and N. For the coil of Example 1, raising the current from 4 A to 6 A raises the torque in the ratio 6 ÷ 4: 0.144 × 6 ÷ 4 = 0.216 N·m. Halving the field would halve it again. Reversing the current (swapping the battery leads) reverses the torque, so the motor turns the other way.
Class activity
Paper only; never open or build a motor. Draw the top view of a rectangular coil between a north pole on the left and a south pole on the right. Label the sides ab and cd, draw the current directions as arrows, and use the left-hand rule to draw the force on each side. Then draw the split ring with its two brushes. Repeat the drawing after turning the battery leads round, and say what changed.
Practice
1
A coil of N = 20 turns has sides ℓ = 0.1 m inside a field B = 0.5 T and carries I = 3 A. Find the force on one side.
F = N·B·I·ℓ = 20 × 0.5 × 3 × 0.1 = 3 N.
2
The same coil is 0.04 m wide and its plane is parallel to the field. Find the torque.
τ = F × width = 3 × 0.04 = 0.12 N·m.
3
The field points north and the current in a side flows west. Which way does the force on that side point?
Forefinger north, middle finger west: the thumb of the left hand points down. With the current to the east the force would be up, so west gives down.
4
Why is the iron core of a motor made of thin insulated sheets and not of one solid piece?
In a solid core the changing flux would drive large eddy currents, which only heat the iron and waste energy. Thin insulated sheets keep these currents small.
