A loop rotating in a magnetic field (How electric power generator works!)

Fu-Kwun Hwang

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1 A loop rotating in a magnetic field (How electric power generator works!)

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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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Full screen applet or Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list
Press the Alt key and the left mouse button to drag the applet off the browser and onto the desktop. This work is licensed under a Creative Commons Attribution 2.5 Taiwan License

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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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Fu-Kwun Hwang

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2 Re: A loop rotating in a magnetic field (How electric power generator 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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3 Re: A loop rotating in a magnetic field (How electric power generator works!)

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

Quote from: Fu-Kwun Hwang 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 math_failure (math_image_error): \omega t )
The magnetic flux in the loop will be math_failure (math_image_error): \Psi =\int \vec{B}\cdot d\vec{A}= B A cos(\omega t)
So the induced emf math_failure (math_image_error): = -\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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4 Re: A loop rotating in a magnetic field (How electric power generator works!)

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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 math_failure (math_image_error): \Psi =\int \vec{B}\cdot d\vec{A}= \vec{B}\cdot \vec{A(t)} =B d h \cos\omega t

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


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lookang

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5 Re: A loop rotating in a magnetic field (How electric power generator works!)

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Reply #4 on: December 12, 2009, 09:41:48 am » posted from:Singapore,,Singapore

Quote from: Fu-Kwun Hwang on December 12, 2009, 09:12:40 am

Assume the width and height of the loop are d and h.
The magnetic flux in the loop is math_failure (math_image_error): \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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Fu-Kwun Hwang

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6 Re: A loop rotating in a magnetic field (How electric power generator works!)

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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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