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Author Topic: A loop rotating in a magnetic field (How electric power generator works!)  (Read 43633 times)
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Fu-Kwun Hwang
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on: February 02, 2009, 05:31:29 pm »

When a loop rotating in a uniform magnetic field.
The magnetic flux is changing which will induce electric field along the loop (produce voltage).
If the loop is rotating with constant angular velocity \omega . (The angle is \omega t)
The magnetic flux in the loop will be \Psi =\int \vec{B}\cdot d\vec{A}= B A cos(\omega t)
So the induced emf = -\frac {d\Psi}{dt}=B A \omega sin(\omega t)

The following simulation show how it works dynamically.
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So a loop rotating in a uniform magnetic field will generate electric power.
Reverse the process, a current in the uniform magnetic field will rotate (it become a motor).
Please check out Current Loop in magnetic field (How motor works!)
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Reply #1 on: October 03, 2009, 04:44:22 pm » posted from:Taipei,T\'ai-pei,Taiwan

Another similar one:

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lookang
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Reply #2 on: November 03, 2009, 01:30:16 pm » posted from:Singapore,,Singapore

When a loop rotating in a uniform magnetic field.
The magnetic flux is changing which will induce electric field along the loop (produce voltage).
If the loop is rotating with constant angular velocity \omega . (The angle is \omega t)
The magnetic flux in the loop will be \Psi =\int \vec{B}\cdot d\vec{A}= B A cos(\omega t)
So the induced emf = -\frac {d\Psi}{dt}=B A \omega sin(\omega t)


I suspect your model in the ejs codes is not reflecting the correct physics phenomena.
i change it to
"E0*cos(2*pi*x/(5*R))" for the Z in the analytic3D curve which symbolize the emf generated
and
E0*Math.cos(c2*pi2) for the Z in the emfparticle3D point which symbolize the instantaneous emf generated.
is my interpretation of the code in the model correct?

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Fu-Kwun Hwang
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Reply #3 on: December 12, 2009, 09:12:40 am » posted from:Taipei,T\'ai-pei,Taiwan

Assume the width and height of the loop are d and h.
The magnetic flux in the loop is \Psi =\int \vec{B}\cdot d\vec{A}= \vec{B}\cdot \vec{A(t)} =B d h \cos\omega t

So the induced emf \epsilon= -\frac {d\Psi}{dt}=B (dh) \omega \sin\omega t=BA_{max} \omega \sin\omega t
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Reply #4 on: December 12, 2009, 09:41:48 am » posted from:Singapore,,Singapore

Assume the width and height of the loop are d and h.
The magnetic flux in the loop is \Psi =\int \vec{B}\cdot d\vec{A}= \vec{B}\cdot \vec{A(t)} =B d h \cos\omega t

So the induced emf \epsilon= -\frac {d\Psi}{dt}=B (dh) \omega \sin\omega t=BA_{max} \omega \sin\omega t

should be
is \Psi =\int \vec{B}\cdot d\vec{A}= \vec{B}\cdot \vec{A(t)} =B d h \sin\omega t
I check by sub t=0 ,  $\Psi =0.
The initial condition is causing the error in your equation. Right ?
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Reply #5 on: December 12, 2009, 11:46:40 am » posted from:Taipei,T\'ai-pei,Taiwan

Yes. There was a bug.
I am sorry that : I use different symbol in the simulation and in the web page.
The cta in EJS code was not the same \theta in the above equation, and the direction for rotation was positive for colckwise rotation.

I have modified the whole code and change all the convention back to normal.

cta in the EJS code is the same as \theta now (i.e. maximum area mean cta=0;)
THank you for the bug report. It is fixed and the first simulation has been updated!

P.S. another bug was fixed, too. (the check box for the Efield show be showE instead of showBF)
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