I'm looking at the following problem from "Physics 3" by Halliday, Resnick and Krane (4th edition):

The armature of a motor has 97 turns each of area 190 cm² and rotates in a uniform magnetic field of 0.33 T. A potential difference of 24 V is applied. If no load is attached and friction is neglected, find the rotational speed at equilibrium.

My attempt is as follows:

Initially, the potential difference is 24 V. Because of this potential difference, current starts flowing through the motor. Then, because of the 0.33 T magnetic field, the motor starts rotating. The rotational motion changes the magnetic flux through the motor's coil, which produces an induced EMF (electromotive force) that opposes the original EMF. If we call N the number of turns, A the area of the motor's coil, and θ the angle from the magnetic field vector B to the vector normal to the plane of the coil, then the magnetic flux through the coil is:


So, the induced EMF is:

$\varepsilon_{ind}=-\frac{\mathrm{d} }{\mathrm{d} t}(NAB\cos{\theta})=NAB\frac{\mathrm{d} \theta }{\mathrm{d} t}\sin{\theta}$

If we call ω the angular velocity as a function of time ($\frac{\mathrm{d} \theta }{\mathrm{d} t}$), then the net EMF as a function of time is:

$\varepsilon = 24 - \varepsilon_{ind} = 24- NAB\omega\sin\theta$

Is this correct so far? I'm not sure how to proceed from here. Anyway, the answer given in the back of the book (39.5 rad/s or 6.3 rev/s) seems to suggest that the correct way of finding ω is by using $\varepsilon=NBA\omega$ and plugging in the values, that is, ε = 24 V, N = 97, B = 0.33 T and A = 0.0190 m². This gives the correct result of 6.3 revolutions per second. But I'm not sure how to obtain $\varepsilon=NAB\omega$ from the expression that I found above ($\varepsilon = 24 - NAB\omega\sin\theta$).

Edit: Based on the suggestions given in the comments to the answer below, I understand that the "equilibrium" that this question is referring to is when the torque on the coil reaches the value zero; this happens when the total EMF ($\varepsilon = 24 - NAB\omega\sin\theta$) on the coil equals zero. That is:

$\omega = \frac{24}{NAB\sin\theta}$

To reach the desired result, $\sin\theta$ should equal $1$; but I don't understand why that is the case here. Is this reasoning correct? How should I proceed from here?


3 Answers 3


Equilibrium will be reached when the net torque on the armature is zero. Since, as we will see below, it will be impossible to have the net torque vanish over an extended interval of time, we'll look for the situation when the torque averaged over time vanishes.

The Lorentz force law tells us that for a wire of length vector $\vec{l}$ carrying current $I$ in a uniform magnetic field $\vec{B}$, the force $\vec{F}$ is

\begin{equation} F = I \, \vec{l} \times \vec{B} \; \end{equation}

It follows that the strength of the torque $\tau$ on an armature with $N$ turns and cross-sectional area $A$ is given by

\begin{equation} \tau = N\, I \,A \,B \,\sin \theta \end{equation}

where $\theta$ is the direction between the normal vector of the armature cross-section and the magnetic field. (The geometry is actually slightly more involved than you might imagine at first - I recommend studying the diagram found in the bottom panel of this page).

We now want to know what happens if we take the time average of the torque over some time interval that spans many revolutions of the motor. $N$, $A$, and $B$ are all constant in this problem (although see the discussion at the very end of this answer concerning $B$), so the only quantities that can vanish in the average are $\langle I \rangle$ and $\langle \sin \theta \rangle$.

If this motor possesses no commutator, then the torque will switch directions every half cycle and it will subsequently be difficult for the armature to reach any appreciable speed at all. I will assume that the motor possesses a commutator that switches the current every half cycle. In this case, $\langle \sin \theta \rangle \not = 0 $ and so we must have $\langle I \rangle = 0$ at equilibrium.

In order to achieve $\langle I \rangle = 0$, it must be the case that $\langle \epsilon_\mathrm{net} \rangle = 0$ where $\epsilon_\mathrm{net}$ is the net emf and contains contributions from the 24 volt potential difference and the magnetic induction due to the rotation of the armature (but see the final note at the end of the answer for more details). As already pointed out in the statement of the question,

\begin{equation} \epsilon_\mathrm{net} = 24 \mathrm {\; V } - N \, A \, B \sin \theta \frac{d\theta}{dt}\end{equation}

and then taking the time average, for equilibrium to be achieved,

\begin{equation} \langle \sin \theta \frac{d\theta}{dt} \rangle = \frac{24 \mathrm {\; V }}{N \, A \, B}\end{equation}

Strictly speaking, $\langle \sin \theta \frac{d\theta}{dt} \rangle \not= \langle \sin \theta \rangle \langle\frac{d\theta}{dt} \rangle$ because $\frac{d\theta}{dt}$ will vary with time in a manner correlated to $\sin \theta$ (it is important to remember that we are actually averaging over time, not directly over $\theta$). A lower bound for $\frac{d\theta}{dt}$ is then the answer given in the back of the book. A more correct answer would be larger than that. In the approximation that $\frac{d\theta}{dt}$ is a constant in time (not actually true but perhaps quite a good approximation), then the answer would be multiplied by a factor of $\pi/2 \approx 1.57$.

In summary, I think the question was poorly constructed and you should not fret, in this case, about matching the book's answer exactly.

Final note concerning magnetic field from self-induction

There is an additional magnetic field generated from the current through the wires in the armature. As the current changes in time, the flux of this field through the armature will vary in time. This additional changing flux should in principle affect both the emf in the circuit and the torque on the armature.

However, these effects will be negligible if the magnetic field generated by the coils is negligible compared to the 0.33 Tesla magnetic field already in place. I urge you to calculate what is roughly the maximum current that could flow through the wires before the assumption that the corresponding field is negligible breaks down (I checked, and the current would have to become quite large indeed - but you should still check youself).

  • $\begingroup$ Thank you for the answer. But I have another question: when you wrote that, if the motor has a commutator, $\langle \sin \theta \rangle \not = 0$, did you mean to say that only non-negative values of sine will be used, because torque doesn't change direction? $\endgroup$
    – favq
    Commented Aug 10, 2012 at 19:36
  • $\begingroup$ The angle brackets $\langle \, \rangle$ are meant to denote an average over time. As you say, the commutator acts so that only non-negative values of sine are used. This means that over many cycles, rather than having $\sin \theta$ average to zero as would be the case if there were no commutator, $\sin \theta$ will average to some positive value. $\endgroup$ Commented Aug 10, 2012 at 19:54
  • $\begingroup$ Thank you for your help. This solution makes sense. As you said, this question seems to be poorly constructed, giving rise to many interpretation doubts. Therefore, many assumptions which are not stated in the question were needed in order to reach a solution that is close to the book's answer. $\endgroup$
    – favq
    Commented Aug 10, 2012 at 21:39

Yes you are right. $\varepsilon$ is a function of both $\omega$ and $\sin\theta$. But at equilibrium $\omega$ will become constant by definition. Since $\varepsilon$ can't become negative, thus minimum value it can achieve is 0. And that happens when $NAB\omega\sin\theta$ becomes $0$. To do that, $\sin\theta$ becomes $1$. Solving for $\omega$ we get $$\omega = \frac{24}{NAB}$$

Note $\varepsilon$ will vary as function of $\theta$

Maybe this link can be helpful. http://www.pa.msu.edu/courses/2000spring/phy232/lectures/induction/rotatingcoil.html

  • $\begingroup$ But $\sin \theta$ can't stay equal to 1 as the coil continues to rotate. What is confusing about this question is that it seems to require a time-dependent emf to be set equal to a static emf. But I suppose the question would be more clear, and your answer entirely correct, if the question asked for the rotational speed when the coil reaches instantaneous equilibrium $\endgroup$ Commented Aug 6, 2012 at 20:18
  • $\begingroup$ @kleingordon I'm not sure. I studied this long back. You can look into DC motor. They have a concept of achieving a fixed frequency after some time and allowing variable current flow inside. $\endgroup$
    – Ankush
    Commented Aug 6, 2012 at 20:41
  • $\begingroup$ It now occurs to me that one must also consider the emf generated from the self-induction of the coil, which is non-zero for variable currents. It still doesn't seem to me that the torque can vanish over an extended period of time, it will only vanish instantaneously from time to time. $\endgroup$ Commented Aug 6, 2012 at 21:49
  • $\begingroup$ Thank you for the answer. But I don't quite see why $\sin \theta$ would necessarily stay equal to 1 in this case, because the motor continues to rotate. $\endgroup$
    – favq
    Commented Aug 8, 2012 at 0:17
  • $\begingroup$ @anonymous I never said that. It will vary. The thing is $\omega$ will reach constant value at equilibrium. $\endgroup$
    – Ankush
    Commented Aug 8, 2012 at 5:43

By kirchoff's law, the voltage drop across the motor from resistance and EMF has to be the same as the applied 24V.
$\epsilon = 0 = 24V - \epsilon_{ind}$ The current is in fact oscillatory. DC motors use a commutator to switch the direction of the current after each 180 degrees of rotation. Then the coil wrappings on a motor's armature are staggered across 90 degree so that the $sin(\theta)$ becomes the angle is averaged over all the wrappings. So $sin(\theta) \rightarrow 2/\pi$. This speeds up the motor. But using that would make the motor run faster than the book's answer.


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