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 4.8 m/s perpendicular to a 0.65-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 1.5 m. A 0.61- 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 0.18 s. (c) Find the electrical energy dissipated in the resistor in 0.18 s.

Explanation / Answer

Voltage induced = B . v . l = 0.65 * 4.8 * 1.5 = about 4.68 V

Current = V /R = 4.68 / 0.61 = about 7.67 amps
F = B * i * L = 0.65 * 7.67 * 1.5 = about 7.48 N
Since the speed is constant , this force must balance the gravity force on the rod
a)M . g = 3.1
M = 7.48 / 9.8 = about 0.7633 kg

b) PE lost = mgh = 7.48 * 4.8 * 0.18 = about 6.463 J

c) energy dissipated = V * i * t = 4.68 * 7.48 * 0.18 = about 6.30 J

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