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Net force on a loop of current

  • Thread starter jegues
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  • #1
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Homework Statement



See figure attached.

Homework Equations





The Attempt at a Solution



I'm confused as how to calculate the net magnetic force that acts on the loop.

Do I choose a point somewhere and calculate the magnetic force with respect to that? Because I don't quite understand how to calculate the magnetic force with respect to an entire loop.

Any ideas?

Thanks again!
 

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Answers and Replies

  • #2
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can you find mag field at any point due to a straight wire?
and also force on a straight wire due to mag field?
 
  • #3
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can you find mag field at any point due to a straight wire?
and also force on a straight wire due to mag field?
The magnetic field due to current in a straight wire is simply,

[tex]B = \frac{\mu_{0}i}{2 \pi R}[/tex]

The force on a straight wire would be,

[tex]\vec{F_{b}} = i\vec{L}\times \vec{B}[/tex]

but I'm still not seeing how to obtain the net force on the loop???
 
  • #4
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i am guessing that you have problem with finding the force on the vertical wires of the loop ...
well for that, consider any element dx at distance of x from wire on both vertical wires of loop ... as you can see that the force equal will be opposite on the 2 dx elements ... same will happen on all the elements on the vertical wire ... so net force on the vertical wires of loop wil be ... ???
 
  • #5
1,097
2
i am guessing that you have problem with finding the force on the vertical wires of the loop ...
well for that, consider any element dx at distance of x from wire on both vertical wires of loop ... as you can see that the force equal will be opposite on the 2 dx elements ... same will happen on all the elements on the vertical wire ... so net force on the vertical wires of loop wil be ... ???
Zero. When I originally attempted this question I did this as well but I got the wrong answer.
 
  • #6
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isnt the answer d?
 
  • #7
1,097
2
i am guessing that you have problem with finding the force on the vertical wires of the loop ...
well for that, consider any element dx at distance of x from wire on both vertical wires of loop ... as you can see that the force equal will be opposite on the 2 dx elements ... same will happen on all the elements on the vertical wire ... so net force on the vertical wires of loop wil be ... ???
How come the wire with the 12A running through it doesn't have any effect on the vertical wires?
 
  • #8
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it has effect ... but its opposite on 2 wires ... thats why is gets cancelled and you can ignore it.
 
  • #9
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it has effect ... but its opposite on 2 wires ... thats why is gets cancelled and you can ignore it.
How can I verify this using right hand rule I always get confused...

How am I suppose to orient my fingers? (I'm using my index finger, my middle finger, and my thumb, where the thumb is the resultant vector of the cross product of the index finger and middle finger.)

Thanks again!

EDIT: Is the magnetic field always perpendicular to the current in the wire?

Then I would point my index finger in the direction of the current, my middle finger perpendicular to my index finger and my thumb out to the side.

Is this correct?
 
Last edited:
  • #10
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How can I verify this using right hand rule I always get confused...
How am I suppose to orient my fingers? (I'm using my index finger, my middle finger, and my thumb, where the thumb is the resultant vector of the cross product of the index finger and middle finger.)
for Fleming's Rule(which include middle finger, index finger, thumb) -- like for AXB ... the index finger should be along A and midddle finger along B ... then thumb will give direction of the resultant
Note: if it was BXA ... index finger will be along B and middle finger along A
consider this: http://cyclo.mit.edu/~schol/802x/howtos/rhr_howto.pdf [Broken]


however i always prefer using right hand thumb rule ...
why i like it is because it eliminate both right and left rules (which include middle finger, index finger, thumb)

you know how to use it to find the direction of mag field due to straight current ...
for cross product ...
open your hand ... for AXB the four fingers should be pointing towards A and when you curl your hand ... fingers should go towards B ... the thumb will give the direction of force

How can I verify this using right hand rule I always get confused...
Is the magnetic field always perpendicular to the current in the wire?
which wire are we talking about?
 
Last edited by a moderator:
  • #11
1,097
2
for Fleming's Rule(which include middle finger, index finger, thumb) -- like for AXB ... the index finger should be along A and midddle finger along B ... then thumb will give direction of the resultant
Note: if it was BXA ... index finger will be along B and middle finger along A
consider this: http://cyclo.mit.edu/~schol/802x/howtos/rhr_howto.pdf [Broken]


however i always prefer using right hand thumb rule ...
why i like it is because it eliminate both right and left rules (which include middle finger, index finger, thumb)

you know how to use it to find the direction of mag field due to straight current ...
for cross product ...
open your hand ... for AXB the four fingers should be pointing towards A and when you curl your hand ... fingers should go towards B ... the thumb will give the direction of force



which wire are we talking about?
Well because we are looking at the force,

[tex]\vec{F_{b}} = i\vec{L}\times \vec{B}[/tex]


I know that the L vector points along the direction of current, so I orient my index finger in that direction, but how do I know what direction B points towards? I need to oriente my middle finger in this direction I can determine the direction of the resultant force.

Can you clarify? I know how to use right hand rule, I just don't see how to get the direction of B.
 
Last edited by a moderator:
  • #12
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from thumb rule you know that field is circular with wire at center ...

so for any plane ,,, field will be perpendicular to the plane ,, and in this case as you case ... it will vertically downwards at the loop

but i will still advice you to use thumb law
 

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