Magnetic Field of wire on rectangular loop

  • Thread starter BrettB
  • Start date
  • #1
7
0

Homework Statement


I have attached a diagram. In case you can't view it, it shows an infintely long wire [itex] I_i = 5.00 [/itex] A on the positive y direction. 0.100 m to the right, there is a rectangular loop of dimensions 0.150 m x 0.450 m, the long side is parallel to the infinitely long wire to its left. The current is [itex] I_2 = 10.0 [/itex] A and also flowing up (in the position closest to the infinitely long wire).

a) Find the magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire.

b) Find the force on the top, horizontal segment of the loop. Calculate this three ways: net force, average force, and the force on the midpoint.


Homework Equations



Bio-Savaart equation.

The Attempt at a Solution



I didn't have any trouble with (a). The force on the top and bottom horizontal segments just cancel, so I can disregard them. The rest is pretty simple.

Part (b) is giving me lots of trouble. I am not 100% sure I am calculating the magnetic force of the infinitely long wire on the horizontal part of the loop properly, and I really am not certain how to handle the vertical segments of the loop. These are not infinite, so there must be some error introduced in this case, but I have no idea how to approximate it. Would the errors just cancel, and the net force of these also be zero?

Here is what I have done so far, I hope this is ok:

For the net force:

[tex] B = \int_{0.100}^{0.250} \frac{(5.00)(10.0)(\mu_0)}{2\pi x}\,dx = 1.00268(\ln(x))_{0.100}^{0.250} = 9.19 \times 10^{-6} [/tex]

For the average force, I just used the mean value theorem, and multiplied the above by 1/(0.250-0.100) to get 6.12 x 10^{-15}.

To calculate the force on the midpoint, I first calculate the force from the vertical wire:

[tex] \frac{\mu_0(5.00)}{2\pi(0.100+\frac{0.150}{2})}\hat{i} = 5.73\times 10^{-6}\hat{i} [/tex]

Then the other force:

[tex] \frac{\mu_0(10.0)}{2\pi(0.100 + \frac{0.150}{2})}\hat{j} = 1.15 \times 10^{-5}\hat{j} [/tex]

Since these are perpendicular, I got the rest as

[tex] \sqrt{(5.73\times 10^{-6})^2 + (1.15 \times 10^{-5})^2} = 1.28 \times 10^{-5} [/tex]

Intuitively, I would have expected closer values with all these results, so I am surprised they are so far apart, and that makes me question my work.

Any advice gratefully accepted.

Thanks,
Brett
 

Attachments

Last edited:

Answers and Replies

  • #2
7
0
Wow, nobody? Well, I fixed a couple of typos, but I am no closer to certainty. If I've violated some rule, I'm sorry. I'm new here :-) But I would really appreciate some help.

Thanks,
Brett
 
  • #3
hage567
Homework Helper
1,509
2
You haven't done anything wrong, I think you've done a good job of presenting your problem. I just honestly am not sure I can help!
 
  • #4
Doc Al
Mentor
45,017
1,291
b) Find the force on the top, horizontal segment of the loop. Calculate this three ways: net force, average force, and the force on the midpoint.
I'm a bit puzzled by this question. I would interpret it as: Calculate the force exerted on the top horizontal segment by the magnetic field created by the wire in three different ways and compare.

For the "net force" you integrated to find the actual force from the wire on the segment. OK. (Don't really know what "net force" means in this context.)

By "average force", I would have calculated the force per unit length at each end of the segment, averaged them, then multiplied by the length of the segment.

For "force at midpoint", I would have calculated the force per unit length at the midpoint, then multiplied by the length of the segment.

Grasping at straws here, since the question isn't clear to me.
 
  • #5
hage567
Homework Helper
1,509
2
I can point out a couple of things for you to double check. I think you are getting B and F mixed up when you are using the equations. Not sure, just a thought.

Homework Statement


I have attached a diagram. In case you can't view it, it shows an infintely long wire [itex] I_i = 5.00 [/itex] A on the positive y direction. 0.100 m to the right, there is a rectangular loop of dimensions 0.150 m x 0.450 m, the long side is parallel to the infinitely long wire to its left. The current is [itex] I_2 = 10.0 [/itex] A and also flowing up (in the position closest to the infinitely long wire).

a) Find the magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire.

b) Find the force on the top, horizontal segment of the loop. Calculate this three ways: net force, average force, and the force on the midpoint.


Homework Equations



Bio-Savaart equation.

The Attempt at a Solution



I didn't have any trouble with (a). The force on the top and bottom horizontal segments just cancel, so I can disregard them. The rest is pretty simple.

Part (b) is giving me lots of trouble. I am not 100% sure I am calculating the magnetic force of the infinitely long wire on the horizontal part of the loop properly, and I really am not certain how to handle the vertical segments of the loop. These are not infinite, so there must be some error introduced in this case, but I have no idea how to approximate it. Would the errors just cancel, and the net force of these also be zero?

Here is what I have done so far, I hope this is ok:

For the net force:

[tex] B = \int_{0.100}^{0.250} \frac{(5.00)(10.0)(\mu_0)}{2\pi x}\,dx = 1.00268(\ln(x))_{0.100}^{0.250} = 9.19 \times 10^{-6} [/tex]
This is for F, though, you have B there. Just a typo maybe?

For the average force, I just used the mean value theorem, and multiplied the above by 1/(0.250-0.100) to get 6.12 x 10^{-15}.

To calculate the force on the midpoint, I first calculate the force from the vertical wire:

[tex] \frac{\mu_0(5.00)}{2\pi(0.100+\frac{0.150}{2})}\hat{i} = 5.73\times 10^{-6}\hat{i} [/tex]
OK, I don't think this is for F, you are actually calculating B. Check the units.

Then the other force:

[tex] \frac{\mu_0(10.0)}{2\pi(0.100 + \frac{0.150}{2})}\hat{j} = 1.15 \times 10^{-5}\hat{j} [/tex]
B again, not F.

Since these are perpendicular, I got the rest as

[tex] \sqrt{(5.73\times 10^{-6})^2 + (1.15 \times 10^{-5})^2} = 1.28 \times 10^{-5} [/tex]

Intuitively, I would have expected closer values with all these results, so I am surprised they are so far apart, and that makes me question my work.

Any advice gratefully accepted.

Thanks,
Brett
 
  • #6
7
0
Thank you both, VERY much. Hage567, yes, I was calculating the completely wrong thing, thank you.

I think I still must be doing something wrong, since I get answers that are very different, or I don't understand why they should be this different. I'm also wondering why using the mean-value theorem wouldn't be the right way to calculate the "average" value in this case.

Using Doc Al's suggestion to compute the average, I get for the left end of the horiztonal wire:

[tex] \frac{(1.26 \times 10^{-6})(5.00)(10.0)}{2\pi (0.100)} = 1.00268 \times 10^{-4} [/tex]

For the right end:

[tex] \frac{(1.26 \times 10^{-6})(5.00)(10.0)}{2\pi (0.250)} = 4.01070 \times 10^{-5} [/tex]

These values would be the force per meter, I believe.

Taking the average, and multiplying by the length of the wire, I get 1.05 x 10^-5.

For the force/m at the midpoint:

[tex] \frac{(1.26\times 10^{-6})(5.00)(10.0)}{2\pi(0.100 + \frac{0.150}{2})} = 5.72958 \times 10^{-5} [/tex]

and multiplying by 0.150 (the length of the wire), I get 8.59 x 10^-6, which is closer to the total force. But not very.

Is this the right way to go about it? The only conclusion I can come to from this is that, since the force doesn't vary linearly (if you look at the integral, it is a logarithmic function), using these other ways is not the right way to estimate the force on a perpendicular wire.

Thanks again for your help, and any additional help or comments would be really appreciated.

Thanks!
Brett
 

Related Threads on Magnetic Field of wire on rectangular loop

Replies
0
Views
4K
Replies
8
Views
4K
Replies
2
Views
11K
  • Last Post
Replies
5
Views
1K
Replies
1
Views
2K
Replies
9
Views
53K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
4K
Replies
1
Views
6K
Top