A long, straight wire is fixed horizontally, carrying a current of 50.0 A . A second wire having a linear mass density of 1.0 10⁻⁴ kg m ⁻¹ is placed parallel to and directly above this wire at a separation of 5.0 mm . What current must this second wire carry so that the magnetic repulsion balances its weight?

Step-by-step solution

Let I₁ be the current in the fixed wire and I₂ be the current in the second wire. The magnetic force per unit length between two parallel current-carrying wires is given by: F L = ₀ I₁ I₂ 2 d For the second wire to remain suspended, the magnetic repulsion per unit length must balance its weight per unit length ( g ): ₀ I₁ I₂ 2 d = g Substituting the given values ( I₁ = 50.0 A , d = 5.0 10⁻³ m , = 1.0 10⁻⁴ kg m ⁻¹ , g = 9.8 m s ⁻² ): 4 10⁻⁷ 50.0 I₂ 2 5.0 10⁻³ = 1.0 10⁻⁴ 9.8 2 10⁻⁷ 50.0 I₂ 5.0 10⁻³ = 9.8 10⁻⁴ 2 10⁻³