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MIT 8.02 Electricity and Magnetism, Spring 2002
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in lecture 14 ,how dose B=UoI/2piR come out ? thanks
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don't use Ampere'rule,but use Biot-Savart Law in integral
\[\int\limits_{}^{}dB = \int\limits_{}^{}\mu I(dl \times r )/(4\pi \times r^{2})\] i can not find the way to write vector. but here it is: dl is in vector, r ( r^ is the unit vector of r with the length of 1). you will always have dl perpendicular to r^, so dl x r^ = dl (this time dl is not vector anymore). you then take the intergral along the closed circle for dl, and the result is 2piR, pi and R will be cancelled, the you got the final result
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