Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 4.2.5

Interactive: DC Motor and Commutator

Problem 4.2.5

DC motor

DC Motor: $B = B_{y} a_{y}$ . The rectangular loop rotates around the $x$ -axis. Assignment: How can we make this rotate infinitely?

Try it yourself first.

Strategy

Put an angle on it: describe the position of the loop by its rotation angle $α$ about the shaft, write $m (α)$ , and use $T = m \times B$ of Problem 4.2.4. The sign of the torque as a function of $α$ answers the assignment.

1. The moment as a function of the angle

Let the loop have the area $S$ and let $α$ be the angle of rotation about the $x$ -axis measured from the position in which the loop lies in the $x y$ -plane with $a_{n} = a_{z}$ . Rotating $a_{z}$ about $a_{x}$ by $α$ gives $a_{n} = cos α a_{z} - sin α a_{y}$ , so

m = I S (cos α a_{z} - sin α a_{y}) .

2. Torque

T = m \times B = I S B_{y} cos α (a_{z} \times a_{y}) - I S B_{y} sin α (a_{y} \times a_{y}) = - I S B_{y} cos α a_{x} .

Answer

T_{x} = - I S B_{y} cos α

3. Reading the result

$∣ T ∣$ is largest at $α = 0$ , where $m ⊥ B$ and the loop lies along the field (the "start" position of the figure).
$T = 0$ at $α = 9 0^{\circ}$ and at $α = 27 0^{\circ}$ , the two dead points. For $- 9 0^{\circ} < α < 9 0^{\circ}$ the torque is negative, so from $α = 0$ the loop turns towards smaller $α$ , that is towards $α = 27 0^{\circ}$ ( $\equiv - 9 0^{\circ}$ ), where $m = + I S a_{y}$ is parallel to $B$ : the stable position of least energy. At the other dead point, $9 0^{\circ}$ , the moment $m = - I S a_{y}$ is antiparallel to $B$ , an unstable position.
The loop coasts through $27 0^{\circ}$ on its inertia, and there $cos α$ changes sign: the torque reverses and now brakes. With a fixed current the loop cannot complete the turn. It oscillates about $α = 27 0^{\circ}$ and settles there (Figure 4.2.20).

4. Answer to the assignment

Reverse the current every half turn, at the dead points, so that $m$ flips whenever $cos α$ changes sign. Then $T_{x} = - I S B_{y} ∣ cos α ∣$ always has the same sign and the loop keeps turning, in the sense of decreasing $α$ . Mechanically this is the split-ring commutator with its brushes, drawn on the shaft in the figure: the two half-rings exchange their brushes every half turn. Electronically it is the inverter bridge of a brushless DC motor, switched by a position sensor.

Problem 4.2.5. The torque over one turn. With a fixed current (solid) it reverses every half turn and the loop only oscillates. With the current reversed at the dead points (dashed) the torque keeps its sign, pulsing between 0 and ISB_y. — **Figure 4.2.20.** Problem 4.2.5. The torque over one turn. With a fixed current (solid) it reverses every half turn and the loop only oscillates. With the current reversed at the dead points (dashed) the torque keeps its sign, pulsing between $0$ and $I S B_{y}$ .

Run it without a commutator and see the loop settle at the stable dead point, then switch the commutator on.

InteractiveLoading simulation…

Check

At $α = 0$ the result reduces to Problem 4.2.4 with $B_{x} = B_{z} = 0$ and $m = + I S a_{z}$ (the current here circulates anticlockwise seen from $+ z$ ), $T = m \times B = I S B_{y} a_{z} \times a_{y} = - I S B_{y} a_{x}$ , as found. One loop still pulses between zero and $I S B_{y}$ . Real armatures carry many loops and many commutator segments, which fill in the dips, and the mean torque is proportional to $I$ : this is the motor constant $k_{T}$ of the DC machine.

Problem 4.2.5: DC motor