A straight wire of uniform cross-section is placed horizontally in a uniform horizontal magnetic field of magnitude 1 T which is perpendicular to the length of the wire. An electric field of 0.1 V/m is maintained along the wire. If the wire remains suspended in mid-air in equilibrium, the resistivity of the material of the wire is x 10⁻⁶ m . The value of x is _____. (Given: Density of wire material = 5000 kg/m ^3 , g

Step-by-step solution

For the wire to remain suspended in mid-air, the magnetic force must balance the weight of the wire. F_m = mg BIL = mg We know that current I = JA and mass m = dAL , where J is the current density, A is the cross-sectional area, L is the length, and d is the density of the material. B(JA)L = (dAL)g BJ = dg Using the microscopic form of Ohm's law, J = E , where E is the electric field and is the resistivity: B ( E ) = dg = BE dg Substituting the given values: = 1 0.1 5000 10 = 0.1 50000 = 2 10⁻⁶ m Thus, x = 2 . Answ