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

[cookiemnstr510510]

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

F_wire=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.

F_{wire d}=I₁LxB₁=I₁LB₁(i hat direction)
F_{wire 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!

 

[haruspex]

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?

 

[cookiemnstr510510]

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 F_wire=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

 

[cookiemnstr510510]

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:

F_{wire b}=I₂LxB₁=I₂LB₁ (i hat)
F_{wire d}=I₂LxB₁=I₂LB₁ (i hat)

F_total=2(I₂LB₁) (i hat)

B₁=B_wire=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)

F_total=(2I₂Lμ₀I₁)/(2πd) (i hat) = (I₂I₁Lμ₀)/(πd) (i hat)

 

[haruspex]

I believe I've figured out what you are saying, attempt below:

F_{wire b}=I₂LxB₁=I₂LB₁ (i hat)
F_{wire d}=I₂LxB₁=I₂LB₁ (i hat)

F_total=2(I₂LB₁) (i hat)

B₁=B_wire=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)

F_total=(2I₂Lμ₀I₁)/(2πd) (i hat) = (I₂I₁Lμ₀)/(πd) (i hat)

Yes, that's what I think they want.

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?

 

[cookiemnstr510510]

Right. So what torque do these forces exert about the central axis?

τ=F_{wire b}(R) + F_{wire 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...

 

[haruspex]

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?

 

[cookiemnstr510510]

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)

 

[haruspex]

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?

 

[cookiemnstr510510]

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?

 

[haruspex]

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,

the line of action goes directly through the the axis

So how much does it miss by?

 

[cookiemnstr510510]

So how much does it miss by?

It misses it by 0?

 

[haruspex]

It misses it by 0?

Right. So what can you say about the torque it exerts about that axis?

 

[cookiemnstr510510]

Right. So what can you say about the torque it exerts about that axis?

That there is no torque about the axis?

 

[haruspex]

That there is no torque about the axis?

Yes!

 

[cookiemnstr510510]

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?

 

[haruspex]

since there is no torque about the axis this means that there is no rotation, correct?

Right.

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.

 

[cookiemnstr510510]

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!