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

AI Thread Summary
Walter Lewin's explanation of Ampere's Law emphasizes the importance of understanding the magnetic field around current-carrying wires. The discussion highlights the application of the law using Amperian loops, where the magnetic field is calculated based on the enclosed current. Participants confirm that while the calculations are correct, the configuration may not be perfectly symmetrical due to gaps between wires. The consensus is that the derived equations are valid for the problem at hand, despite minor inaccuracies in real-world applications. Overall, the thread reinforces the practical use of Ampere's Law in electromagnetism.
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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

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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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