# 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.......!!

Roshan Mohanty
64 Points
12 years ago

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

Kushal Patel
13 Points
7 years ago
Here,the following force balance existsmagnetic repulsive force between the two wires per unit length = weight per unit length of one wire orμ0I1.I2 / 2πx = W/ = (m/L).gso,current on wire 2I2 = [2πx (m/L).g] / μ0I1we havex = 5mm = 0.005mm/L = 10-4 kg/mI1 = 50 Aso,I2 = [2 x 3.14 x (10-4) x 9.81] / 2x10-7 x 50thus, by solving, we getI2 = 30.77 A