Section 4.1 · Worked examples · 4 of 4
Worked example 4.1.4: Force and torque on a DC-motor armature
About the worked examples of Section 4.1
Four short calculations tied to the applications. They use only , its wire form and the long-wire field of Problem 4.1.1, with no new physics. Constants used: H/m ( H/m) and C. A good way to use them: read the given quantities, produce the formula from the Lorentz force or from Problem 4.1.1 yourself, and only then put the numbers in.
Force and torque on a DC-motor armature
#cf. Application 4.1.2
Given. An armature coil of turns, long sides cm, width cm, carries A in the gap field T. The motor runs at rpm.
Find. (a) the force on one long side, for one turn and for the coil, (b) the torque when the coil plane is parallel to , (c) the mean torque over a revolution with a commutator, (d) the mechanical power.
On each long side with , so . The current runs opposite ways along the two sides, so the two forces are equal and opposite: no net force, but a couple about the shaft, with and the angle between and the normal of the coil, Eq. (4.1.24).
One turn: N. The sides of the coil lie side by side, so N on each long side, in opposite directions.
The two forces are cm apart and perpendicular to the coil plane ():
the same by both routes.
As the coil turns, would reverse every half turn. The split-ring commutator reverses exactly when changes sign, so the torque is rectified, , and its mean over a revolution is
rad/s and W.
The defaults are this example: read , , the mean torque and at 1500 rpm.
Twenty newtons, the weight of kg, on a cm side from A in T: a hand-sized motor of about W. A real armature has many coils spaced round the rotor, each commutated at its own moment, so the total torque stays close to instead of pulsating. The brushless motor replaces the commutator by electronic switching but uses the same .