# A rod weighing 150 g, lying across two horizontal rails, the distance between which is 30 cm, flows a current of 50 A

**A rod weighing 150 g, lying across two horizontal rails, the distance between which is 30 cm, flows a current of 50 A. The coefficient of sliding friction between the rod and the rails is 0.6. Find the magnetic induction at which the conductor begins to slide along the rails. The lines of force of the field are directed vertically.**

m = 150 g = 0.15 kg. l = 30 cm = 0.3 m. I = 50 A. μ = 0.6. ∠α = 90 “. G = 9.8 m / s ^ 2. B -? In order for the rod to start moving along the rails, it is necessary that the Ampere force Famp, which acts on it, be balanced by the friction force Ftr: Famp = Ftr. The Ampere force is determined by the formula: Famp = I * l * B * sinα. The friction force is determined by the formula: Ftr = μ * m * g. I * l * B * sinα = μ * m * g. B = μ * m * g / I * l * sinα. B = 0.6 * 0.15 kg * 9.8 m / s ^ 2/50 A * 0.3 m * sin90 “= 0.0588 T.

Answer: magnetic field induction B = 0.0588 T.