Magnetic Force Experienced by a Loop of Wire

In summary: B.In summary, the homework statement is that you need to solve for the magnetic force along the inverted parabola. You figured out the equation for the inverted parabola and found that y=-ax2+h. You also realized that you will need to integrate the magnetic force along the entire parabola.
  • #1
acedeno
36
4

Homework Statement


The problem is shown in the link below:

http://postimage.org/image/do2b7xz0b/

Homework Equations



Biot-Savart Law's and dF=Idl x B

The Attempt at a Solution



My attempt is really just a mess. I essentially want to break it up into 2 segments. (1) Being the straight horizontal segment from -w to w and (2) the semi-oval segment. For the straight segment, I get dl x B to be ydx*z_hat and when I take the integrals in the equation dF=Idl x B, I get a Force of 2wIy*z_hat and since y=0 all along this point I get a Force of 0 due to the line segment at the origin. As for the oval segment I'm probably only correct as far as y=h-x^2 is what needs to be integrated. Any help would be appreciated!
 
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  • #2
Well, you're right about dividing the problem into two sections, and you figured out the straight (base of the inverted parabola) right.

So the next step is to write the equation for the inverted parabola: y = y(x).

I also suggest "going advanced" and using bold lower-case letters for the unit vectors, thus i_hat = i etc. In fact, use bold for all vectors, e.g dF = i*ds x B etc.
 
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  • #3
Yeah, isn't that what I have? y=h-x2
 
  • #4
That ca'nt be the right equation, can it? What happened to w?
Let x = +w or -w, what do you get for y?
 
  • #5
well in that case just y=h-w^2

but that doesn't make much sense to me, I was assuming something along the lines:

y=-a(w+x)(w-x) ??
 
  • #6
Ouf, nevermind, x=w and y=h-w2 make sense to me now. I reviewed a little bit of my high school quadratic forms! :P
 
  • #7
acedeno said:
Ouf, nevermind, x=w and y=h-w2 make sense to me now. I reviewed a little bit of my high school quadratic forms! :P

Well, no, you need y as a function of x.

Hint: what's the standard formula for a parabola with the base at (0,0)?
 
  • #8
y=x2
 
  • #9
Make it y = ax2 and you got a deal.

Now, llok at that graph y = ax2. Picture what you have to do to that graph to make it look like your loop (minus the straight section).
 
  • #10
rude man said:
Make it y = ax2 and you got a deal.

Now, llok at that graph y = ax2. Picture what you have to do to that graph to make it look like your loop (minus the straight section).

This is what led me to believe that it was y=h-x2. Do you want me to say y=-ax2+h ? or am I completely approaching this incorrectly?
 
  • #11
You're getting warm. Of course, you realize that it can't be right:
plug x = +w or x = -w into it, and what do you get for y?

Look at your last formula: y = h - ax2. a is not part of your figure, so how about solving for a by forcing y = 0 when x = + or -w?

I'm wondering if nowadays they don't teach enough analytic geometry before embarking on calculus. In my day, analyt came first, and I'm talking ivy league college, not high school. You have to master analyt first!
 
  • #12
hmm..

y=h(1-x2/w2) ??
 
  • #13
Cigars!

Now, you will need to integrate the magnetic force along the entire parabola. You have already stated that dF = i*ds x B, which is right. You know i and B, what do you do about ds?
 
  • #14
I'm not sure, I feel as if I need to express ds in such a way that it takes into account both the change in y and the change in x. Possibly a ratio?
 
  • #15
I take the derivative of y with respect to x and voila?!
 
  • #16
You're on the right track.

Draw an element of ds on your parabola. What would ds be in terms of x and y, and, remembering it's a vector, using unit vectors i and j?

Is ds different on the left vs. the right-hand part of the parabola?

BTW let's call the current I instead of i so as not to confuse it with the unit vector.

Also, note that I is defined couterclockwise.
 
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  • #17
I want to say di/dj = ds and so:

y=h-hx2/w2

di/dj = dy/dx = -2hx/w2
 
  • #18
hmm actually,

ds = dxi + dyj

dy=-2hx/w2 * dx

dx/dy=w2/-2hx

dx= -w2/2hx *dy

therefore,

ds = (-w2/2hx*dy)i + (-2hx/w2*dx)j
 
  • #19
acedeno said:
hmm actually,

ds = dxi + dyj

Almost. Draw a small piece ds along the rigth-hand part of the parabola. Which way does it point? (Remeber the direction of I).

[/QUOTE]

I have to leave for a few hrs. Try to get ds for both sides, remembering the direction of I. Also realize that you will eventually need to do an integration, so you need ds x B to be a function of one variable only.
 
  • #20
How about. ds x B = (wkydx - 2khx^2/w^2)k

should i sub in my equation for y in for y?
 
  • #21
Start with ds, you need to get this first in terms of x and y, then in terms of x. Only then can you take the cross-product with B. B will also first have to be in terms of x before you can do the cross-multiplication to get dFand the eventual integration to get F.
 

FAQ: Magnetic Force Experienced by a Loop of Wire

What is the magnetic force experienced by a loop of wire?

The magnetic force experienced by a loop of wire is the force exerted on the wire due to the interaction between the magnetic field and the current flowing through the wire.

How is the magnetic force calculated?

The magnetic force experienced by a loop of wire can be calculated using the formula F = I * L * B * sin(theta), where I is the current flowing through the wire, L is the length of the wire, B is the strength of the magnetic field, and theta is the angle between the wire and the magnetic field.

What factors affect the magnetic force on a loop of wire?

The magnetic force on a loop of wire is affected by the strength of the magnetic field, the current flowing through the wire, the length of the wire, and the angle between the wire and the magnetic field. The force is directly proportional to the current and the length of the wire, and is also affected by the angle between the wire and the magnetic field, with the maximum force occurring when the wire is perpendicular to the field.

What is the direction of the magnetic force on a loop of wire?

The direction of the magnetic force on a loop of wire is perpendicular to both the direction of the current flowing through the wire and the direction of the magnetic field. This means that the force will be either towards or away from the center of the loop, depending on the direction of the current and the direction of the magnetic field.

How does the magnetic force on a loop of wire affect the motion of the wire?

The magnetic force on a loop of wire can cause the wire to rotate around its own axis, if the current is allowed to flow freely. If the wire is fixed in place, the force will cause the wire to move in a circular path around the magnetic field lines. Additionally, the force can also cause the wire to experience a torque, which can result in rotational or translational motion depending on the wire's orientation and the strength of the magnetic force.

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