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

[flyingpig]

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

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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

 

[ideasrule]

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.

 

[flyingpig]

What do you mean not symmetrical? I used a circle as my closed path...circles are very very symmetric!

 

[ideasrule]

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.