a long straight wire is fixed horizontally and carries a current of 50A.... a second wire having linear mass density of 1X10^-4kg/m......is placed parallel and above the first wire at a separation of 5mm.....what current must be flowing in the second wire such that the magnetic repulsion can balance its weight....???

PLS ANSWER....>ELABORATE iT SHOWING THE WORKING AND THE LOGIC......

ANSWER WILLL SURELY BE APPROVED.......!!

First concentrate at the lower wire..
magnetic field due to it will be  B=u.I/2*pie*d(where u=4pie * 10^-7 SI,d=5*10^-3 m)------Biot Savarts Law(i)
 force on the upper wire will be F=I(L X B)
L=length of the upper wire and B is the magnetic field
force will be in upward direction
so F should be equal to mg
F=mg
ILB=10^-4 * L * g
=>IB=10^-4 * g
=>I*(u.I/2*pie*d)=10^-4 * g-------From (i)
so from here u can easily find I(current)
*If you hav any doubt in the procedure..do tell
*Thank You