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Quiz Time!

July 10th, 2008 · 3 Comments

I gave this quiz to my Physics 208 class this Monday, modified slightly from one of the textbook homework problems. I was going to work it out here, but I think I’ll leave the solution as a challenge for you.

A straight conducting wire of mass M and length L is placed on a frictionless inclined plane tilted at an angle θ from the horizontal. There is a uniform, vertical magnetic field B at all points. To keep the wire from sliding down the incline, a voltage source is attached to the ends of the wire. When just the right amount of current flows through the wire, the wire remains at rest. Determine the magnitude and direction of the current in the wire that will cause the wire to remain at rest.

Force on a straight current-carrying wire in a magnetic field:

Where L is the length of the wire in the direction of current flow. I realize “direction” is a little ambiguous since I haven’t included a diagram, so don’t worry too much about that part of the problem. Hints below:

It’s just like a box on an inclined plane. You have a gravitational force straight downwards and a magnetic force pushing straight sideways. All that remains is to use trig (that’s the slightly tricky part) to figure out the components of those forces along the inclined place. Set them equal, solve for I. With a little luck, someone will beat me to it and post the answer in the comments, but if not I’ll put up the final result in a few days.

Tags: College Physics 101 · Worked Problems

3 responses so far ↓

  • 1 CCPhysicist // Jul 11, 2008 at 12:50 pm

    Nice problem. I may have to give it a go this fall.

    Did you keep any statistics on performance and what they did wrong? I would be interested to know what fraction of them drew a free-body diagram as part of their solution.

    If they did, I would expect most errors to result from (a) just using ILB for the force along the plane or (b) getting the components wrong.

  • 2 andy.s // Jul 13, 2008 at 7:36 am

    I suppose the wire is lying perpendicular to the slope.

    Force downslope would be -mg sin theta; B-force would be IBL, the upslope component would be IBL cos theta.

    So the current wold be mg tan theta/BL.

    Incidentally, does this site do LaTeX, like Cosmic Variance? Let’s find out: [tex]\LaTeX[/tex]

  • 3 andy.s // Jul 14, 2008 at 7:37 am

    Is this Wordpress? Let’s see if this works.
    $latex \LaTeX$

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