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JEE MainPhysicsMagnetic Effects of Current

A straight horizontal metallic rod of mass 0.8 kg and length 0.5 m is suspended by two identical light vertical conducting wires. A uniform vertical magnetic field of 2 T is present in the region. When a steady current of 6 A is passed through the rod, the suspension wires swing and the rod reaches a new equilibrium position. The tension in each suspension wire in this equilibrium position is (Take g = 10 m s ⁻² )

Options

  1. A5 N
  2. B10 N
  3. C7 N
  4. D4 N

Correct answer

A. 5 N

Step-by-step solution

The magnetic force on the current-carrying rod is given by F_m = ILB . Since the rod is horizontal and the magnetic field is vertical, they are perpendicular to each other. The magnetic force will be horizontal. F_m = 6 0.5 2 = 6 N The gravitational force acting downwards on the rod is: W = mg = 0.8 10 = 8 N In the equilibrium position, the total tension in the two suspension wires balances the resultant of these two mutually perpendicular forces. Resultant force R = F_m^2 + W^2 = 6^2 + 8^2 = 36 + 64 = 10 N Since t

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