Dallas Jett Armstrong v (Student - section: 730) PHY184B, Spring 2024 - Physics for Scientists and Engineers II Messages Courses Help Logout Main Menu Contents | Grades Chat Syllabus Course Contents > Chapter 28 > 28.41 * Timer Notes Evaluate Feedback Print h Oil - pi The figure shows the cross section through three long wires with a linear mass distribution of 139 9/m. They carry currents /1, /2, and is in the directions shown. Wires 2 and 3 are 10.3 cm apart and are attached to a vertical surface, and each carries a current of 511 A. What current, /1, will allow wire 1 to "float" at a perpendicular distance d = 10.9 cm from the vertical surface? (Neglect the thickness of the wires.) 916.18 A Submit Answer Incorrect. Tries 3/100 Previous Triesthe weight of i wire fu pulls it downward - - so it should move downw- ma and Now the magnetic force due to i z and is wire on the in wive in f , and F2 and its Component along vertical direction is F so if the wive have to float then this vertical component Shoud be equal to the Weight of the in wire, now here we assume that the length of the wireis L and mass per unit length of the wife is on so total man of the with in and as we Know F = Bid for iz wire F, = B , ind so I have done a mis take in this F part so it should he rather than so , actual form is :- F, Coso + F2 Coso= mdg. 51 1 x 21 1 + 5 1 1 x 21 -12-05 X 162 X COSO = 139 xiox 12.05 X 152