# Rotational speed of a coil in a magnetic field

1. Jun 24, 2012

### pc2-brazil

This is for self-study.

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution
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 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:
$$\Phi=NAB\cos{\theta}$$
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, and consider that the angle in t = 0 is 0º, we have that θ = ωt, so the net EMF in function of time is:
$$\varepsilon = 24- NAB\omega\sin\omega t$$
Is this correct so far? I'm not sure how to proceed from here, but I think that the rotational speed in equilibrium is reached when the angular acceleration α = 0.

2. Jun 24, 2012

### TSny

(1) For equilibrium, how should the average induced emf compare to the applied voltage?

(2) Note that your expression for the induced emf would average to zero due to the switch in sign of the sine function every half turn. However, a DC motor has a commutator that effectively switches the connection of the coil to the outside circuit every half turn. This will make the output emf generated by Faraday's law always have the same sign. So, you can get the average induced emf by averaging your expression over half a turn.

3. Jul 3, 2012

### pc2-brazil

Thank you for the suggestion.
I think the motor in this question doesn't involve a commutator, because this wasn't mentioned in the book in which I found it ("Physics" by Halliday, Resnick & Krane, 4th edition).

Anyway, the answer given in the back (39.4 rad/s or 6.3 rev/s) seems to suggest that the value of ω is obtained by using $$\varepsilon=NBA\omega$$ and plugging in ε = 24 V, N = 97, B = 0.33 T and A = 0.0190 m².
But I'm not sure how to obtain $\varepsilon=NAB\omega$ from the original expression that I wrote, $\varepsilon = 24- NAB\omega\sin\omega t$. Any hints?

4. Jul 3, 2012

### CWatters

If I've understood the question the bits you are missing are..

a) NABωsinωt is at a maximium when sinωt=1

b) Once upto speed the back emf (aka induced voltage) = the applied voltage.

5. Jul 10, 2012

### pc2-brazil

Thank you for the response. But this raises a doubt:

Since the induced voltage changes with time, and the applied voltage is constant (24 V), how does this system reach equilibrium? Does the maximum induced voltage have to be equal to the applied voltage? If so, why exactly is that true?