Homework Help: Magnetic Field of Three Circular loops

1. Jul 6, 2011

arl146

1. The problem statement, all variables and given/known data
The radius of the circular loop is R = 21.2 cm and the wire carries the same electric current I = 19.3 A in all three cases. In the first and in the second case the plane of the circle is parallel to the straight wire segments, in the third case the plane of the circle is perpendicular to the straight segments. The straight segments are very long in all three cases in both directions.

(i.) What is the size of the magnetic field at the center of the circle in the first case?
(ii.) 2nd case
(iii.) 3rd case

2. Relevant equations
For the first one I used B= (mu*i) / (2*pi*R) for the straight part and B= (mu*i) / (2R) for the loop. Which at the end I added the two together.

3. The attempt at a solution

First answer is 7.54×10^(-5) T. I got that correct. The 2nd I'm confused about because of the break in the loop and the 3rd I just don't know what to do with it since the loop is positioned differently. For the second wouldn't it be really similar to the first?

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Last edited: Jul 6, 2011
2. Jul 6, 2011

rl.bhat

In the second case, find the direction of the field due to the straight conductor and the loop?
Are they in the same direction?
similarly, what are the directions of the fields in the third case?

3. Jul 6, 2011

arl146

In the 2nd case, yea, wouldn't they be the same? It would be pointing out of the page?
So, in all 3 cases, the straight part's field is pointing out of the page? And in the 3rd case, the loop part is towards like the +y direction?

I ask those all as questions because I'm not very confident in those answers but that's what I'm getting with the right hand rule

4. Jul 7, 2011

rl.bhat

In the 2nd case field due to wire goes in to the page and due to the loop fiedl comes out of the page.
In all the case, the field due to the wire is out of the page on left side of the wire and that into the page towards right of the wire.

5. Jul 7, 2011

arl146

I don't get it? how does the 2nd case of the straight wire go into the page?

6. Jul 7, 2011

rl.bhat

If you candider the gap is very small, the direction of the current in the vertical wire is in upward direction.
If you hold it in your right hand pointing your thumb in the direction of the current, field is going into the page in the right side of the wire.

7. Jul 7, 2011

arl146

okay i guess i get that, but how come on the loop you wrap your fingers in the direction of the current but on the straight part you use your thumb? just because its in one direction unlike the loop in which is curving? and can you explain this quote a little more i dont understand the left and right side part, is it because the current would go up the wire and come back down it? that confuses me

8. Jul 8, 2011

arl146

can someone help, im confused on this ?

9. Jul 9, 2011

rl.bhat

In fact you can use the right hand rule for both, straight conductor and the loop.
Consider the straight conductor and the loop separately close to each other. Apply the right hand rule to the points on the straight conductor and loop close to each other. You can see that the fields dut to these points at the center are in the opposite direction.

10. Jul 10, 2011

arl146

im so confused. can you start from the beginning and be like real clear because im really not following ..

11. Jul 11, 2011

rl.bhat

You can find the field at any point due to a current carrying straight wire by using right hand clasp rule. According to that, thumb indicates the direction of the current in the conductor and the finger tips indicate the direction of the field.
In the current loop, finger tips of the right hand indicate the direction of the current and thumb indicates the direction of the field.
Second method is identical to the first method if you apply the first method to a small section of the loop which you can treat as a straight conductor and sum it up to get the total field at the center.

12. Jul 11, 2011

arl146

oh okay, that makes sense, i thought thats what you were getting at. i still dont understand why you said this "In all the case, the field due to the wire is out of the page on left side of the wire and that into the page towards right of the wire." how can the wire have a different field on the left side and right side, how would you know that, you were talking about the straight wire right? how can it have it both ways? so what now?

13. Jul 11, 2011

arl146

okay, disregard that last post. i completely understand everything now. http://www.matter.org.uk/schools/content/magneticfields/fields_2.html [Broken] this website helped a bit and reading back over what you said, made a lot of sense. So, in the 2nd case, we said that the B field due to the straight wire is both into and out of the page depending on the side of the wire. We also said for the loop on the 2nd case, that the B field is out of the page. So, the 3rd case, same thing for straight wire but the loop is going straight up. I got all that now. So what do I do next then?

Last edited by a moderator: May 5, 2017
14. Aug 9, 2011

arl146

Can someone please help me with this problem??? much appreciated !! The gap in the second one confuses me a little and the third one it throws me off because of the axis it lays on!

15. Aug 9, 2011

rl.bhat

In the second case, if the straight conductor is very long, you can neglect the gap. And field due to it at the center of the circular coil is μο*I/(2πr). The direction is into the page.
The magnetic field at the center of the circular loop carrying current will be μο*I/(2r) and the direction is out of the page. So the net field will be difference of these fields.
In the third case, the same two fields mentioned above are perpendicular to each other. So find the resultant field.

16. Aug 9, 2011

arl146

I got the second one, stupid mistake that I was doing ! Yea thats where I'm confused, what does that mean if the fields are perpendicular? I wouldn't add/subract them right so what would I do? multiply them or something?

17. Aug 9, 2011

rl.bhat

If two vectors X and Y are perpendicular to each other then, the resultant vector R will be written as
R^2 = X2 + Y2

18. Aug 9, 2011

arl146

ohh, yea duhh ... i always over think this stuff. thanks so much, got it !