What is the net magnetic field at different points between two parallel wires?

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Discussion Overview

The discussion revolves around calculating the net magnetic field at various points between two parallel wires carrying current. Participants explore the application of the right-hand rule and the correct approach to summing magnetic fields produced by each wire at specified locations.

Discussion Character

  • Homework-related, Technical explanation, Debate/contested

Main Points Raised

  • One participant presents a method for calculating the net magnetic field using the equation B = (u*I)/(2*pi*r) and proposes specific calculations for different points relative to the wires.
  • Another participant questions whether the calculations should consider vector directions or just magnitudes, suggesting that the approach taken may lead to incorrect results for certain points.
  • Some participants emphasize the importance of using the right-hand rule to determine the direction of the magnetic fields produced by each wire, which affects how the fields should be combined.
  • A participant suggests a convention for denoting the direction of the magnetic field based on the current's direction in the wires, leading to a revised calculation for the magnetic fields at different points.
  • There is a discussion about the implications of using different conventions for calculating the magnetic fields, particularly how it affects the results for part (a) compared to parts (b) and (c).
  • One participant expresses uncertainty about their previous calculations and seeks clarification on whether their revised approach makes sense.
  • Another participant confirms that the revised calculations appear correct, highlighting the utility of the right-hand rule in resolving such problems.

Areas of Agreement / Disagreement

Participants exhibit disagreement regarding the correct application of the right-hand rule and the treatment of magnetic field directions. While some agree on the revised calculations, others point out potential inconsistencies in the approach taken for different parts of the problem.

Contextual Notes

There are unresolved issues regarding the assumptions made about the direction of the magnetic fields and how they should be combined, particularly in relation to the right-hand rule and the sign conventions used in calculations.

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Homework Statement



Find the net mag field at the points asked for. It is just a matter of getting the distances (r values) correct based on the points asked for relative for each wire.

http://img4.imageshack.us/img4/5906/picf.th.jpg

Homework Equations



I think the relevant equation is:

B = (u*I)/(2*pi*r) where u is the permeability of free space.

The Attempt at a Solution



Assume Wire 1 is to the right, and Wire 2 is the left one.

For part (a) -- directly between the point of wires, the net magnetic field is Bnet = B1+B2

B1 = (u*5A)/(2*pi*5cm) + (u*5A)/(2*pi*5cm)

For part (b) -- 10cm to the right of the right wire, the net magnetic field is Bnet = B1+B2, where B1 and B2 are as follows:

B1 = (u*5A)/(2*pi*10cm)
B2 = (u*5A)/(2*pi*20cm)For part (c) -- 20cm left of the wire on the left, the net magnetic field is Bnet = B1+B2, where B1 and B2 are as follows:

B1 = (u*5A)/(2*pi*30cm)
B2 = (u*5A)/(2*pi*20cm)

---

Does my logic/work seem correct? I believe it is.. looking for your input. Thanks!
 
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You need to make sure wheter you're calculating with vectors or magnitudes. If you're using magnitudes here, you're wrong in b and c. Is the point in part a between the wires?
 
Kruum said:
You need to make sure wheter you're calculating with vectors or magnitudes. If you're using magnitudes here, you're wrong in b and c. Is the point in part a between the wires?

Hi, I'm sorry I forgot to put the subquestions! :(

I placed them in. I am looking for the net magnitude only. Do my answers make sense now? :p
 
fizzziks said:
I am looking for the net magnitude only. Do my answers make sense now?

Your answer for a is correct. But check the signs in b and c by using the right-hand rule. :wink:
 
Kruum said:
Your answer for a is correct. But check the signs in b and c by using the right-hand rule. :wink:

I thought the signs were independent on each other? Don't you have to calculate the independent magnetic field produced by each wire first at that particular point then add them two?

If not, then I must've made a mistake in part a, no? That's what I did for part a.
 
fizzziks said:
I thought the signs were independent on each other? Don't you have to calculate the independent magnetic field produced by each wire first at that particular point then add them two?

If you use the right-hand rule, you see that the magnetic fields are going into different directions, so you would have wrong answer if you just slapped the numbers into your equations.

If not, then I must've made a mistake in part a, no? That's what I did for part a.

Again, right-hand rule says the magnetic fields are going in the same directions so here you add the magnitudes.
 
I see. So, in the case for b and c, I have to denote one convention to say if the current is going down the wire, it will produce a negative magnetic field, B. So, let's say the current down wire 2 (the wire on the left) will produce negative current.

Hence, for part (b)

B1 = (u*5A)/(2*pi*10cm)
B2 = -(u*5A)/(2*pi*20cm)

For part (c)

B1 = (u*5A)/(2*pi*30cm)
B2 = -(u*5A)/(2*pi*20cm)

Is that better? :)
 
fizzziks said:
Hence, for part (b)

B1 = (u*5A)/(2*pi*10cm)
B2 = -(u*5A)/(2*pi*20cm)

For part (c)

B1 = (u*5A)/(2*pi*30cm)
B2 = -(u*5A)/(2*pi*20cm)

Is that better? :)

For parts b and c yes, but that convention would contradict with part a where you now are subrtacting the magnetic fields instead of adding them. Only way to solve these kind of problems are using the http://en.wikipedia.org/wiki/Right_hand_rule" .
 
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I see. Thanks for the help. I think I will stick to that convention I just mentioned.

Hence, for part (a)

Bnet = (u*5A)/(2*pi*5cm) - (u*5A)/(2*pi*5cm) = 0I think that looks good right?
 
  • #10
fizzziks said:
Bnet = (u*5A)/(2*pi*5cm) - (u*5A)/(2*pi*5cm) = 0
I think that looks good right?

Nope. For part a Bnet = 2(u*5A)/(2*pi*5cm)
 
  • #11
Hm. I will have to go back and revise the RHR. Hence, the answers IF ASSUMING if the current going down in wire 2 is negative, I will get:

(a)
Bnet = 2(u*5A)/(2*pi*5cm)

(b)
B1 = (u*5A)/(2*pi*10cm)
B2 = -(u*5A)/(2*pi*20cm)

(c)
B1 = (u*5A)/(2*pi*30cm)
B2 = -(u*5A)/(2*pi*20cm)

I hope this make sense.
 
  • #12
fizzziks said:
(a)
Bnet = 2(u*5A)/(2*pi*5cm)

(b)
B1 = (u*5A)/(2*pi*10cm)
B2 = -(u*5A)/(2*pi*20cm)

(c)
B1 = (u*5A)/(2*pi*30cm)
B2 = -(u*5A)/(2*pi*20cm)

I hope this make sense.

Yep, that looks right. And the right-hand rule really can save your butt in many situations. :biggrin:
 

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