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ExamsJEE AdvancedPhysics

A conducting rod of mass m carrying a current i rests on two frictionless rails fixed on an inclined plane that makes an angle theta with the horizontal. A uniform magnetic field B is applied normal (perpendicular) to the plane of the inclined rails. What is the condition for the rod to remain in equilibrium on the incline?

  1. B*i*l = m*g*tan(theta)
  2. B*i*l = m*g*sin(theta)
  3. B*i*l = m*g*cos(theta)
  4. Equilibrium cannot be reached

Correct answer: B*i*l = m*g*sin(theta)

Solution

The current flows along the rod, which lies in the inclined plane; B is perpendicular to that plane. The magnetic force F = i*l x B therefore lies in the plane and is directed up the slope (perpendicular to the rod, along the incline). For equilibrium, this in-plane force must balance the component of gravity acting down the slope, which is m*g*sin(theta). Hence B*i*l = m*g*sin(theta). (Contrast: if B were vertical, the answer would be B*i*l = m*g*tan(theta).)

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