Question

in lecture 14 ,how dose B=UoI/2piR come out ?
thanks

Answer 1

don't use Ampere'rule,but use Biot-Savart Law in integral

Answer 2

\[\int\limits_{}^{}dB = \int\limits_{}^{}\mu I(dl \times r )/(4\pi \times r^{2})\]
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