Question

1) In a region of space with an external magnetic field of 0.55-T in the vertical direction, two metal rails of negligible resistance rest 32 cm apart on a 6° ramp. The rails are joined at the bottom by a 0.60-Ω resistor. At the top of the rails, a copper bar of mass 0.040 kg and negligible resistance is laid across the rails and released from rest. When the copper rod reaches a steady speed, what will be (a) the current across the rod due to the motional emf, (b) the force exerted on the the rod by the magnetic field, and (c) the terminal (steady) velocity of the rod as it slides without friction down the rails?

Answer 1

The induced emf generated across the rodis Here, B is the magnetic field,l is length of conductor and v is velocity of conductor The induced current across the rod due to motional emf is , Blv cos θ Here, R is the resistance of the conductor The magnetic field force acting on the conductor is FBil cose Blv If the bar attains terminal velocity, the net force is zero, that means the magnetic field force is equal to component of the weight along the rails -mg sin θ yngR sin θ B414 cos (0.040kg)(9.8 m/s2)(06Ω)sin(8) (0.55 T) (0.32 m) cos (6" 0.80m/s (a) Magnitude of current in the r od is, (0.55 T) (0.32 m) (0.80m/s) cos (6*) 0.6Ω 0.23A b)Force exerted on the rodby the magnetic field is F-Bil cos θ (0.55 T) (0.233A) (0.32 m)co(6) 0.041N c) As shown from calculations in part(a),terminal speed is, v0.80m/s