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JEE MainPhysicsMagnetic Effects of Current

A metallic rod of mass 0.6 kg and length 1 m lies horizontally on a rough inclined plane which makes an angle of 45^ with the horizontal. The coefficient of static friction between the rod and the plane is 0.5 . A uniform vertical magnetic field of 0.5 T is present. A current flows through the rod such that the resulting magnetic force acts horizontally, pushing the rod against the incline. The minimum current requir

Options

  1. A6 A
  2. B4 A
  3. C12 A
  4. D36 A

Correct answer

B. 4 A

Step-by-step solution

Let the current in the rod be I . The magnetic force on the rod is horizontal and is given by F_m = ILB . This force has two components relative to the inclined plane: 1. Up the incline: F_m 45^ = ILB 45^ 2. Perpendicular to the incline (pressing into it): F_m 45^ = ILB 45^ The gravitational force mg also has two components: 1. Down the incline: mg 45^ 2. Perpendicular to the incline (pressing into it): mg 45^ The normal reaction N provided by the incline is: N = mg 45^ + ILB 45^ The maximum static friction availab

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