What is the force of attraction between a north and south pole separated by air? The flux density in the air is 0.84 Tesla and the rectangular poles measure 2cm by 3 cm. (168N) F = B^2/2u0 0.84^2/2 x 1.257 x 10^-6 Not getting 168Newtons, can anyone please give me a hand here?
Could you show your work? With units? If F = B^2/(2μ0) What are the units on B? What are the units on μ0? What are the units when you divide the first by (twice) the second?
Well, since the OP can't be troubled to answer these questions, I guess I will. The units on B are tesla, which is (Ns)/(Cm) The units on μ0 are H/m or N/(A^2) Thus, the units of B^2/(2μ0) are [(Ns)^2/(Cm)^2]*[(A^2)/N] which reduce to N/(m^2) Thus, a simple dimensional analysis reveals not only that the equation given simply cannot yield the desired answer, but gives a really strong indication of exactly what needs to be done in order to correct it. But, hey, units are just a troublesome bother that only waste time.
You are right. That was an editing typo when doing a final fix-up of the LaTeX script. Thanks for catching it!