ydoaPs Posted June 9, 2005 Share Posted June 9, 2005 a magnetic field is located north to south accross the z axis. a wire runs accross the x axis and is attatched to the source of the magnetism in such a way that the wire will not bend. a current is applied to the wire which makes a force on the y axis from the equation [imath]\mathbb{F}=\mathbb{B}{I}l[/imath]. if the wire cannot bend, will the force move the magnet as well or will it just try to bend the wire? Link to comment Share on other sites More sharing options...
Meir Achuz Posted June 9, 2005 Share Posted June 9, 2005 If the system is rigidly attached, nothing will move because there is lno esternal force. The B field will just try to bend the wire. A bending stress will be set up in the wire. Link to comment Share on other sites More sharing options...
ydoaPs Posted June 9, 2005 Author Share Posted June 9, 2005 thanx Link to comment Share on other sites More sharing options...
ydoaPs Posted June 27, 2005 Author Share Posted June 27, 2005 what is there were springs on the y axis? would the entire device move? Link to comment Share on other sites More sharing options...
DQW Posted June 27, 2005 Share Posted June 27, 2005 Yes, the conductor and the magnet would push away from each other, stretching the spring between them. Link to comment Share on other sites More sharing options...
labview1958 Posted June 27, 2005 Share Posted June 27, 2005 Let's say I have a magnet on the edge of a table. I bring a rotating copper disc near to the edge of the magnet on the table. What will happen? a. the magnet stays put. b. the magnet lifts of c. the magnet is pushed away from the edge but is still on the table. Here's the diagram rotating copper disc .........magnet on edge of table magnet height is 5 cm and the base 1cm X 1cm Link to comment Share on other sites More sharing options...
ydoaPs Posted June 27, 2005 Author Share Posted June 27, 2005 Yes, the conductor and the magnet would push away from each other, stretching the spring between them. i was thinking of running the current the other way. the force moves the wire toward the magnet and the collision pushes it. the spring moves the wire back and it can start again. to work best, the current should probably be in pulses. is it possible? Link to comment Share on other sites More sharing options...
ydoaPs Posted June 27, 2005 Author Share Posted June 27, 2005 Let's say I have a magnet on the edge of a table. I bring a rotating copper disc near to the edge of the magnet on the table. What will happen? a. the magnet stays put. b. the magnet lifts of c. the magnet is pushed away from the edge but is still on the table. Here's the diagram rotating copper disc .........magnet on edge of table magnet height is 5 cm and the base 1cm X 1cm the disc gets a current Link to comment Share on other sites More sharing options...
swansont Posted June 27, 2005 Share Posted June 27, 2005 i was thinking of running the current the other way. the force moves the wire toward the magnet and the collision pushes it. the spring moves the wire back and it can start again. to work best, the current should probably be in pulses. is it possible? Momentum is conserved, so absent any other forces this will not work. You could, however, possibly use the friction of the surface to your advantage. Link to comment Share on other sites More sharing options...
ydoaPs Posted June 27, 2005 Author Share Posted June 27, 2005 o well, would have been cool, though. Link to comment Share on other sites More sharing options...
ydoaPs Posted June 27, 2005 Author Share Posted June 27, 2005 i haven't done collisions since first semester, so i'm not sure which equation to use. i used m1v1=m2v2 and got the velocity to be [math]v=\frac{{\mathbb{B}}Il}{\Delta{t}m_{magnet}}[/math]. i don't think i used the right one, but i can't remember the other one. help would be appreciated [edit] crap, i forgot about the spring anyway. what is it? [math]F=kx[/math]?[/edit] Link to comment Share on other sites More sharing options...
labview1958 Posted July 2, 2005 Share Posted July 2, 2005 Can the following setup measure drag force? I hope the image comes? Link to comment Share on other sites More sharing options...
DQW Posted July 4, 2005 Share Posted July 4, 2005 What is the direction of magnetization of the magnet ? Link to comment Share on other sites More sharing options...
labview1958 Posted July 4, 2005 Share Posted July 4, 2005 The North pole is facing the rotatating hollow copper ring. The rotating hollow copper ring is 6 cm in diameter and has a thickness of 1mm. Link to comment Share on other sites More sharing options...
DQW Posted July 5, 2005 Share Posted July 5, 2005 I do not understand the origin of this "drag force". Could you please derive it or show me an argument for a force along the [imath] \hat {\phi} [/imath] direction ? I'm a little confused right now... Link to comment Share on other sites More sharing options...
labview1958 Posted July 5, 2005 Share Posted July 5, 2005 When a magnet approaches a spinning copper disc it experiences both electromagnetic lift and drag forces. It is mentioned here. http://www.superlife.info/en/book/victor/levitation.htm Link to comment Share on other sites More sharing options...
DQW Posted July 5, 2005 Share Posted July 5, 2005 I could not open that link. Will try again later. Link to comment Share on other sites More sharing options...
Dave Posted July 5, 2005 Share Posted July 5, 2005 Just as a very quick aside, you might want to consider using \mathbf in your LaTeX strings instead of \mathbb. The blackboard fonts are intended for use in naming sets and other mathematical objects (as far as I'm aware). Whilst I don't have anything against you using them, I think you want to represent vectors (normally done in a couple of ways); [imath]\mathbf{F} = \mathbf{B}Il[/imath] being the most common, I've also seen [imath]\vec{F} = \vec{B}Il[/imath] Link to comment Share on other sites More sharing options...
Klaynos Posted July 5, 2005 Share Posted July 5, 2005 Just on a little side note when talking about vectors the true equation is: [math]\bold {F} = \bold {B} \times I \bold {l}[/math] When B is always perpendicular to l then: F=BIl is true. Link to comment Share on other sites More sharing options...
labview1958 Posted July 6, 2005 Share Posted July 6, 2005 I have a new link. Try this! http://zone.ni.com/devzone/conceptd.nsf/webmain/DA3932D02D45350786256854005646F3 Link to comment Share on other sites More sharing options...
DQW Posted July 6, 2005 Share Posted July 6, 2005 But surely, you see how that geometry is different from yours. Link to comment Share on other sites More sharing options...
ydoaPs Posted July 7, 2005 Author Share Posted July 7, 2005 Just as a very quick aside, you might want to consider using \mathbf in your LaTeX strings instead of \mathbb. The blackboard fonts are intended for use in naming sets and other mathematical objects (as far as I'm aware). Whilst I don't have anything against you using them, I think you want to represent vectors (normally done in a couple of ways); [imath]\mathbf{F} = \mathbf{B}Il[/imath] being the most common, I've also seen [imath']\vec{F} = \vec{B}Il[/imath] ok, i'm used to writing vectors with broad backs or two lines. i'll do it that way from now on, thanx Link to comment Share on other sites More sharing options...
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