How Does Walter Lewin Explain Ampere's Law in Electromagnetism?

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Homework Help Overview

The discussion revolves around the application of Ampere's Law in electromagnetism, specifically in relation to a problem involving the calculation of the magnetic field around current-carrying wires. The original poster references a video by Walter Lewin and presents their attempts to solve the problem using the law.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply Ampere's Law to different configurations of current-carrying wires, questioning the symmetry of their chosen Amperian loop and the implications for their calculations. They express confidence in their results but seek validation from others.

Discussion Status

Some participants provide feedback on the correctness of the original poster's calculations, noting that while the answers are generally correct, there are nuances regarding symmetry and the influence of wire configurations. The discussion reflects a mix of agreement and clarification on the assumptions made in the problem.

Contextual Notes

Participants discuss the potential impact of the non-symmetrical arrangement of wires on the magnetic field calculations, highlighting the importance of considering the actual physical setup versus the idealized assumptions made in the problem.

flyingpig
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Homework Statement



http://img18.imageshack.us/img18/8196/ampere.th.png

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The Attempt at a Solution



I've watched Walter Lewin's vid http://www.youtube.com/watch?v=sxCZnb-EMtk&feature=relmfu like five times and he seemed pretty angry with the way books explain this...

Anyways

First and foremost

\oint \vec{B} \cdot \vec{ds} = \mu_0 I

(a) my Amperian loop encloses no current, so it is 0

(b)http://img716.imageshack.us/img716/339/ampb.th.png

Uploaded with ImageShack.us

By the right hand rule, my thumb points into the page and my B-field vector points as in the picture.

So I get

\oint \vec{B} \cdot \vec{ds} = \mu_0 I_1 N_1

B(2\pi b) = \mu_0 I_1 N_1

B = \frac{ \mu_0 I_1 N_1}{2\pi b}

Now for c, I got to enlarge my Amperian loop

http://img17.imageshack.us/img17/6403/ampc.th.png

Uploaded with ImageShack.us

Since the currents are in opposite direction, I must take their absolute value difference. So I have something like this

\oint \vec{B} \cdot \vec{ds} = \mu_0 NI

B(2\pi c) = \mu_0 \left |N_1I_1 - N_2I_2 \right|

B = \frac{\mu_0 \left |N_1I_1 - N_2I_2 \right|}{2\pi c}

I am actually pretty confident about this, but I just started this today, so I need a thumbs up from an expert
 
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Your answers are correct. Note, though, that they are not exact, because the configuration is not exactly symmetrical. For instance, the field very near a wire will be dominated by that wire, while the field far from any wire will be close to the answers you got.
 
What do you mean not symmetrical? I used a circle as my closed path...circles are very very symmetric!
 
Yes, but the actual wires do not form a perfect circle, because there are gaps in between. However, I'm very sure that they don't matter for the purposes of this question, and that your answers are the intended ones.
 

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