# DC motor turning at constant speed eventually

1. Sep 22, 2012

### sgstudent

Hi, I have a question related to the DC motor that i hope you geniuses can help me with.

By Fleming's Left hand rule, there is an induced forced (IF in diagram) at the downwards position at the armature AB. Thus, by Fleming's Right hand rule there is an induced current (IC in diagram) which opposes the current that is supplied by the battery (C in the diagram). As the induced force is directly proportional to the 'net' current flowing through AB, the induced force slowly decreases. This is because the induced current will keep increasing since there is a net moment about P which means there is an angular acceleration. After a while, there will be no more net moment as the induced current keeps on increasing which in turn causes the induced force to decrease. Since net moment about P=IF*l-friction*l, so IF*l >0 hence, the induced current is always lesser than the current supplied by the battery. This causes the 'net' current flowing across the circuit to be smaller than the inital current when the induced current in the opposite direction has not increased yet.

The image: http://i.imgur.com/2pkmq.png

Is this correct? Thanks for the help

2. Sep 23, 2012

### Philip Wood

You have the right idea. Here are some nitpicks:
• 'Induced force' isn't the usual term. More usual names would be 'motor effect force' or '(magnetic) Lorentz force'.
• It's more usual in Physics to think of a single current in the coil, not a battery current and an induced current flowing in opposite directions at the same time (which, though no doubt mathematically equivalent, isn't what's going on). Instead, we consider that, as it moves, the magnetic flux linked with the coil changes, inducing an opposing emf of magnitude εind. The net emf in the circuit is therefore (εbatt – εind). Thus the current in the coil is (εbatt – εind)/R.
• Ideally, before your third sentence, you'd explain that because the coil is acted upon by the force (strictly by a combination of two such forces, giving a torque) it rotates faster and faster, increasing εind), and so decreasing the current.
Sorry – I've just re-read your fourth sentence, which says pretty much the same thing, though, as I've explained, I think you should keep off 'induced current'.

3. Sep 23, 2012

### sgstudent

Hi Mr Wood thanks for the help, I'll take note of the mistakes. That being said, will this be similar to the AC generator whereby instead of the induced emf, there is a motor effect force that opposes your hand motion. So eventually, the hand turns with constant speed. That bing said, will the induced emf's peak voltage increase at a decreasing rate until it stabilises? Thanks for the help

4. Sep 23, 2012

### Philip Wood

Edited (in fact rewritten so it makes some sense – I hope)...

The opposing (motor effect) torque is proportional to the speed of turning, if the motor coil is in a circuit of constant resistance, R. So for a fixed applied torque, the motor will indeed reach a speed at which (applied torque – motor effect torque) just balances the frictional torque. If we neglect the frictional torque then the governing equation is
$$G_{app} - G_{mot} = I\frac{d\omega}{dt}$$
in which I is the moment of inertial of the rotating assembly, and ω is its angular velocity.

If we assume that the B field is radial then $G_{mot} = {K\omega}$

in which, we can show, $K = \frac{(BAn)^2}{R}$

Thus $G_{app} - K\omega = I\frac{d\omega}{dt}$

As you can check by substitution, the solution to this (assuming ω=0 at t=0) is
$$\omega = \left(\frac{G_{app}}{K}\right)\left[1 - exp\left(\frac{-Kt}{I}\right)\right]$$.
Thus ω approaches a final speed, more and more slowly, just as you say.
The final speed is $\omega = \left(\frac{G_{app}}{K}\right)$
But note: you won't necessarily apply a constant torque with your hand.

Last edited: Sep 23, 2012
5. Sep 23, 2012

### sgstudent

Hi Mr Wood, I'm not very sure what you mean by εind because in the AC generator won't there be a motor effect force rather than a induced emf? Thanks again for the help!

6. Sep 23, 2012

### Philip Wood

sgstudent: You're absolutely right not to be sure what I meant: it was rubbish. I'm editing it!

7. Sep 23, 2012

### sgstudent

haha thanks for the help Mr Wood!