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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.......!!
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 fieldforce will be in upward directionso F should be equal to mgF=mgILB=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
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