Forces of a straight wire on a semi-circular wire loop (Magnetism)

In summary: Am I wrong?In summary, the conversation discusses an ideal, infinitely long wire carrying a current and partially surrounded by a loop of current. The force exerted on the loop is discussed and calculated, and the question of whether this force causes rotation of the loop is raised. The solution involves breaking the semicircular wire into segments, calculating the magnetic field at each segment, and finding the total force exerted on the loop. The concept of torque is then introduced to consider whether the force causes rotation around the central axis.
  • #1
cookiemnstr510510
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Homework Statement


An ideal, infinitely long straight wire carries a current I1 and is partially surrounded by a loop of current as shown below (image attached). The loop has length L, parallel to the infinitely long wire. The loop carries current I2.
A) Calculate the magnitude and direction of the force exerted on the entire loop.
B) Does this force tend to cause rotation of the loop?

Homework Equations


Fwire=ILxB= ILBsinθ (The F, L, and B are all vectors, once you do the cross product they are no longer vectors)

The Attempt at a Solution


A)
Each wire is creating its own magnetic field at the location of the other wire. I broke the semicircular curved wire into 4 segments, labeled them a-d (segments.jpg).
The magnetic field from the infinitely long wire on segment "a" and "c" of the curved semicircular wire loop is everywhere parallel or anti parallel to the current traveling through those segments. Therefore the force on those wires is zero. The only segments that are affected by the infinitely long wire are segments "b" and "d" of our loop. Segment "b" feels a force towards the center of its circular shape (to the right) and segment "d" feels a force to the right.

Fwire d=I1LxB1=I1LB1(i hat direction)
Fwire b=I1LxB1=I1LB1(i hat direction)

So the force exerted on the entire loop is
2[I1LB1(i hat direction)]

B) Not sure how to approach part b, any advice?

Thanks in advanced! As always I appreciate the help!
 

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  • #2
For part a, the givens are I1, I2, not I1, B1.

For part b, what do you think? How does a force on an object cause it to rotate about an axis? What axis do you think they mean?
 
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  • #3
haruspex said:
For part a, the givens are I1, I2, not I1, B1.

For part b, what do you think? How does a force on an object cause it to rotate about an axis? What axis do you think they mean?
Hmmm, okay.
So are you saying since it gave me I1 and I2 I need to write my magnetic field in terms of I2?
The equation I am tempted to use if the one in my above post which states Fwire=ILxB. So are you saying write that B in terms of known variables?

The reason I used B1 is because the infinite wire with current I1 creates a magnetic field (which makes sense to label B1) at the position of the loop.

The only thing I can think about rotational motion would be some sort of Torque equation, would this be the right avenue to explore?

Thanks
 
  • #4
haruspex said:
For part a, the givens are I1, I2, not I1, B1.

For part b, what do you think? How does a force on an object cause it to rotate about an axis? What axis do you think they mean?
I believe I've figured out what you are saying, attempt below:

Fwire b=I2LxB1=I2LB1 (i hat)
Fwire d=I2LxB1=I2LB1 (i hat)

Ftotal=2(I2LB1) (i hat)

B1=Bwire=Magnetic field that wire with current I1 is creating at location of wire 2 (the loop) with current
I2= (μ0I1)/(2πd) (d being distance from one wire to the other)

Ftotal=(2I20I1)/(2πd) (i hat) = (I2I10)/(πd) (i hat)
 
  • #5
cookiemnstr510510 said:
I believe I've figured out what you are saying, attempt below:

Fwire b=I2LxB1=I2LB1 (i hat)
Fwire d=I2LxB1=I2LB1 (i hat)

Ftotal=2(I2LB1) (i hat)

B1=Bwire=Magnetic field that wire with current I1 is creating at location of wire 2 (the loop) with current
I2= (μ0I1)/(2πd) (d being distance from one wire to the other)

Ftotal=(2I20I1)/(2πd) (i hat) = (I2I10)/(πd) (i hat)
Yes, that's what I think they want.
cookiemnstr510510 said:
The only thing I can think about rotational motion would be some sort of Torque equation
Right. So what torque do these forces exert about the central axis?
 
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  • #6
haruspex said:
Right. So what torque do these forces exert about the central axis?
τ=Fwire b(R) + Fwire d (R), R=lever arm. I am stuck here.

This is a 3-D shape, both forces are in the positive "i" direction. I would think that at least a component of some force would have to be in the negative j direction (picture attached of my coordinates) for there to be any rotation around the axis, that's just my first thought...
 

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  • #7
cookiemnstr510510 said:
R=lever arm
The lever arm is the perpendicular distance from the line of action of the force to the axis, i.e., if you follow the line of action of the force, how much does it miss the axis by at its closest approach?
 
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  • #8
haruspex said:
The lever arm is the perpendicular distance from the line of action of the force to the axis, i.e., if you follow the line of action of the force, how much does it miss the axis by at its closest approach?
I feel like the line of action goes directly through the the axis (axis meaning the infinitely long wire)
 
  • #9
cookiemnstr510510 said:
I feel like the line of action goes directly through the the axis (axis meaning the infinitely long wire)
Right. So how long does that make the lever arm, and what does that tell you about torque?
 
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  • #10
haruspex said:
Right. So how long does that make the lever arm, and what does that tell you about torque?
The lever arms length would be the radius of curvature, R. But there is no component of force that is perpendicular to the lever arm, so it won't rotate?
 
  • #11
cookiemnstr510510 said:
The lever arms length would be the radius of curvature, R
No. I asked: "if you follow the line of action of the force, how much does it miss the axis by at its closest approach?", and you answered, correctly,
cookiemnstr510510 said:
the line of action goes directly through the the axis
So how much does it miss by?
 
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  • #12
haruspex said:
So how much does it miss by?
It misses it by 0?
 
  • #13
cookiemnstr510510 said:
It misses it by 0?
Right. So what can you say about the torque it exerts about that axis?
 
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  • #14
haruspex said:
Right. So what can you say about the torque it exerts about that axis?
That there is no torque about the axis?
 
  • #15
cookiemnstr510510 said:
That there is no torque about the axis?
Yes!
 
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  • #16
haruspex said:
Yes!
Cool!
So since there is no torque about the axis this means that there is no rotation, correct?
From looking at the problem initially my intuition was that since the force was only in one direction how could there be rotation? Does that logic make sense?
 
  • #17
cookiemnstr510510 said:
since there is no torque about the axis this means that there is no rotation, correct?
Right.
cookiemnstr510510 said:
initially my intuition was that since the force was only in one direction how could there be rotation?
No, that does not follow. Rotations are always in the context of a specified axis. If the net force acts through that axis then there is no torque about it.
But consider observing someone push a trolley past you (from rest). If you take yourself as axis then the force exerted on the trolley does have a torque about you, and indeed the trolley does acquire an angular velocity about you. You might not think of that as rotating about you, because the distance is changing; but technically it does acquire a rotation about you, i.e. a tangential motion, as well as a radial motion.
 
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  • #18
haruspex said:
No, that does not follow. Rotations are always in the context of a specified axis. If the net force acts through that axis then there is no torque about it.
But consider observing someone push a trolley past you (from rest). If you take yourself as axis then the force exerted on the trolley does have a torque about you, and indeed the trolley does acquire an angular velocity about you. You might not think of that as rotating about you, because the distance is changing; but technically it does acquire a rotation about you, i.e. a tangential motion, as well as a radial motion.
Okay, thanks for the insight!
Greatly appreciated!
 

Related to Forces of a straight wire on a semi-circular wire loop (Magnetism)

1. What is the relationship between the forces of a straight wire and a semi-circular wire loop?

The forces between a straight wire and a semi-circular wire loop are based on the principles of magnetism. A current passing through a straight wire creates a magnetic field around it, and when this wire is placed near a semi-circular wire loop, the magnetic field exerts a force on the loop, causing it to move.

2. How does the direction of current affect the forces between the wires?

The direction of current in the straight wire determines the direction of the magnetic field it creates. The force between the wires is dependent on the relative direction of the magnetic field and the direction of the current in the loop. If the current and magnetic field are aligned, the force will be attractive, but if they are in opposite directions, the force will be repulsive.

3. What factors affect the strength of the force between the wires?

The strength of the force between the straight wire and the semi-circular wire loop is affected by several factors, including the current in the wire, the distance between the wires, and the magnetic properties of the materials used. A higher current, shorter distance, and stronger magnetic materials will result in a stronger force between the wires.

4. Can the forces between the wires be calculated?

Yes, the forces between the straight wire and the semi-circular wire loop can be calculated using the formula F = I * L * B, where F is the force in Newtons, I is the current in Amperes, L is the length of the wire, and B is the magnetic field strength. This formula is known as the Lorentz force law and is used to calculate the force between any two conductors carrying an electric current.

5. What practical applications do the forces between the wires have?

The forces between a straight wire and a semi-circular wire loop have practical applications in many devices, including electric motors, generators, and speakers. These forces are also used in magnetic levitation technology, where the repulsive force between two magnets is used to suspend an object in mid-air. Understanding these forces is crucial in the development of various electromechanical devices.

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