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GR9277 #54 |
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Problem
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\prob{54}
A rectangular loop of wire with dimensions shown above is coplanar with a long wire carrying current I. The distance between the wire and the left side of the loop is r. The loop is pulled to the right as indicated.
What are the directions of the induced current in the loop and the magnetic forces on the left and the right sides of the loop as the loop is pulled?
Induced Current ... Force on Left Side ... Force on Right Side
- Counterclockwise ... To the left ... To the right
- Counterclockwise ... To the left ... To the left
- Counterclockwise ... To the right ... To the left
- Clockwise ... To the right ... To the left
- Clockwise ... To the left ... To the right
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Electromagnetism }Faraday Law
The induced current would act, according to Lenz Law, to oppose the change. In this case, since the field is decreasing (the wire is being pulled away from the field), the induced current would act to increase the field. On the side closest to the long wire, it would thus point in the same direction as the current from the long-wire. This eliminates all but choices (D) and (E).
Now, since the rectangular loop wire cannot induce a force on itself, the force is due to the field from the long wire. To the left of the loop, the long wire has a field pointing into the page, and thus the force there is left-wards. One can check again that choice (E) is right by right-hand-ruling the field on the right side of the loop. Since the field due to the long wire is again into the page, the force here is towards the right (since the current runs down the page on the right side of the loop).
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Alternate Solutions |
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Comments |
Rune
2007-10-28 15:52:10 | To determine direction, here's another way if you hate the right hand rule:
if you have two wires, with current in the same direction, they attract. If the current is in the opposite direction they repell. So if you consider the wire, and the side of the loop, they have current in the same direction, so they should attract, and therefore the force on that segment should be to the left. On the other hand, the right side has current that is going opposite of the main wire, so the wires repel, a force to the right. |  | Andresito
2006-03-27 16:04:03 | [Request]
Express in the solution that Force = I x B. Helps on recalling and does not make the solution considerably longer.
Thank you
agaliarept
2006-12-01 17:36:42 |
My thoughts exactly. Just needed a simple clarification. Thank you Andresito.
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| | jax
2005-12-05 15:33:28 | I have always hated the right hand rule. I keep doing it and I'm getting the force either up or down (I know that makes no sense and I guess somehow I am doing the right hand rule wrong)
Here's what I'm doing. thumb goes in the direction of the velocity of the loop (to the right). My fingers are pointing down because the field on the right side of the long wire points down. Now apparently my palm is supposed to point in the direction of the force. My palm is facing forward which seems to be the equivalent of 'up' in the direction of the current. What am I doing wrong? Maybe I have a deformed hand. :(
At least I get the right answer if I actually compute a cross product...
jax
2005-12-05 17:44:24 |
Actually after putting some more thought into this, I still cant get the right direction of the force when doing a cross product, but I'm convinced that I am doing something wrong. I was trying to use the Lorentz force law  but that doesn't seem to work.
I found some old undergrad notes where the prof wrote that two wires with current traveling in the same direction attracted each other (thus opposite currents repel) and that's as much explanation as I can find. Can anybody elaborate?
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gatboja
2005-12-06 14:54:49 |
Hey jax. I'm thinking if your thumb goes in the direction of the current instead of the velocity of the loop, that should work. If you're dealing with a single moving charge, then the thumb is in the direction of the velocity (which is the dir'n of the current).
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yosun
2005-12-06 19:25:50 |
jax: there's another right-hand rule for currents.  , where dl is the length element of the current. in this case, since the problem asks for the force on the loop, B is due to the long wire and dl is the length/direction of the current in the loop.
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yosun
2005-12-06 20:14:41 |
jax: the right-hand-rule basically determines which unique direction the third axis should point in. so, if you like coordinate axes better, you can use that idea in lieu of the RHR. Example: suppose you have  , where  points in the x direction,  points in the y direction, then the cross product points in the +z direction. (exactly as 3 coordinate axes would. however, if you have A pointing in the y direction, B in the x direction, then their cross product points in the negative z direction. etc)
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FortranMan
2008-10-01 12:12:00 |
you're probably no longer reading this site jax, but just stay with the right hand rule for a sec. Start from between the wires and move away from them.
For currents moving in the same direction, the field lines cancel in the middle, but conjoin away from the middle of the wires. For currents moving in opposite directions, the field lines conjoin in the middle, but cancel each other everywhere outside the wires.
Now think in terms of air pressure. In the former case, the lack of magnetic field on the inside compared to the outside forces the wires together, where as in the latter case the opposite happens. This is very similar to what happens with a balloon, but with air.
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