NTNUJAVA Virtual Physics LaboratoryEnjoy the fun of physics with simulations! Backup site http://enjoy.phy.ntnu.edu.tw/ntnujava/
April 19, 2021, 06:34:51 am
 Welcome, Guest. Please login or register.Did you miss your activation email? 1 Hour 1 Day 1 Week 1 Month Forever Login with username, password and session length

 Home Help Search Login Register
Ask questions. Follow-up questions. ...Wisdom

 Pages: [1]   Go Down
 Author Topic: A loop rotating in a magnetic field (How electric power generator works!)  (Read 58088 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
Hero Member

Offline

Posts: 3086

 « Embed this message 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.
-*-

Embed a running copy of this simulation

Embed a running copy link(show simulation in a popuped window)
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
• Please feel free to post your ideas about how to use the simulation for better teaching and learning.
• Post questions to be asked to help students to think, to explore.
• Upload worksheets as attached files to share with more users.
Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!

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!)
 Logged
Fu-Kwun Hwang
Hero Member

Offline

Posts: 3086

 « Embed this message Reply #1 on: October 03, 2009, 04:44:22 pm »

Another similar one:

Embed a running copy of this simulation

Embed a running copy link(show simulation in a popuped window)
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
• Please feel free to post your ideas about how to use the simulation for better teaching and learning.
• Post questions to be asked to help students to think, to explore.
• Upload worksheets as attached files to share with more users.
Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!
 Logged
lookang
Hero Member

Offline

Posts: 1796

http://weelookang.blogspot.com

 « Embed this message Reply #2 on: November 03, 2009, 01:30:16 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 $= -\frac {d\Psi}{dt}=B A \omega sin(\omega t)$

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?

Embed a running copy of this simulation

Embed a running copy link(show simulation in a popuped window)
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
• Please feel free to post your ideas about how to use the simulation for better teaching and learning.
• Post questions to be asked to help students to think, to explore.
• Upload worksheets as attached files to share with more users.
Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!
 *** There are 1 more attached files. You need to login to acces it! Logged
Fu-Kwun Hwang
Hero Member

Offline

Posts: 3086

 « Embed this message Reply #3 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 $\Psi =\int \vec{B}\cdot d\vec{A}= \vec{B}\cdot \vec{A(t)} =B d h \cos\omega t$

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

Offline

Posts: 1796

http://weelookang.blogspot.com

 « Embed this message Reply #4 on: December 12, 2009, 09:41:48 am »

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 ?
 Logged
Fu-Kwun Hwang
Hero Member

Offline

Posts: 3086

 « Embed this message Reply #5 on: December 12, 2009, 11:46:40 am »

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)
 Logged
 Pages: [1]   Go Up
Ask questions. Follow-up questions. ...Wisdom