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# 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.......!!

9 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

4 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