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A wire having mass per unit length of 0.440 g/cm carries a 2.80 A current horizontally to the south. What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward?

  • physics -

    B*I*L = M g = (mass per length)*L*g

    L cancels out.

    Solve for B and use the right hand rule for the direction of the magnetic force (up)

    B = (mass per length)*g/I
    = 4.40*10^-2 kg/m*9.8 m/s^2/2.80 A
    = ___ Tesla

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