Finding average back emf

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A motor with a brush-and-commutator arrangement has a circular coil with radius 2.5 cm and 150 turns of wire. The magnetic field has magnitude 0.06 T and the coil rotates at 440 rev/min.


What is the average back emf?

i have tried this

Flux=Magnetix Field*Area
Flux=0.06*pi*r^2
Here r=2.5cm=2.5*10^-2m
Avarage Emf=N*Flux/Time
Time=1/Speed
Speed=440 rpm=440/60=7.3333
Average Emf=Flux*150/Time

so average Emf = 0.06*phi*r^2*150 / 0.136363 = 0.1295 V

but, my answer is wrong..

please advise!
 

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  • #2
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help please.... :cry:
 
  • #3
Redbelly98
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A motor with a brush-and-commutator arrangement has a circular coil with radius 2.5 cm and 150 turns of wire. The magnetic field has magnitude 0.06 T and the coil rotates at 440 rev/min.


What is the average back emf?

i have tried this

Flux=Magnetix Field*Area
Flux=0.06*pi*r^2
Here r=2.5cm=2.5*10^-2m
Avarage Emf=N*Flux/Time
Time=1/Speed
Speed=440 rpm=440/60=7.3333
Average Emf=Flux*150/Time

so average Emf = 0.06*phi*r^2*150 / 0.136363 = 0.1295 V

but, my answer is wrong..

please advise!
Your "time" is the time, in seconds, for one complete revolution of the coil.

Note, however, that flux does not simply increase from zero to 0.06*pi*r^2 during this time. You'll have to think more carefully about how the flux varies during a revolution of the coil.

p.s.
Welcome to PF :smile:
 
  • #4
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Your "time" is the time, in seconds, for one complete revolution of the coil.

Note, however, that flux does not simply increase from zero to 0.06*pi*r^2 during this time. You'll have to think more carefully about how the flux varies during a revolution of the coil.
why it's not increase from zero to 0.06*pi*r^2 ????
i cant think other method :(
 
  • #5
lanedance
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think about the projection of the coil onto a plane perp to mag field, this feeds into the area used to calculate flux & so the instantaneous emf

then consider the average of a sinuosoid...
 
  • #6
Redbelly98
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think about the projection of the coil onto a plane perp to mag field, this feeds into the area used to calculate flux & so the instantaneous emf
Another way to think about it, flux depends on the angle between the rotating coil and the magnetic field. It may help for you to draw a figure, or look at a figure in your text book.
 
  • #7
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Another way to think about it, flux depends on the angle between the rotating coil and the magnetic field. It may help for you to draw a figure, or look at a figure in your text book.
the angle is 90 degree rite??

so,

Emf = NBA.sin(theta).w

after did the calculation, i ended up Emf = 0.81 V

but, my answer is also wrong...
 
  • #8
lanedance
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the angle is continuously changing with time at a constant rate. Need to think how this affects EMF - it will be a function of time. Then how to average the produced EMF

so - peak emf is at 90deg, but the angle varies with time
 
  • #9
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the angle is continuously changing with time at a constant rate. Need to think how this affects EMF - it will be a function of time. Then how to average the produced EMF

so - peak emf is at 90deg, but the angle varies with time
i don't understand :cry:
 
  • #10
lanedance
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you've written the EMF is

EMF = N.B.A.cos(theta)

theta is the angle bewteen a vector perpindicluar to the coil plane & field. As the coil is rotating theta varies linearly with t

so let
theta(t) = w.t

where, w is the angular frequency

so as a function of t
EMF(t) = N.B.A.cos(w.t)

so when you average over a full period of rotation, what is the average of the cos(w.t) term?
 
  • #11
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you've written the EMF is

EMF = N.B.A.cos(theta)

theta is the angle bewteen a vector perpindicluar to the coil plane & field. As the coil is rotating theta varies linearly with t

so let
theta(t) = w.t

where, w is the angular frequency

so as a function of t
EMF(t) = N.B.A.cos(w.t)

so when you average over a full period of rotation, what is the average of the cos(w.t) term?
is it..

d (cos wt) / dt ?????

so, -w sin wt ???
 
  • #12
lanedance
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why are you differentiating? a derivative gives you an instantaneous rate of change

what gives you an average, do you know about rms?
 
  • #13
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why are you differentiating? a derivative gives you an instantaneous rate of change

what gives you an average, do you know about rms?
rms?
i just how to find Vmax with rms..
Vrms = 0.707 * Vo
the rest i know nothing :shy:
 
Last edited:
  • #14
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please help me...
i'm so clueless...
and really dunno how to do.. :cry::cry::cry:
 
  • #15
Redbelly98
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the angle is 90 degree rite??
No, the angle is changing, since the magnet is rotating. Sometimes it is 90 degrees, but it keeps changing.

so,

Emf = NBA.sin(theta).w
That's almost right. Here are a few questions to get on track:

1. What units must w have in this equation?
2. What does the brush-and-commutator do? (This will modify the sin(theta) term somewhat.)
3. Have you taken calculus yet?
 
  • #16
lanedance
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sorry i have re-read the posting and nee to make some corrections, so summarising where we are at:

The flus through the coil is given by
[tex]\Phi = NBACos(wt)[/tex]

Then the induced EMF is by the reat of change of flux
[tex]EMF= -\frac{d \Phi}{dt} = -\frac{d}{dt} (NBA.Cos(\omega t)) = \omega NBA.Sin(\omega t) [/tex]

you were correct to differentiate, do you know what [tex](\omega t) [/tex] is?

Now you must find the average of the EMF. As mentioned by Redbelly the commutator arrangement will change the EMF slightly.

Think about what the sin function look like:
- It starts at 0, rises to 1 then drops back to 0
- then drops to -1 then and rise back to 0, then repeats
So the EMF will continuosly vary from positive to negative unless we do soemthing about it. Think about the affect of the commutator and brush arrangement here as Redbelly suggested.
 
  • #17
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hi all..
i managed to do this..
thanks a lot for your explanations...
 
  • #18
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That's almost right. Here are a few questions to get on track:

1. What units must w have in this equation?
2. What does the brush-and-commutator do? (This will modify the sin(theta) term somewhat.)
3. Have you taken calculus yet?
for point 2
do you mean that the sin wave will be rectified because of the use of commutator?
then the average value will be 1/%pi of the maximum value.
 
  • #19
Redbelly98
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Hi, welcome to PF :smile:

for point 2
do you mean that the sin wave will be rectified because of the use of commutator?
Yes, exactly.

then the average value will be 1/%pi of the maximum value.
Hmmm, not quite ... care to recheck your calculation?
 

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