Line Integral Help
- Thread starter aprilcm
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topsquark
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- Aug 27, 2012
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This assumes that B is not a function of s! It also assumes that C is a circle centered about a line source of current. Please make sure to post the whole question! (For those that didn't catch that, this is a well known problem about the magnetic field around a line current. The problem uses Ampere's Law.)
\(\displaystyle \int_C \dfrac{\vec{B} \cdot \vec{B}}{B}~ds = \int_C \dfrac{B^2}{B}~ds = \int_C B~ds = B \int_C ds\) as I've mentioned elsewhere.
Now, your contour C is a circle of radius r, and \(\displaystyle C = 2 \pi r\), thus \(\displaystyle ds = dC = 2 \pi dr\).
I haven't left much, but can you finish on your own?
-Dan
\(\displaystyle \int_C \dfrac{\vec{B} \cdot \vec{B}}{B}~ds = \int_C \dfrac{B^2}{B}~ds = \int_C B~ds = B \int_C ds\) as I've mentioned elsewhere.
Now, your contour C is a circle of radius r, and \(\displaystyle C = 2 \pi r\), thus \(\displaystyle ds = dC = 2 \pi dr\).
I haven't left much, but can you finish on your own?
-Dan