A conducting rod slides down between two frictionless vertical copper tracks at

Question

A conducting rod slides down between two frictionless vertical copper tracks at a constant speed of 4.3 m/s perpendicular to a 0.55-T magnetic field. The resistance of the rod and tracks is negligible. The rod maintains electrical contact with the tracks at all times and has a length of 1.2 m. A 0.81-? resistor is attached between the tops of the tracks. (a) What is the mass of the rod? (b) Find the change in the gravitational potential energy that occurs in a time of 0.20 s.
(c) Find the electrical energy dissipated in the resistor in 0.20 s.

Explanation / Answer

Voltage induced = B . v . l = 0.55 . 4.3 . 1.5 = 2.36 V ( I don't have a calculator ) Current = V /R = 2.36/ 0.81 = 2.92 amps F = B . i . L = 0.55*2.92 .*1.3 = 2.08 N Since the speed is constant , this force must balance the gravity force on the rod M . g = 2.08/9.8 M = 3.1 / 9.8 = 0.212 kg b) PE lost = mgh = 2.08 . 4.3 . 0.2 = 2.683 J c) energy dissipated = V . i . t = 2.3 . 4.3 . 0.2 = about 1.978 J

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