- #1
flyingpig
- 2,579
- 1
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
[tex]\oint \vec{B} \cdot \vec{ds} = \mu_0 I[/tex]
(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
[tex]\oint \vec{B} \cdot \vec{ds} = \mu_0 I_1 N_1 [/tex]
[tex]B(2\pi b) = \mu_0 I_1 N_1[/tex]
[tex]B = \frac{ \mu_0 I_1 N_1}{2\pi b}[/tex]
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
[tex]\oint \vec{B} \cdot \vec{ds} = \mu_0 NI[/tex]
[tex]B(2\pi c) = \mu_0 \left |N_1I_1 - N_2I_2 \right|[/tex]
[tex]B = \frac{\mu_0 \left |N_1I_1 - N_2I_2 \right|}{2\pi c}[/tex]
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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