A long, straight wire is fixed horizontally and carries a current of 50.0 A. A second wire having linear mass `10^(-4) Kg` directly above this wire at a separation of 5.0 mm. What current should this second wire carry such that the magnetic repulsion can balance its weight?

Explanation

According to question `(mu_oi_1i_2)/(2pid) = (dF)/(dl)`
`(2xx10^(-7) xx 50xxi_2)/(5xx10^(-3) = 1xx10^(-4) xx9.8`
`i_2= (9.8xx10^(-7))/(20xx10^(-7)A`
`= 0.49A in opposite direction.